编程语言 on Clarity

1 Array Creation Routines

Sun, 24 May 2026 00:00:00 +0000

title: "Array creation routines" date: 2026-05-24T13:51:21Z

Array creation routines

Ones and zeros

1import numpy as np

Create a new array of 2*2 integers, without initializing entries.

array([[0, 0],
       [0, 0]])

Let X = np.array([1,2,3], [4,5,6], np.int32). Create a new array with the same shape and type as X.

1X = np.array([[1,2,3], [4,5,6]], np.int32)

array([[1, 2, 3],
       [4, 5, 6]])

Create a 3-D array with ones on the diagonal and zeros elsewhere.

array([[ 1.,  0.,  0.],
       [ 0.,  1.,  0.],
       [ 0.,  0.,  1.]])

array([[ 1.,  0.,  0.],
       [ 0.,  1.,  0.],
       [ 0.,  0.,  1.]])

Create a new array of 3*2 float numbers, filled with ones.

1 Array Creation Routines Solution

Sun, 24 May 2026 00:00:00 +0000

title: "Array creation routines" date: 2026-05-24T13:51:24Z

Array creation routines

Ones and zeros

1import numpy as np

Create a new array of 2*2 integers, without initializing entries.

1np.empty([2,2], int)

array([[0, 0],
       [0, 0]])

Let X = np.array([1,2,3], [4,5,6], np.int32). Create a new array with the same shape and type as X.

1X = np.array([[1,2,3], [4,5,6]], np.int32)
2np.empty_like(X)

array([[1, 2, 3],
       [4, 5, 6]])

Create a 3-D array with ones on the diagonal and zeros elsewhere.

1np.eye(3)

array([[ 1.,  0.,  0.],
       [ 0.,  1.,  0.],
       [ 0.,  0.,  1.]])

1np.identity(3)

array([[ 1.,  0.,  0.],
       [ 0.,  1.,  0.],
       [ 0.,  0.,  1.]])

Create a new array of 3*2 float numbers, filled with ones.

1.numpy Beginner

Sun, 24 May 2026 00:00:00 +0000

title: "适合初学者快速入门的Numpy实战全集" date: 2026-05-24T13:49:57Z

适合初学者快速入门的Numpy实战全集

本文由光城同学整理

Numpy是一个用python实现的科学计算的扩展程序库，包括：

1、一个强大的N维数组对象Array；
2、比较成熟的（广播）函数库；
3、用于整合C/C++和Fortran代码的工具包；
4、实用的线性代数、傅里叶变换和随机数生成函数。numpy和稀疏矩阵运算包scipy配合使用更加方便。

NumPy（Numeric Python）提供了许多高级的数值编程工具，如：矩阵数据类型、矢量处理，以及精密的运算库。专为进行严格的数字处理而产生。多为很多大型金融公司使用，以及核心的科学计算组织如：Lawrence Livermore，NASA用其处理一些本来使用C++，Fortran或Matlab等所做的任务。

本文目录

1.Numpy基本操作

1.1 列表转为矩阵

1.2 维度

1.3 行数和列数()

1.4 元素个数

2.Numpy创建array

2.1 一维array创建

2.2 多维array创建

2.3 创建全零数组

2.4 创建全1数据

2.5 创建全空数组

2.6 创建连续数组

2.7 reshape操作

2.8 创建连续型数据

2.9 linspace的reshape操作

3.Numpy基本运算

3.1 一维矩阵运算

3.2 多维矩阵运算

3.3 基本计算

4.Numpy索引与切片

5.Numpy array合并

5.1 数组合并

5.2 数组转置为矩阵

5.3 多个矩阵合并

5.4 合并例子2

6.Numpy array分割

6.1 构造3行4列矩阵

6.2 等量分割

6.3 不等量分割

6.4 其他的分割方式

7.Numpy copy与 =

7.1 =赋值方式会带有关联性

7.2 copy()赋值方式没有关联性

8.广播机制

9.常用函数

1.Numpy基本操作

1.1 列表转为矩阵

1import numpy as np
2array = np.array([
3    [1,3,5],
4    [4,6,9]
5])
6
7print(array)

[[1 3 5]
 [4 6 9]]

1.2 维度

1print('number of dim:', array.ndim)

number of dim: 2

1.3 行数和列数()

1print('shape:',array.shape)

shape: (2, 3)

1.4 元素个数

1print('size:',array.size)

size: 6

2.Numpy创建array

2.1 一维array创建

1import numpy as np
2# 一维array
3a = np.array([2,23,4], dtype=np.int32) # np.int默认为int32
4print(a)
5print(a.dtype)

[ 2 23  4]
int32

2.2 多维array创建

1# 多维array
2a = np.array([[2,3,4],
3              [3,4,5]])
4print(a) # 生成2行3列的矩阵

[[2 3 4]
 [3 4 5]]

2.3 创建全零数组

1a = np.zeros((3,4))
2print(a) # 生成3行4列的全零矩阵

[[0. 0. 0. 0.]
 [0. 0. 0. 0.]
 [0. 0. 0. 0.]]

2.4 创建全1数据

1# 创建全一数据，同时指定数据类型
2a = np.ones((3,4),dtype=np.int)
3print(a)

[[1 1 1 1]
 [1 1 1 1]
 [1 1 1 1]]

2.5 创建全空数组

1# 创建全空数组，其实每个值都是接近于零的数
2a = np.empty((3,4))
3print(a)

[[0. 0. 0. 0.]
 [0. 0. 0. 0.]
 [0. 0. 0. 0.]]

2.6 创建连续数组

1# 创建连续数组
2a = np.arange(10,21,2) # 10-20的数据，步长为2
3print(a)

[10 12 14 16 18 20]

2.7 reshape操作

1# 使用reshape改变上述数据的形状
2b = a.reshape((2,3))
3print(b)

[[10 12 14]
 [16 18 20]]

2.8 创建连续型数据

1# 创建线段型数据
2a = np.linspace(1,10,20) # 开始端1，结束端10，且分割成20个数据，生成线段
3print(a)

[ 1.          1.47368421  1.94736842  2.42105263  2.89473684  3.36842105
  3.84210526  4.31578947  4.78947368  5.26315789  5.73684211  6.21052632
  6.68421053  7.15789474  7.63157895  8.10526316  8.57894737  9.05263158
  9.52631579 10.        ]

2.9 linspace的reshape操作

1# 同时也可以reshape
2b = a.reshape((5,4))
3print(b)

[[ 1.          1.47368421  1.94736842  2.42105263]
 [ 2.89473684  3.36842105  3.84210526  4.31578947]
 [ 4.78947368  5.26315789  5.73684211  6.21052632]
 [ 6.68421053  7.15789474  7.63157895  8.10526316]
 [ 8.57894737  9.05263158  9.52631579 10.        ]]

3.Numpy基本运算

3.1 一维矩阵运算

1import numpy as np
2# 一维矩阵运算
3a = np.array([10,20,30,40])
4b = np.arange(4)
5print(a,b)

[10 20 30 40] [0 1 2 3]

1c = a - b
2print(c)

[10 19 28 37]

1print(a*b) # 若用a.dot(b),则为各维之和

[  0  20  60 120]

1# 在Numpy中，想要求出矩阵中各个元素的乘方需要依赖双星符号 **，以二次方举例，即：
2c = b**2
3print(c)

[0 1 4 9]

1# Numpy中具有很多的数学函数工具
2c = np.sin(a)
3print(c)

[-0.54402111  0.91294525 -0.98803162  0.74511316]

1print(b<2)

[ True  True False False]

1a = np.array([1,1,4,3])
2b = np.arange(4)
3print(a==b)

[False  True False  True]

3.2 多维矩阵运算

1a = np.array([[1,1],[0,1]])
2b = np.arange(4).reshape((2,2))
3print(a)

[[1 1]
 [0 1]]

1print(b)

[[0 1]
 [2 3]]

1# 多维度矩阵乘法
2# 第一种乘法方式:
3c = a.dot(b)
4print(c)

[[2 4]
 [2 3]]

1# 第二种乘法:
2c = np.dot(a,b)
3print(c)

[[2 4]
 [2 3]]

1# 多维矩阵乘法不能直接使用'*'号
2
3a = np.random.random((2,4))
4
5print(np.sum(a))

3.825517216750851

1print(np.min(a))

0.09623355767721398

1print(np.max(a))

0.7420428188342583

1print("a=",a)

a= [[0.48634962 0.74204282 0.09623356 0.69074812]
 [0.60218881 0.52734181 0.41434585 0.26626662]]

如果你需要对行或者列进行查找运算，

1.scipy Intro

Sun, 24 May 2026 00:00:00 +0000

title: "SciPy-数值计算库" date: 2026-05-24T14:11:00Z

1import numpy as np
2import pylab as pl

1import matplotlib as mpl
2mpl.rcParams['font.sans-serif'] = ['SimHei']

SciPy-数值计算库

1import scipy
2scipy.__version__

'1.0.0'

常数和特殊函数

1from scipy import constants as C
2print (C.c) # 真空中的光速
3print (C.h) # 普朗克常数

299792458.0
6.62607004e-34

1C.physical_constants["electron mass"]

(9.10938356e-31, 'kg', 1.1e-38)

1# 1英里等于多少米, 1英寸等于多少米, 1克等于多少千克, 1磅等于多少千克
2print(C.mile)
3print(C.inch)
4print(C.gram)
5print(C.pound)

1609.3439999999998
0.0254
0.001
0.45359236999999997

1import scipy.special as S

1print (1 + 1e-20)
2print (np.log(1+1e-20))
3print (S.log1p(1e-20))

1.0
0.0
1e-20

1m = np.linspace(0.1, 0.9, 4)
2u = np.linspace(-10, 10, 200)
3results = S.ellipj(u[:, None], m[None, :])
4
5print([y.shape for y in results])

[(200, 4), (200, 4), (200, 4), (200, 4)]

1#%figonly=使用广播计算得到的`ellipj()`返回值
2fig, axes = pl.subplots(2, 2, figsize=(12, 4))
3labels = ["$sn$", "$cn$", "$dn$", "$\phi$"]
4for ax, y, label in zip(axes.ravel(), results, labels):
5    ax.plot(u, y)
6    ax.set_ylabel(label)
7    ax.margins(0, 0.1)
8    
9axes[1, 1].legend(["$m={:g}$".format(m_) for m_ in m], loc="best", ncol=2);

10 Minutes To Pandas

Sun, 24 May 2026 00:00:00 +0000

title: "十分钟入门pandas" date: 2026-05-24T14:08:58Z

十分钟入门pandas

1#coding:utf8
2%matplotlib inline

这个一篇针对pandas新手的简短入门，想要了解更多复杂的内容，参阅Cookbook

通常，我们首先要导入以下几个库：

1import pandas as pd
2import numpy as np
3import matplotlib.pyplot as plt

创建对象

通过传递一个list来创建Series，pandas会默认创建整型索引：

1s = pd.Series([1,3,5,np.nan,6,8])
2s

0    1.0
1    3.0
2    5.0
3    NaN
4    6.0
5    8.0
dtype: float64

通过传递一个numpy array，日期索引以及列标签来创建一个DataFrame：

1dates = pd.date_range('20130101', periods=6)
2dates

DatetimeIndex(['2013-01-01', '2013-01-02', '2013-01-03', '2013-01-04',
               '2013-01-05', '2013-01-06'],
              dtype='datetime64[ns]', freq='D')

1df = pd.DataFrame(np.random.randn(6,4), index=dates, columns=list('ABCD'))
2df

10 Random Sampling

Sun, 24 May 2026 00:00:00 +0000

title: "Random Sampling" date: 2026-05-24T13:50:59Z

Random Sampling

1import numpy as np

1np.__version__

'1.11.2'

1__author__ = 'kyubyong. longinglove@nate.com'

Simple random data

Q1. Create an array of shape (3, 2) and populate it with random samples from a uniform distribution over [0, 1).

array([[ 0.13879034,  0.71300174],
       [ 0.08121322,  0.00393554],
       [ 0.02349471,  0.56677474]])

Q2. Create an array of shape (1000, 1000) and populate it with random samples from a standard normal distribution. And verify that the mean and standard deviation is close enough to 0 and 1 repectively.

10 Random Sampling Solutions

Sun, 24 May 2026 00:00:00 +0000

title: "Random Sampling" date: 2026-05-24T13:51:01Z

Random Sampling

1import numpy as np

1np.__version__

'1.11.2'

1__author__ = 'kyubyong. longinglove@nate.com'

Simple random data

Q1. Create an array of shape (3, 2) and populate it with random samples from a uniform distribution over [0, 1).

1np.random.rand(3, 2) 
2# Or np.random.random((3,2))

array([[ 0.13879034,  0.71300174],
       [ 0.08121322,  0.00393554],
       [ 0.02349471,  0.56677474]])

10.scipy Spatial

Sun, 24 May 2026 00:00:00 +0000

title: "10.scipy-spatial" date: 2026-05-24T14:11:03Z

1import numpy as np
2import pylab as pl
3from scipy import spatial

1import matplotlib as mpl
2mpl.rcParams['font.sans-serif'] = ['SimHei']

空间算法库-spatial

计算最近旁点

1x = np.sort(np.random.rand(100))
2idx = np.searchsorted(x, 0.5)
3print (x[idx], x[idx - 1]) #距离0.5最近的数是这两个数中的一个

0.5244435681885733 0.4982156075770372

1from scipy import spatial
2np.random.seed(42)
3N = 100
4points = np.random.uniform(-1, 1, (N, 2))
5kd = spatial.cKDTree(points)
6
7targets = np.array([(0, 0), (0.5, 0.5), (-0.5, 0.5), (0.5, -0.5), (-0.5, -0.5)])
8dist, idx = kd.query(targets, 3)

1r = 0.2
2idx2 = kd.query_ball_point(targets, r)
3idx2

array([list([48]), list([37, 78]), list([22, 79, 92]), list([6, 35, 58]),
       list([7, 42, 55, 83])], dtype=object)

1idx3 = kd.query_pairs(0.1) - kd.query_pairs(0.08)
2idx3

{(1, 46),
 (3, 21),
 (3, 82),
 (3, 95),
 (5, 16),
 (9, 30),
 (10, 87),
 (11, 42),
 (11, 97),
 (18, 41),
 (29, 74),
 (32, 51),
 (37, 78),
 (39, 61),
 (41, 61),
 (50, 84),
 (55, 83),
 (73, 81)}

 1#%figonly=用cKDTree寻找近旁点
 2x, y = points.T
 3colors = "r", "b", "g", "y", "k"
 4
 5fig, (ax1, ax2, ax3) = pl.subplots(1, 3, figsize=(12, 4))
 6
 7for ax in ax1, ax2, ax3:
 8    ax.set_aspect("equal")
 9    ax.plot(x, y, "o", markersize=4)
10    
11for ax in ax1, ax2:
12    for i in range(len(targets)):
13        c = colors[i]
14        tx, ty = targets[i]
15        ax.plot([tx], [ty], "*", markersize=10, color=c)
16
17for i in range(len(targets)):
18    nx, ny = points[idx[i]].T
19    ax1.plot(nx, ny, "o", markersize=10, markerfacecolor="None", 
20             markeredgecolor=colors[i], markeredgewidth=1)
21    
22    nx, ny = points[idx2[i]].T
23    ax2.plot(nx, ny, "o", markersize=10, markerfacecolor="None", 
24             markeredgecolor=colors[i], markeredgewidth=1)
25    
26    ax2.add_artist(pl.Circle(targets[i], r, fill=None, linestyle="dashed"))
27    
28for pidx1, pidx2 in idx3:
29    sx, sy = points[pidx1]
30    ex, ey = points[pidx2]
31    ax3.plot([sx, ex], [sy, ey], "r", linewidth=2, alpha=0.6)
32    
33ax1.set_xlabel(u"搜索最近的3个近旁点")
34ax2.set_xlabel(u"搜索距离在0.2之内的所有近旁点")
35ax3.set_xlabel(u"搜索所有距离在0.08到0.1之间的点对");

100 Numpy Exercises

Sun, 24 May 2026 00:00:00 +0000

title: "Alternative solution, in Python 3.5 and above" date: 2026-05-24T13:50:37Z

100 numpy exercises

This is a collection of exercises that have been collected in the numpy mailing list, on stack overflow and in the numpy documentation. The goal of this collection is to offer a quick reference for both old and new users but also to provide a set of exercises for those who teach.

If you find an error or think you've a better way to solve some of them, feel free to open an issue at https://github.com/rougier/numpy-100

100 Numpy Exercises No Solution

Sun, 24 May 2026 00:00:00 +0000

title: "100 numpy exercises" date: 2026-05-24T13:50:39Z

100 numpy exercises

If you find an error or think you've a better way to solve some of them, feel free to open an issue at https://github.com/rougier/numpy-100

100 Numpy Exercises With Hint

Sun, 24 May 2026 00:00:00 +0000

title: "100 numpy exercises with hint" date: 2026-05-24T13:50:42Z

100 numpy exercises with hint

If you find an error or think you've a better way to solve some of them, feel free to open an issue at https://github.com/rougier/numpy-100

11 Set Routines

Sun, 24 May 2026 00:00:00 +0000

title: "Set routines" date: 2026-05-24T13:51:04Z

Set routines

1import numpy as np

1np.__version__

'1.11.2'

1author = 'kyubyong. longinglove@nate.com'

Making proper sets

Q1. Get unique elements and reconstruction indices from x. And reconstruct x.

1x = np.array([1, 2, 6, 4, 2, 3, 2])

unique elements = [1 2 3 4 6]
reconstruction indices = [0 1 4 3 1 2 1]
reconstructed = [1 2 6 4 2 3 2]

Boolean operations

Q2. Create a boolean array of the same shape as x. If each element of x is present in y, the result will be True, otherwise False.

11 Set Routines Solutions

Sun, 24 May 2026 00:00:00 +0000

title: "Set routines" date: 2026-05-24T13:51:07Z

Set routines

1import numpy as np

1np.__version__

'1.11.2'

1author = 'kyubyong. longinglove@nate.com'

Making proper sets

Q1. Get unique elements and reconstruction indices from x. And reconstruct x.

1x = np.array([1, 2, 6, 4, 2, 3, 2])
2out, indices = np.unique(x, return_inverse=True)
3print "unique elements =", out
4print "reconstruction indices =", indices
5print "reconstructed =", out[indices]

unique elements = [1 2 3 4 6]
reconstruction indices = [0 1 4 3 1 2 1]
reconstructed = [1 2 6 4 2 3 2]

Boolean operations

Q2. Create a boolean array of the same shape as x. If each element of x is present in y, the result will be True, otherwise False.

12 Sorting Searching And Counting

Sun, 24 May 2026 00:00:00 +0000

title: "Soring, searching, and counting" date: 2026-05-24T13:51:10Z

Soring, searching, and counting

1import numpy as np

1np.__version__

'1.11.2'

1author = 'kyubyong. longinglove@nate.com'

Sorting

Q1. Sort x along the second axis.

1x = np.array([[1,4],[3,1]])

[[1 4]
 [1 3]]

Q2. Sort pairs of surnames and first names and return their indices. (first by surname, then by name).

1surnames =    ('Hertz',    'Galilei', 'Hertz')
2first_names = ('Heinrich', 'Galileo', 'Gustav')

[1 2 0]

Q3. Get the indices that would sort x along the second axis.

1x = np.array([[1,4],[3,1]])

[[0 1]
 [1 0]]

Q4. Create an array such that its fifth element would be the same as the element of sorted x, and it divide other elements by their value.

12 Sorting Searching And Counting Solutions

Sun, 24 May 2026 00:00:00 +0000

title: "Soring, searching, and counting" date: 2026-05-24T13:51:13Z

Soring, searching, and counting

1import numpy as np

1np.__version__

'1.11.2'

1author = 'kyubyong. longinglove@nate.com'

Sorting

Q1. Sort x along the second axis.

1x = np.array([[1,4],[3,1]])
2out = np.sort(x, axis=1)
3x.sort(axis=1)
4assert np.array_equal(out, x)
5print out

[[1 4]
 [1 3]]

Q2. Sort pairs of surnames and first names and return their indices. (first by surname, then by name).

1surnames =    ('Hertz',    'Galilei', 'Hertz')
2first_names = ('Heinrich', 'Galileo', 'Gustav')
3print np.lexsort((first_names, surnames))

[1 2 0]

Q3. Get the indices that would sort x along the second axis.

13 Statistics

Sun, 24 May 2026 00:00:00 +0000

title: "Statistics" date: 2026-05-24T13:51:15Z

Statistics

1__author__ = "kyubyong. kbpark.linguist@gmail.com"

1import numpy as np

1np.__version__

'1.11.3'

Order statistics

Q1. Return the minimum value of x along the second axis.

1x = np.arange(4).reshape((2, 2))
2print("x=\n", x)

x=
 [[0 1]
 [2 3]]
ans=
 [0 2]

Q2. Return the maximum value of x along the second axis. Reduce the second axis to the dimension with size one.

1x = np.arange(4).reshape((2, 2))
2print("x=\n", x)

x=
 [[0 1]
 [2 3]]
ans=
 [[1]
 [3]]

Q3. Calcuate the difference between the maximum and the minimum of x along the second axis.

13 Statistics Solutions

Sun, 24 May 2026 00:00:00 +0000

title: "Statistics" date: 2026-05-24T13:51:18Z

Statistics

1__author__ = "kyubyong. kbpark.linguist@gmail.com"

1import numpy as np

1np.__version__

'1.11.3'

Order statistics

Q1. Return the minimum value of x along the second axis.

1x = np.arange(4).reshape((2, 2))
2print("x=\n", x)
3print("ans=\n", np.amin(x, 1))

x=
 [[0 1]
 [2 3]]
ans=
 [0 2]

Q2. Return the maximum value of x along the second axis. Reduce the second axis to the dimension with size one.

1x = np.arange(4).reshape((2, 2))
2print("x=\n", x)
3print("ans=\n", np.amax(x, 1, keepdims=True))

x=
 [[0 1]
 [2 3]]
ans=
 [[1]
 [3]]

Q3. Calcuate the difference between the maximum and the minimum of x along the second axis.

2 Array Manipulation Routines

Sun, 24 May 2026 00:00:00 +0000

title: "Array manipulation routines" date: 2026-05-24T13:51:27Z

Array manipulation routines

1import numpy as np

1np.__version__

'1.11.2'

Q1. Let x be a ndarray [10, 10, 3] with all elements set to one. Reshape x so that the size of the second dimension equals 150.

[[ 1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.
   1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.
   1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.
   1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.
   1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.
   1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.
   1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.
   1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.
   1.  1.  1.  1.  1.  1.]
 [ 1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.
   1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.
   1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.
   1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.
   1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.
   1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.
   1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.
   1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.
   1.  1.  1.  1.  1.  1.]]

Q2. Let x be array [[1, 2, 3], [4, 5, 6]]. Convert it to [1 4 2 5 3 6].

2 Array Manipulation Routines Solutions

Sun, 24 May 2026 00:00:00 +0000

title: "Array manipulation routines" date: 2026-05-24T13:51:30Z

Array manipulation routines

1import numpy as np

1np.__version__

'1.11.2'

Q1. Let x be a ndarray [10, 10, 3] with all elements set to one. Reshape x so that the size of the second dimension equals 150.

1x = np.ones([10, 10, 3])
2out = np.reshape(x, [-1, 150])
3print out
4assert np.allclose(out, np.ones([10, 10, 3]).reshape([-1, 150]))

[[ 1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.
   1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.
   1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.
   1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.
   1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.
   1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.
   1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.
   1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.
   1.  1.  1.  1.  1.  1.]
 [ 1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.
   1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.
   1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.
   1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.
   1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.
   1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.
   1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.
   1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.  1.
   1.  1.  1.  1.  1.  1.]]

Q2. Let x be array [[1, 2, 3], [4, 5, 6]]. Convert it to [1 4 2 5 3 6].

2.numpy

Sun, 24 May 2026 00:00:00 +0000

title: "1.Numpy简易入门" date: 2026-05-24T13:49:59Z

1.Numpy简易入门

1.1 认识NumPy数组对象

Numpy是一个用python实现的科学计算的扩展程序库，包括：

1、一个强大的N维数组对象Array；
2、比较成熟的（广播）函数库；
3、用于整合C/C++和Fortran代码的工具包；
4、实用的线性代数、傅里叶变换和随机数生成函数。numpy和稀疏矩阵运算包scipy配合使用更加方便。

1import numpy as np                     # 导入NumPy工具包

1data = np.arange(12).reshape(3, 4)  # 创建一个3行4列的数组

1data

array([[ 0,  1,  2,  3],
       [ 4,  5,  6,  7],
       [ 8,  9, 10, 11]])

1type(data)

numpy.ndarray

1data.ndim         # 数组维度的个数，输出结果2，表示二维数组

1data.shape        # 数组的维度，输出结果（3，4），表示3行4列

(3, 4)

1data.size         # 数组元素的个数，输出结果12，表示总共有12个元素

1data.dtype # 数组元素的类型，输出结果dtype('int64'),表示元素类型都是int64

dtype('int32')

1.2 创建NumPy数组

1import numpy as np

1data1 = np.array([1, 2, 3])                   # 创建一个一维数组

1data1

array([1, 2, 3])

1data2 = np.array([[1, 2, 3], [4, 5, 6]])   # 创建一个二维数组

1data2

array([[1, 2, 3],
       [4, 5, 6]])

1np.zeros((3, 4))#创建一个全0数组

array([[0., 0., 0., 0.],
       [0., 0., 0., 0.],
       [0., 0., 0., 0.]])

1np.ones((3, 4))#创建全一数组

array([[1., 1., 1., 1.],
       [1., 1., 1., 1.],
       [1., 1., 1., 1.]])

1np.empty((5, 2))# 创建全空数组，其实每个值都是接近于零的数

array([[ 6.95312756e-310,  2.12199579e-314],
       [ 2.12199579e-314,  4.94065646e-324],
       [ 0.00000000e+000, -7.06252554e-311],
       [ 0.00000000e+000, -8.12021073e-313],
       [ 1.29923372e-311,  2.07507571e-322]])

1np.arange(1, 20, 5)

array([ 1,  6, 11, 16])

1np.array([1, 2, 3, 4], float)

array([1., 2., 3., 4.])

1np.ones((2, 3), dtype='float64')

array([[1., 1., 1.],
       [1., 1., 1.]])

1.3 ndarry对象的数据类型

1.3.1 查看数据类型

1data_one = np.array([[1, 2, 3], [4, 5, 6]])

1data_one.dtype.name

'int32'

1.3.2 转换数据类型

1data = np.array([[1, 2, 3], [4, 5, 6]])

1data.dtype

dtype('int32')

1float_data = data.astype(np.float64) # 数据类型转换为float64

1float_data.dtype

dtype('float64')

1float_data = np.array([1.2, 2.3, 3.5])

1float_data

array([1.2, 2.3, 3.5])

1int_data = float_data.astype(np.int64) # 数据类型转换为int64

1int_data

array([1, 2, 3], dtype=int64)

1str_data = np.array(['1', '2', '3'])

1int_data = str_data.astype(np.int64)

1int_data

array([1, 2, 3], dtype=int64)

1.4 数组运算

1.4.1向量化运算

1import numpy as np

1data1 = np.array([[1, 2, 3], [4, 5, 6]])

1data2 = np.array([[1, 2, 3], [4, 5, 6]])

1data1 + data2        # 数组相加

array([[ 2,  4,  6],
       [ 8, 10, 12]])

1data1 * data2        # 数组相乘

array([[ 1,  4,  9],
       [16, 25, 36]])

1data1 - data2        # 数组相减

array([[0, 0, 0],
       [0, 0, 0]])

1data1 / data2       # 数组相除

array([[1., 1., 1.],
       [1., 1., 1.]])

1.4.2 数组广播

numpy数组间的基础运算是一对一，也就是a.shape==b.shape，但是当两者不一样的时候，就会自动触发广播机制，如下例子：

2.scipy Optimize

Sun, 24 May 2026 00:00:00 +0000

title: "f计算方程组的误差，[1,1,1]是未知数的初始值" date: 2026-05-24T14:11:06Z

1import pylab as pl
2import numpy as np

1import matplotlib as mpl
2mpl.rcParams['font.sans-serif'] = ['SimHei']

拟合与优化-optimize

非线性方程组求解

 1from math import sin, cos
 2from scipy import optimize
 3
 4def f(x): #❶
 5    x0, x1, x2 = x.tolist() #❷
 6    return [
 7        5*x1+3,
 8        4*x0*x0 - 2*sin(x1*x2),
 9        x1*x2 - 1.5
10    ]
11
12# f计算方程组的误差，[1,1,1]是未知数的初始值
13result = optimize.fsolve(f, [1,1,1]) #❸
14print (result)
15print (f(result))

[-0.70622057 -0.6        -2.5       ]
[0.0, -9.126033262418787e-14, 5.329070518200751e-15]

 1def j(x):  #❶
 2    x0, x1, x2 = x.tolist()
 3    return [[0, 5, 0],
 4            [8 * x0, -2 * x2 * cos(x1 * x2), -2 * x1 * cos(x1 * x2)],
 5            [0, x2, x1]]
 6
 7
 8result = optimize.fsolve(f, [1, 1, 1], fprime=j)  #❷
 9print(result)
10print(f(result))

[-0.70622057 -0.6        -2.5       ]
[0.0, -9.126033262418787e-14, 5.329070518200751e-15]

最小二乘拟合

 1import numpy as np
 2from scipy import optimize
 3
 4X = np.array([ 8.19,  2.72,  6.39,  8.71,  4.7 ,  2.66,  3.78])
 5Y = np.array([ 7.01,  2.78,  6.47,  6.71,  4.1 ,  4.23,  4.05])
 6
 7def residuals(p): #❶
 8    "计算以p为参数的直线和原始数据之间的误差"
 9    k, b = p
10    return Y - (k*X + b)
11
12# leastsq使得residuals()的输出数组的平方和最小，参数的初始值为[1,0]
13r = optimize.leastsq(residuals, [1, 0]) #❷
14k, b = r[0]
15print ("k =",k, "b =",b)

k = 0.6134953491930442 b = 1.794092543259387

 1#%figonly=最小化正方形面积之和（左），误差曲面（右）
 2scale_k = 1.0
 3scale_b = 10.0
 4scale_error = 1000.0
 5
 6def S(k, b):
 7    "计算直线y=k*x+b和原始数据X、Y的误差的平方和"
 8    error = np.zeros(k.shape)
 9    for x, y in zip(X, Y):
10        error += (y - (k * x + b)) ** 2
11    return error
12
13ks, bs = np.mgrid[k - scale_k:k + scale_k:40j, b - scale_b:b + scale_b:40j]
14error = S(ks, bs) / scale_error
15
16from mpl_toolkits.mplot3d import Axes3D
17from matplotlib.patches import Rectangle
18
19fig = pl.figure(figsize=(12, 5))
20
21ax1 = pl.subplot(121)
22
23ax1.plot(X, Y, "o")
24X0 = np.linspace(2, 10, 3)
25Y0 = k*X0 + b
26ax1.plot(X0, Y0)
27
28for x, y in zip(X, Y):
29    y2 = k*x+b
30    rect = Rectangle((x,y), abs(y-y2), y2-y, facecolor="red", alpha=0.2)
31    ax1.add_patch(rect)
32
33ax1.set_aspect("equal")
34
35
36ax2 = fig.add_subplot(122, projection='3d')
37
38ax2.plot_surface(
39    ks, bs / scale_b, error, rstride=3, cstride=3, cmap="jet", alpha=0.5)
40ax2.scatter([k], [b / scale_b], [S(k, b) / scale_error], c="r", s=20)
41ax2.set_xlabel("$k$")
42ax2.set_ylabel("$b$")
43ax2.set_zlabel("$error$");

3 String Operations

Sun, 24 May 2026 00:00:00 +0000

title: "3_String_operations" date: 2026-05-24T13:51:32Z

String operations

1from __future__ import print_function
2import numpy as np

1author = "kyubyong. https://github.com/Kyubyong/numpy_exercises"

1np.__version__

'1.11.3'

Q1. Concatenate x1 and x2.

1x1 = np.array(['Hello', 'Say'], dtype=np.str)
2x2 = np.array([' world', ' something'], dtype=np.str)

['Hello world' 'Say something']

Q2. Repeat x three time element-wise.

1x = np.array(['Hello ', 'Say '], dtype=np.str)

['Hello Hello Hello ' 'Say Say Say ']

Q3-1. Capitalize the first letter of x element-wise.
Q3-2. Lowercase x element-wise.
Q3-3. Uppercase x element-wise.
Q3-4. Swapcase x element-wise.
Q3-5. Title-case x element-wise.

3 String Operations Solutions

Sun, 24 May 2026 00:00:00 +0000

title: "compare" date: 2026-05-24T13:51:35Z

String operations

1from __future__ import print_function
2import numpy as np

1author = "kyubyong. https://github.com/Kyubyong/numpy_exercises"

1np.__version__

'1.11.3'

Q1. Concatenate x1 and x2.

1x1 = np.array(['Hello', 'Say'], dtype=np.str)
2x2 = np.array([' world', ' something'], dtype=np.str)
3out = np.char.add(x1, x2)
4print(out)

['Hello world' 'Say something']

Q2. Repeat x three time element-wise.

1x = np.array(['Hello ', 'Say '], dtype=np.str)
2out = np.char.multiply(x, 3)
3print(out)

['Hello Hello Hello ' 'Say Say Say ']

Q3-1. Capitalize the first letter of x element-wise.
Q3-2. Lowercase x element-wise.
Q3-3. Uppercase x element-wise.
Q3-4. Swapcase x element-wise.
Q3-5. Title-case x element-wise.

3.scipy Linalg

Sun, 24 May 2026 00:00:00 +0000

title: "3.scipy-linalg" date: 2026-05-24T14:11:09Z

1import pylab as pl
2import numpy as np
3from scipy import linalg

1import matplotlib as mpl
2mpl.rcParams['font.sans-serif'] = ['SimHei']

线性代数-linalg

解线性方程组

 1import numpy as np
 2from scipy import linalg
 3
 4m, n = 500, 50
 5A = np.random.rand(m, m)
 6B = np.random.rand(m, n)
 7X1 = linalg.solve(A, B)
 8X2 = np.dot(linalg.inv(A), B)
 9print (np.allclose(X1, X2))
10%timeit linalg.solve(A, B)
11%timeit np.dot(linalg.inv(A), B)

True
5.38 ms ± 120 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
8.14 ms ± 994 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

1luf = linalg.lu_factor(A)
2X3 = linalg.lu_solve(luf, B)
3np.allclose(X1, X3)

True

 1M, N = 1000, 100
 2np.random.seed(0)
 3A = np.random.rand(M, M)
 4B = np.random.rand(M, N)
 5Ai = linalg.inv(A)
 6luf = linalg.lu_factor(A)   
 7%timeit linalg.inv(A)
 8%timeit np.dot(Ai, B)
 9%timeit linalg.lu_factor(A)    
10%timeit linalg.lu_solve(luf, B)

50.6 ms ± 1.94 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
3.49 ms ± 306 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
20.1 ms ± 1.42 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
4.49 ms ± 65 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

最小二乘解

 1from numpy.lib.stride_tricks import as_strided
 2
 3
 4def make_data(m, n, noise_scale):  #❶
 5    np.random.seed(42)
 6    x = np.random.standard_normal(m)
 7    h = np.random.standard_normal(n)
 8    y = np.convolve(x, h)
 9    yn = y + np.random.standard_normal(len(y)) * noise_scale * np.max(y)
10    return x, yn, h
11
12
13def solve_h(x, y, n):  #❷
14    X = as_strided(
15        x, shape=(len(x) - n + 1, n), strides=(x.itemsize, x.itemsize))  #❸
16    Y = y[n - 1:len(x)]  #❹
17    h = linalg.lstsq(X, Y)  #❺
18    return h[0][::-1]  #❻

1x, yn, h = make_data(1000, 100, 0.4)
2H1 = solve_h(x, yn, 120)
3H2 = solve_h(x, yn, 80)
4
5print("Average error of H1:", np.mean(np.abs(h[:100] - h)))
6print("Average error of H2:", np.mean(np.abs(h[:80] - H2)))

Average error of H1: 0.0
Average error of H2: 0.2958422158342371

 1#%figonly=实际的系统参数与最小二乘解的比较
 2fig, (ax1, ax2) = pl.subplots(2, 1, figsize=(6, 4))
 3ax1.plot(h, linewidth=2, label=u"实际的系统参数")
 4ax1.plot(H1, linewidth=2, label=u"最小二乘解H1", alpha=0.7)
 5ax1.legend(loc="best", ncol=2)
 6ax1.set_xlim(0, len(H1))
 7
 8ax2.plot(h, linewidth=2, label=u"实际的系统参数")
 9ax2.plot(H2, linewidth=2, label=u"最小二乘解H2", alpha=0.7)
10ax2.legend(loc="best", ncol=2)
11ax2.set_xlim(0, len(H1));

4 Numpy Specific Help Functions

Sun, 24 May 2026 00:00:00 +0000

title: "4_Numpy-specific_help_functions" date: 2026-05-24T13:51:38Z

1import numpy as np

1np.__version__

'1.11.2'

Q1. Search for docstrings of the numpy functions on linear algebra.

Search results for 'linear algebra'
-----------------------------------
numpy.linalg.solve
    Solve a linear matrix equation, or system of linear scalar equations.
numpy.poly
    Find the coefficients of a polynomial with the given sequence of roots.
numpy.restoredot
    Restore `dot`, `vdot`, and `innerproduct` to the default non-BLAS
numpy.linalg.eig
    Compute the eigenvalues and right eigenvectors of a square array.
numpy.linalg.cond
    Compute the condition number of a matrix.
numpy.linalg.eigh
    Return the eigenvalues and eigenvectors of a Hermitian or symmetric matrix.
numpy.linalg.pinv
    Compute the (Moore-Penrose) pseudo-inverse of a matrix.
numpy.linalg.LinAlgError
    Generic Python-exception-derived object raised by linalg functions.

Q2. Get help information for numpy dot function.

4 Numpy Specific Help Functions Solutions

Sun, 24 May 2026 00:00:00 +0000

title: "4_Numpy-specific_help_functions_Solutions" date: 2026-05-24T13:51:41Z

1import numpy as np

1np.__version__

'1.11.2'

Q1. Search for docstrings of the numpy functions on linear algebra.

1np.lookfor('linear algebra')

Search results for 'linear algebra'
-----------------------------------
numpy.linalg.solve
    Solve a linear matrix equation, or system of linear scalar equations.
numpy.poly
    Find the coefficients of a polynomial with the given sequence of roots.
numpy.restoredot
    Restore `dot`, `vdot`, and `innerproduct` to the default non-BLAS
numpy.linalg.eig
    Compute the eigenvalues and right eigenvectors of a square array.
numpy.linalg.cond
    Compute the condition number of a matrix.
numpy.linalg.eigh
    Return the eigenvalues and eigenvectors of a Hermitian or symmetric matrix.
numpy.linalg.pinv
    Compute the (Moore-Penrose) pseudo-inverse of a matrix.
numpy.linalg.LinAlgError
    Generic Python-exception-derived object raised by linalg functions.

Q2. Get help information for numpy dot function.

4.scipy Stats

Sun, 24 May 2026 00:00:00 +0000

title: "4.scipy-stats" date: 2026-05-24T14:11:12Z

1import numpy as np
2import pylab as pl
3from scipy import stats

1import matplotlib.pyplot as plt
2import matplotlib as mpl
3mpl.rcParams['font.sans-serif'] = ['SimHei']

统计-stats

连续概率分布

1from scipy import stats
2[k for k, v in stats.__dict__.items() if isinstance(v, stats.rv_continuous)]

['ksone',
 'kstwobign',
 'norm',
 'alpha',
 'anglit',
 'arcsine',
 'beta',
 'betaprime',
 'bradford',
 'burr',
 'burr12',
 'fisk',
 'cauchy',
 'chi',
 'chi2',
 'cosine',
 'dgamma',
 'dweibull',
 'expon',
 'exponnorm',
 'exponweib',
 'exponpow',
 'fatiguelife',
 'foldcauchy',
 'f',
 'foldnorm',
 'weibull_min',
 'weibull_max',
 'frechet_r',
 'frechet_l',
 'genlogistic',
 'genpareto',
 'genexpon',
 'genextreme',
 'gamma',
 'erlang',
 'gengamma',
 'genhalflogistic',
 'gompertz',
 'gumbel_r',
 'gumbel_l',
 'halfcauchy',
 'halflogistic',
 'halfnorm',
 'hypsecant',
 'gausshyper',
 'invgamma',
 'invgauss',
 'invweibull',
 'johnsonsb',
 'johnsonsu',
 'laplace',
 'levy',
 'levy_l',
 'levy_stable',
 'logistic',
 'loggamma',
 'loglaplace',
 'lognorm',
 'gilbrat',
 'maxwell',
 'mielke',
 'kappa4',
 'kappa3',
 'nakagami',
 'ncx2',
 'ncf',
 't',
 'nct',
 'pareto',
 'lomax',
 'pearson3',
 'powerlaw',
 'powerlognorm',
 'powernorm',
 'rdist',
 'rayleigh',
 'reciprocal',
 'rice',
 'recipinvgauss',
 'semicircular',
 'skewnorm',
 'trapz',
 'triang',
 'truncexpon',
 'truncnorm',
 'tukeylambda',
 'uniform',
 'vonmises',
 'vonmises_line',
 'wald',
 'wrapcauchy',
 'gennorm',
 'halfgennorm',
 'crystalball',
 'argus']

1stats.norm.stats()

(array(0.), array(1.))

1X = stats.norm(loc=1.0, scale=2.0)
2X.stats()

(array(1.), array(4.))

1x = X.rvs(size=10000) # 对随机变量取10000个值
2np.mean(x), np.var(x) # 期望值和方差

(1.0048352738823323, 3.9372117720073554)

1stats.norm.fit(x) # 得到随机序列期望值和标准差

(1.0048352738823323, 1.984240855341749)

1pdf, t = np.histogram(x, bins=100, normed=True)  #❶
2t = (t[:-1] + t[1:]) * 0.5  #❷
3cdf = np.cumsum(pdf) * (t[1] - t[0])  #❸
4p_error = pdf - X.pdf(t)
5c_error = cdf - X.cdf(t)
6print ("max pdf error: {}, max cdf error: {}".format(
7    np.abs(p_error).max(),
8    np.abs(c_error).max()))

max pdf error: 0.018998755595167102, max cdf error: 0.018503342378306975


C:\ProgramData\Anaconda3\lib\site-packages\ipykernel_launcher.py:1: VisibleDeprecationWarning: Passing `normed=True` on non-uniform bins has always been broken, and computes neither the probability density function nor the probability mass function. The result is only correct if the bins are uniform, when density=True will produce the same result anyway. The argument will be removed in a future version of numpy.
  """Entry point for launching an IPython kernel.

1#%figonly=正态分布的概率密度函数（左）和累积分布函数（右）
2fig, (ax1, ax2) = pl.subplots(1, 2, figsize=(7, 2))
3ax1.plot(t, pdf, label=u"统计值")
4ax1.plot(t, X.pdf(t), label=u"理论值", alpha=0.6)
5ax1.legend(loc="best")
6ax2.plot(t, cdf)
7ax2.plot(t, X.cdf(t), alpha=0.6);

5 Input And Output

Sun, 24 May 2026 00:00:00 +0000

title: "Input and Output" date: 2026-05-24T13:51:44Z

Input and Output

1from __future__ import print_function
2import numpy as np

1author = "kyubyong. https://github.com/Kyubyong/numpy_exercises"

1np.__version__

'1.12.0'

1from datetime import date
2print(date.today())

2017-04-01

NumPy binary files (NPY, NPZ)

Q1. Save x into temp.npy and load it.

1x = np.arange(10)
2...
3
4# Check if there exists the 'temp.npy' file.
5import os
6if os.path.exists('temp.npy'):
7    x2 = ...
8    print(np.array_equal(x, x2))

True

Q2. Save x and y into a single file 'temp.npz' and load it.

1x = np.arange(10)
2y = np.arange(11, 20)
3...
4
5with ... as data:
6    x2 = data['x']
7    y2 = data['y']
8    print(np.array_equal(x, x2))
9    print(np.array_equal(y, y2))

True
True

Text files

Q3. Save x to 'temp.txt' in string format and load it.

5 Input And Output Solutions

Sun, 24 May 2026 00:00:00 +0000

title: "Input and Output" date: 2026-05-24T13:51:46Z

Input and Output

1from __future__ import print_function
2import numpy as np

1author = "kyubyong. https://github.com/Kyubyong/numpy_exercises"

1np.__version__

'1.12.0'

1from datetime import date
2print(date.today())

2017-04-01

NumPy binary files (NPY, NPZ)

Q1. Save x into temp.npy and load it.

1x = np.arange(10)
2np.save('temp.npy', x) # Actually you can omit the extension. If so, it will be added automatically.
3
4# Check if there exists the 'temp.npy' file.
5import os
6if os.path.exists('temp.npy'):
7    x2 = np.load('temp.npy')
8    print(np.array_equal(x, x2))

True

Q2. Save x and y into a single file 'temp.npz' and load it.

5.scipy Integrate

Sun, 24 May 2026 00:00:00 +0000

title: "调用ode对lorenz进行求解, 用两个不同的初始值" date: 2026-05-24T14:11:14Z

1import pylab as pl
2import numpy as np
3from scipy import integrate
4from scipy.integrate import odeint
5import matplotlib as mpl
6mpl.rcParams['font.sans-serif'] = ['SimHei']

数值积分-integrate

球的体积

1def half_circle(x):
2    return (1-x**2)**0.5

1N = 10000
2x = np.linspace(-1, 1, N)
3dx = x[1] - x[0]
4y = half_circle(x)
52 * dx * np.sum(y) # 面积的两倍

3.1415893269307373

1np.trapz(y, x) * 2 # 面积的两倍

3.1415893269315975

1from scipy import integrate
2pi_half, err = integrate.quad(half_circle, -1, 1)
3pi_half * 2

3.141592653589797

1def half_sphere(x, y):
2    return (1-x**2-y**2)**0.5

1volume, error = integrate.dblquad(half_sphere, -1, 1, 
2        lambda x:-half_circle(x), 
3        lambda x:half_circle(x))
4
5print (volume, error, np.pi*4/3/2)

2.094395102393199 1.0002356720661965e-09 2.0943951023931953

解常微分方程组

 1#%fig=洛伦茨吸引子：微小的初值差别也会显著地影响运动轨迹
 2from scipy.integrate import odeint 
 3import numpy as np 
 4
 5def lorenz(w, t, p, r, b): #❶
 6    # 给出位置矢量w，和三个参数p, r, b计算出
 7    # dx/dt, dy/dt, dz/dt的值
 8    x, y, z = w.tolist()
 9    # 直接与lorenz的计算公式对应 
10    return p*(y-x), x*(r-z)-y, x*y-b*z
11
12t = np.arange(0, 30, 0.02) # 创建时间点 
13# 调用ode对lorenz进行求解, 用两个不同的初始值 
14track1 = odeint(lorenz, (0.0, 1.00, 0.0), t, args=(10.0, 28.0, 3.0)) #❷
15track2 = odeint(lorenz, (0.0, 1.01, 0.0), t, args=(10.0, 28.0, 3.0)) #❸
16#%hide
17from mpl_toolkits.mplot3d import Axes3D
18fig = pl.figure()
19ax = Axes3D(fig)
20ax.plot(track1[:,0], track1[:,1], track1[:,2], lw=1)
21ax.plot(track2[:,0], track2[:,1], track2[:,2], lw=1);

6 Linear Algebra

Sun, 24 May 2026 00:00:00 +0000

title: "Linear algebra" date: 2026-05-24T13:51:49Z

Linear algebra

1import numpy as np

1np.__version__

'1.11.2'

Matrix and vector products

Q1. Predict the results of the following code.

1x = [1,2]
2y = [[4, 1], [2, 2]]
3#print np.dot(x, y)
4#print np.dot(y, x)
5#print np.matmul(x, y)
6#print np.inner(x, y)
7#print np.inner(y, x)

Q2. Predict the results of the following code.

1x = [[1, 0], [0, 1]]
2y = [[4, 1], [2, 2], [1, 1]]
3#print np.dot(y, x)
4#print np.matmul(y, x)

Q3. Predict the results of the following code.

1x = np.array([[1, 4], [5, 6]])
2y = np.array([[4, 1], [2, 2]])
3#print np.vdot(x, y)
4#print np.vdot(y, x)
5#print np.dot(x.flatten(), y.flatten())
6#print np.inner(x.flatten(), y.flatten())
7#print (x*y).sum()

Q4. Predict the results of the following code.

6 Linear Algebra Solutions

Sun, 24 May 2026 00:00:00 +0000

title: "Linear algebra" date: 2026-05-24T13:51:52Z

Linear algebra

1import numpy as np

1np.__version__

'1.11.2'

Matrix and vector products

Q1. Predict the results of the following code.

1x = [1,2]
2y = [[4, 1], [2, 2]]
3print np.dot(x, y)
4print np.dot(y, x)
5print np.matmul(x, y)
6print np.inner(x, y)
7print np.inner(y, x)

[8 5]
[6 6]
[8 5]
[6 6]
[6 6]

Q2. Predict the results of the following code.

1x = [[1, 0], [0, 1]]
2y = [[4, 1], [2, 2], [1, 1]]
3print np.dot(y, x)
4print np.matmul(y, x)

[[4 1]
 [2 2]
 [1 1]]
[[4 1]
 [2 2]
 [1 1]]

Q3. Predict the results of the following code.

6.scipy Signal

Sun, 24 May 2026 00:00:00 +0000

title: "设计一个带通滤波器：" date: 2026-05-24T14:11:17Z

1import pylab as pl
2import numpy as np
3from scipy import signal

1import matplotlib as mpl
2mpl.rcParams['font.sans-serif'] = ['SimHei']

信号处理-signal

中值滤波

 1#%fig=使用中值滤波剔除瞬间噪声
 2t = np.arange(0, 20, 0.1)
 3x = np.sin(t)
 4x[np.random.randint(0, len(t), 20)] += np.random.standard_normal(20)*0.6 #❶
 5x2 = signal.medfilt(x, 5) #❷
 6x3 = signal.order_filter(x, np.ones(5), 2)
 7print (np.all(x2 == x3))
 8pl.plot(t, x, label=u"带噪声的信号")
 9pl.plot(t, x2 + 0.5, alpha=0.6, label=u"中值滤波之后的信号")
10pl.legend(loc="best");

True

滤波器设计

 1sampling_rate = 8000.0
 2
 3# 设计一个带通滤波器：
 4# 通带为0.2*4000 - 0.5*4000
 5# 阻带为<0.1*4000, >0.6*4000
 6# 通带增益的最大衰减值为2dB
 7# 阻带的最小衰减值为40dB
 8b, a = signal.iirdesign([0.2, 0.5], [0.1, 0.6], 2, 40) #❶
 9
10# 使用freq计算滤波器的频率响应
11w, h = signal.freqz(b, a) #❷
12
13# 计算增益
14power = 20*np.log10(np.clip(np.abs(h), 1e-8, 1e100)) #❸
15freq = w / np.pi * sampling_rate / 2

 1#%fig=用频率扫描波测量的频率响应
 2# 产生2秒钟的取样频率为sampling_rate Hz的频率扫描信号
 3# 开始频率为0， 结束频率为sampling_rate/2
 4t = np.arange(0, 2, 1/sampling_rate) #❶
 5sweep = signal.chirp(t, f0=0, t1=2, f1=sampling_rate/2) #❷
 6# 对频率扫描信号进行滤波
 7out = signal.lfilter(b, a, sweep) #❸
 8# 将波形转换为能量
 9out = 20*np.log10(np.abs(out)) #❹
10# 找到所有局部最大值的下标
11index = signal.argrelmax(out, order=3)  #❺
12# 绘制滤波之后的波形的增益
13pl.figure(figsize=(8, 2.5))
14pl.plot(freq, power, label=u"带通IIR滤波器的频率响应") 
15pl.plot(t[index]/2.0*4000, out[index], label=u"频率扫描波测量的频谱", alpha=0.6) #❻
16pl.legend(loc="best")
17#%hide
18pl.title(u"频率扫描波测量的滤波器频谱")
19pl.ylim(-100,20)
20pl.ylabel(u"增益(dB)")
21pl.xlabel(u"频率(Hz)");

7 Discrete Fourier Transform

Sun, 24 May 2026 00:00:00 +0000

title: "cf." date: 2026-05-24T13:51:55Z

1from __future__ import print_function
2import numpy as np
3import matplotlib.pyplot as plt
4%matplotlib inline

1from datetime import date
2date.today()

datetime.date(2017, 11, 2)

1author = "kyubyong. https://github.com/Kyubyong/numpy_exercises"

1np.__version__

'1.13.1'

Complex Numbers

Q1. Return the angle of a in radian.

1a = 1+1j
2output = ...
3print(output)

0.785398163397

Q2. Return the real part and imaginary part of a.

1a = np.array([1+2j, 3+4j, 5+6j])
2real = ...
3imag = ...
4print("real part=", real)
5print("imaginary part=", imag)

real part= [ 1.  3.  5.]
imaginary part= [ 2.  4.  6.]

Q3. Replace the real part of a with 9, the imaginary part with [5, 7, 9].

7 Discrete Fourier Transform Solutions

Sun, 24 May 2026 00:00:00 +0000

title: "cf." date: 2026-05-24T13:51:58Z

1from __future__ import print_function
2import numpy as np
3import matplotlib.pyplot as plt
4%matplotlib inline

1from datetime import date
2date.today()

datetime.date(2017, 11, 2)

1author = "kyubyong. https://github.com/Kyubyong/numpy_exercises"

1np.__version__

'1.13.1'

Complex Numbers

Q1. Return the angle of a in radian.

1a = 1+1j
2output = np.angle(a, deg=False)
3print(output)

0.785398163397

Q2. Return the real part and imaginary part of a.

1a = np.array([1+2j, 3+4j, 5+6j])
2real = a.real
3imag = a.imag
4print("real part=", real)
5print("imaginary part=", imag)

real part= [ 1.  3.  5.]
imaginary part= [ 2.  4.  6.]

Q3. Replace the real part of a with 9, the imaginary part with [5, 7, 9].

7.scipy Interpolate

Sun, 24 May 2026 00:00:00 +0000

title: "X-Y轴分为15*15的网格" date: 2026-05-24T14:11:20Z

1import numpy as np
2import pylab as pl
3from scipy import interpolate

1import matplotlib as mpl
2mpl.rcParams['font.sans-serif'] = ['SimHei']

插值-interpolate

一维插值

WARNING

高次interp1d()插值的运算量很大，因此对于点数较多的数据，建议使用后面介绍的UnivariateSpline()。

 1#%fig=`interp1d`的各阶插值
 2from scipy import interpolate
 3
 4x = np.linspace(0, 10, 11)
 5y = np.sin(x)
 6
 7xnew = np.linspace(0, 10, 101)
 8pl.plot(x, y, 'ro')
 9for kind in ['nearest', 'zero', 'slinear', 'quadratic']:
10    f = interpolate.interp1d(x, y, kind=kind)  #❶
11    ynew = f(xnew)  #❷
12    pl.plot(xnew, ynew, label=str(kind))
13
14pl.legend(loc='lower right')

<matplotlib.legend.Legend at 0x274ccb5f6a0>

外推和Spline拟合

 1#%fig=使用UnivariateSpline进行插值：外推（上），数据拟合（下）
 2x1 = np.linspace(0, 10, 20)
 3y1 = np.sin(x1)
 4sx1 = np.linspace(0, 12, 100)
 5sy1 = interpolate.UnivariateSpline(x1, y1, s=0)(sx1)  #❶
 6
 7x2 = np.linspace(0, 20, 200)
 8y2 = np.sin(x2) + np.random.standard_normal(len(x2)) * 0.2
 9sx2 = np.linspace(0, 20, 2000)
10spline2 = interpolate.UnivariateSpline(x2, y2, s=8)  #❷
11sy2 = spline2(sx2)
12
13pl.figure(figsize=(8, 5))
14pl.subplot(211)
15pl.plot(x1, y1, ".", label=u"数据点")
16pl.plot(sx1, sy1, label=u"spline曲线")
17pl.legend()
18
19pl.subplot(212)
20pl.plot(x2, y2, ".", label=u"数据点")
21pl.plot(sx2, sy2, linewidth=2, label=u"spline曲线")
22pl.plot(x2, np.sin(x2), label=u"无噪声曲线")
23pl.legend()

<matplotlib.legend.Legend at 0x274bb27f588>

8 Logic Functions

Sun, 24 May 2026 00:00:00 +0000

title: "Logic functions" date: 2026-05-24T13:52:01Z

Logic functions

1import numpy as np

1np.__version__

'1.11.2'

Truth value testing

Q1. Let x be an arbitrary array. Return True if none of the elements of x is zero. Remind that 0 evaluates to False in python.

1x = np.array([1,2,3])
2#
3
4x = np.array([1,0,3])
5#

True
False

Q2. Let x be an arbitrary array. Return True if any of the elements of x is non-zero.

1x = np.array([1,0,0])
2#
3
4x = np.array([0,0,0])
5#

True
False

Array contents

Q3. Predict the result of the following code.

8 Logic Functions Solutions

Sun, 24 May 2026 00:00:00 +0000

title: "Logic functions" date: 2026-05-24T13:52:03Z

Logic functions

1import numpy as np

1np.__version__

'1.11.2'

Truth value testing

Q1. Let x be an arbitrary array. Return True if none of the elements of x is zero. Remind that 0 evaluates to False in python.

1x = np.array([1,2,3])
2print np.all(x)
3
4x = np.array([1,0,3])
5print np.all(x)

True
False

Q2. Let x be an arbitrary array. Return True if any of the elements of x is non-zero.

1x = np.array([1,0,0])
2print np.any(x)
3
4x = np.array([0,0,0])
5print np.any(x)

True
False

Array contents

Q3. Predict the result of the following code.

8.scipy Sparse

Sun, 24 May 2026 00:00:00 +0000

title: "8.scipy-sparse" date: 2026-05-24T14:11:23Z

1%matplotlib inline
2import numpy as np
3import pylab as pl
4from scipy import sparse
5from scipy.sparse import csgraph

稀疏矩阵-sparse

稀疏矩阵的储存形式

1from scipy import sparse
2a = sparse.dok_matrix((10, 5))
3
4a[2, 3] = 1.0
5a[3, 3] = 2.0
6a[4, 3] = 3.0
7print(a.keys())
8print(a.values())

dict_keys([(2, 3), (3, 3), (4, 3)])
dict_values([1.0, 2.0, 3.0])

1b = sparse.lil_matrix((10, 5))
2b[2, 3] = 1.0
3b[3, 4] = 2.0
4b[3, 2] = 3.0
5print(b.data)
6print(b.rows)

[list([]) list([]) list([1.0]) list([3.0, 2.0]) list([]) list([]) list([])
 list([]) list([]) list([])]
[list([]) list([]) list([3]) list([2, 4]) list([]) list([]) list([])
 list([]) list([]) list([])]

1row = [2, 3, 3, 2]
2col = [3, 4, 2, 3]
3data = [1, 2, 3, 10]
4c = sparse.coo_matrix((data, (row, col)), shape=(5, 6))
5print (c.col, c.row, c.data)
6print (c.toarray())

[3 4 2 3] [2 3 3 2] [ 1  2  3 10]
[[ 0  0  0  0  0  0]
 [ 0  0  0  0  0  0]
 [ 0  0  0 11  0  0]
 [ 0  0  3  0  2  0]
 [ 0  0  0  0  0  0]]

矩阵向量相乘

1 import numpy as np
2from scipy.sparse import csr_matrix
3A = csr_matrix([[1, 2, 0], [0, 0, 3], [4, 0, 5]])
4v = np.array([1, 0, -1])
5A.dot(v)

array([ 1, -3, -1], dtype=int32)

示例1

构造一个1000x1000 lil_matrix并添加值：

9 Mathematical Functions

Sun, 24 May 2026 00:00:00 +0000

title: "Mathematical functions" date: 2026-05-24T13:52:06Z

Mathematical functions

1import numpy as np

1np.__version__

'1.11.2'

1__author__ = "kyubyong. kbpark.linguist@gmail.com. https://github.com/kyubyong"

Trigonometric functions

Q1. Calculate sine, cosine, and tangent of x, element-wise.

1x = np.array([0., 1., 30, 90])

sine: [ 0.          0.84147098 -0.98803162  0.89399666]
cosine: [ 1.          0.54030231  0.15425145 -0.44807362]
tangent: [ 0.          1.55740772 -6.4053312  -1.99520041]

Q2. Calculate inverse sine, inverse cosine, and inverse tangent of x, element-wise.

1x = np.array([-1., 0, 1.])

inverse sine: [ 1.57079633  0.          1.57079633]
inverse cosine: [ 0.          1.57079633  0.        ]
inverse tangent: [ 0.78539816  0.          0.78539816]

Q3. Convert angles from radians to degrees.

9 Mathematical Functions Solutions

Sun, 24 May 2026 00:00:00 +0000

title: "Mathematical functions" date: 2026-05-24T13:52:09Z

Mathematical functions

1import numpy as np

1np.__version__

'1.11.2'

1__author__ = "kyubyong. kbpark.linguist@gmail.com. https://github.com/kyubyong"

Trigonometric functions

Q1. Calculate sine, cosine, and tangent of x, element-wise.

1x = np.array([0., 1., 30, 90])
2print "sine:", np.sin(x)
3print "cosine:", np.cos(x)
4print "tangent:", np.tan(x)

sine: [ 0.          0.84147098 -0.98803162  0.89399666]
cosine: [ 1.          0.54030231  0.15425145 -0.44807362]
tangent: [ 0.          1.55740772 -6.4053312  -1.99520041]

Q2. Calculate inverse sine, inverse cosine, and inverse tangent of x, element-wise.

1x = np.array([-1., 0, 1.])
2print "inverse sine:", np.arcsin(x2)
3print "inverse cosine:", np.arccos(x2)
4print "inverse tangent:", np.arctan(x2)

inverse sine: [ 1.57079633  0.          1.57079633]
inverse cosine: [ 0.          1.57079633  0.        ]
inverse tangent: [ 0.78539816  0.          0.78539816]

Q3. Convert angles from radians to degrees.

9.scipy Ndimage

Sun, 24 May 2026 00:00:00 +0000

title: "9.scipy-ndimage" date: 2026-05-24T14:11:26Z

1import numpy as np
2import pylab as pl

图像处理-ndimage

形态学图像处理

 1import numpy as np
 2
 3def expand_image(img, value, out=None, size = 10):
 4    if out is None:
 5        w, h = img.shape
 6        out = np.zeros((w*size, h*size),dtype=np.uint8)
 7    
 8    tmp = np.repeat(np.repeat(img,size,0),size,1)
 9    out[:,:] = np.where(tmp, value, out)
10    out[::size,:] = 0
11    out[:,::size] = 0
12    return out
13    
14def show_image(*imgs): 
15    for idx, img in enumerate(imgs, 1):
16        ax = pl.subplot(1, len(imgs), idx)
17        pl.imshow(img, cmap="gray")   
18        ax.set_axis_off()    
19    pl.subplots_adjust(0.02, 0, 0.98, 1, 0.02, 0)

膨胀和腐蚀

 1#%fig=四连通和八连通的膨胀运算
 2from scipy.ndimage import morphology
 3
 4def dilation_demo(a, structure=None):
 5    b = morphology.binary_dilation(a, structure)
 6    img = expand_image(a, 255)
 7    return expand_image(np.logical_xor(a,b), 150, out=img)
 8
 9a = pl.imread("scipy_morphology_demo.png")[:,:,0].astype(np.uint8)
10img1 = expand_image(a, 255)
11
12img2 = dilation_demo(a)
13img3 = dilation_demo(a, [[1,1,1],[1,1,1],[1,1,1]])
14show_image(img1, img2, img3)

1#%fig=不同结构元素的膨胀效果
2img4 = dilation_demo(a, [[0,0,0],[1,1,1],[0,0,0]])
3img5 = dilation_demo(a, [[0,1,0],[0,1,0],[0,1,0]])
4img6 = dilation_demo(a, [[0,1,0],[0,1,0],[0,0,0]])
5show_image(img4, img5, img6)

AsynchronousFileChannel

Sun, 24 May 2026 00:00:00 +0000

异步的文件通道，FileChannel属于"同步非阻塞"，AsynchronousFileChannel才是真正的异步。

从主要方法read/write的返回值为Future可以看出其"异步"的端倪。

open

1public static AsynchronousFileChannel open(Path file,
2                                           Set<? extends OpenOption> options,
3                                           ExecutorService executor,
4                                           FileAttribute<?>... attrs) {
5    FileSystemProvider provider = file.getFileSystem().provider();
6    return provider.newAsynchronousFileChannel(file, options, executor, attrs);
7}

另外一个重载的简化方法声明如下:

1public static AsynchronousFileChannel open(Path file, OpenOption... options) {}

所做的处理便是将可变参数options手动转为Set，executor为null，文件属性为NO_ATTRIBUTES，其实就是一个空的数组。

忽略调用关系，对于Linux来说最终由sun.nio.ch.SimpleAsynchronousFileChannelImpl的open方法实现:

1public static AsynchronousFileChannel open(FileDescriptor fdo,
2                                               boolean reading,
3                                               boolean writing,
4                                               ThreadPool pool) {
5    // Executor is either default or based on pool parameters
6    ExecutorService executor = (pool == null) ? DefaultExecutorHolder.defaultExecutor : pool.executor();
7    return new SimpleAsynchronousFileChannelImpl(fdo, reading, writing, executor);
8}

默认情况下pool参数为null，ThreadPool位于sun.nio.ch包下，其实是对Java线程池ExecutorService的一层包装。DefaultExecutorHolder是SimpleAsynchronousFileChannelImpl的嵌套类，这其实是一个单例模式:

Buffer

Sun, 24 May 2026 00:00:00 +0000

Buffer

说到nio就离不开Buffer体系，类图:

这里以ByteBuffer为例，其实对于布尔类型的其它所有基本类型，都与一种Buffer与之对应，比如还有ShortBuffer。

图中的getXXX指的是获得各种数据类型的便利方法，比如getChar，getInt，putXXX也是一样。asXXX指到其它类型Buffer的转换方法，比如asCharBuffer。

注意，只有ByteBuffer才拥有getXXX, putXXX, asXXX等向其它类型转换的方法，也只有ByteBuffer可以获取direct类型的buffer。

获取

allocate

得到heap buffer，源码:

1public static ByteBuffer allocate(int capacity) {
2    return new HeapByteBuffer(capacity, capacity);
3}

HeapByteBuffer构造器:

1HeapByteBuffer(int cap, int lim) { 	
2    super(-1, 0, lim, cap, new byte[cap], 0);
3}

可以看出，heap buffer的底层其实就是byte数组，构造完成之后各属性的状态如下:

mark: 0
pos: 0
limit: capacity
offset: 0

allocateDirect

获取direct buffer，只有ByteBuffer才可以:

1public static ByteBuffer allocateDirect(int capacity) {
2    return new DirectByteBuffer(capacity);
3}

DirectByteBuffer构造器:

 1DirectByteBuffer(int cap) { 
 2    super(-1, 0, cap, cap);
 3    boolean pa = VM.isDirectMemoryPageAligned();
 4    int ps = Bits.pageSize();
 5    long size = Math.max(1L, (long)cap + (pa ? ps : 0));
 6    Bits.reserveMemory(size, cap);
 7    long base = 0;
 8    try {
 9        base = unsafe.allocateMemory(size);
10    } catch (OutOfMemoryError x) {
11        Bits.unreserveMemory(size, cap);
12        throw x;
13    }
14    unsafe.setMemory(base, size, (byte) 0);
15    if (pa && (base % ps != 0)) {
16        // Round up to page boundary
17        address = base + ps - (base & (ps - 1));
18    } else {
19        address = base;
20    }
21    cleaner = Cleaner.create(this, new Deallocator(base, size, cap));
22    att = null;
23}

内存对齐

VM.isDirectMemoryPageAligned()用以判断JVM是否需要对堆外内存进行对齐，什么是内存页，什么是对齐参考:

BufferedInputStream

Sun, 24 May 2026 00:00:00 +0000

概览

BufferedInputStream的主要方法均为线程安全，其内部的byte数组的最大值为int的最大值减去8，这样做的原因是在Java中数组是一个对象，而一个对象最大拥有8字节的对象头。至于为什么是int的最大值而不是long的最大值，需要同时支持32位和64位虚拟机吧。

字段count的取值范围为[0, buf.length]，意义为[0, count)范围内的字节都是可读的，字段pos的取值范围为[0, count)，即读指针的位置。

read

1public synchronized int read() throws IOException {
2    if (pos >= count) {
3        fill();
4        if (pos >= count)
5            return -1;
6    }
7    return getBufIfOpen()[pos++] & 0xff;
8}

pos >= count说明缓冲区中已没有可读的字节，此时需要调用fill方法从依赖的输入流中真正地读取数据。

 1private void fill() throws IOException {
 2    //如果内部依赖的输入流已经关闭，那么getBufIfOpen会抛出异常
 3    byte[] buffer = getBufIfOpen();
 4    if (markpos < 0)
 5        pos = 0; //markpos为负数，说明我们并未设置标记位置，即不需要再次读取标记处的数据，所以整个缓冲区直接覆盖写入即可
 6    else if (pos >= buffer.length)
 7        if (markpos > 0) {
 8             //markpos为正数，说明缓冲区中尚有可以利用的空间(因为此时pos为buf的长度)，所以这里要做的就是整理缓冲区: 将被标记的数据([markpos, pos))
 9             //拷贝至缓冲区开头
10            int sz = pos - markpos;
11            System.arraycopy(buffer, markpos, buffer, 0, sz);
12            pos = sz;
13            markpos = 0;
14        } else if (buffer.length >= marklimit) {
15            //此时markpos为0，且标记的内容长度超过了marklimit限制，直接丢弃
16            markpos = -1;
17            pos = 0;
18        } else if (buffer.length >= MAX_BUFFER_SIZE) {
19            throw new OutOfMemoryError("Required array size too large");
20        } else {
21             //到达这里的前提是markpos为0，pos已达到缓冲区长度，且缓冲区长度未超过marklimit，即我们已没有可用的空间以供进一步的数据读取
22             //所以需要对缓冲区进行扩容
23            int nsz = (pos <= MAX_BUFFER_SIZE - pos) ?
24                    pos * 2 : MAX_BUFFER_SIZE;
25            //将缓冲区大小降低为marklimit，以减轻内存压力?
26            if (nsz > marklimit)
27                nsz = marklimit;
28            byte nbuf[] = new byte[nsz];
29            System.arraycopy(buffer, 0, nbuf, 0, pos);
30            if (!bufUpdater.compareAndSet(this, buffer, nbuf)) {
31                 //fill方法是在当前线程持有锁的情况下被调用的，所以这里CAS失败的唯一原因就是close方法已经被调用(close方法没有加锁)，这也是对buf的
32                 //操作还需要通过CAS来完成的原因
33                throw new IOException("Stream closed");
34            }
35            buffer = nbuf;
36        }
37    count = pos;
38    //调用底层输入流进行读取
39    int n = getInIfOpen().read(buffer, pos, buffer.length - pos);
40    if (n > 0)
41        count = n + pos;
42}

所以，如果当前底层输入流中没有可读的数据，那么fill方法将会阻塞直到:

C++Primer

Sun, 24 May 2026 00:00:00 +0000

常见类型	函数	面向对象	容器	模板与泛型编程	内存管理	其它
变量字符串与数组	函数	类重载运算与类型转换继承体系	容器容器适配器泛型算法	模板与泛型编程	内存管理	输入输出

变量

字符串与数组
- 1.字符串
- 2.数组

函数

重载运算与类型转换
- 1.重载运算
- 2.二义性类型转换
  - 2.1 转换二义性
  - 2.2 避免转换出现二义性

继承体系

容器

容器适配器
- 1.通用的容器适配器操作
- 2.三个顺序容器适配器

泛型算法

模板与泛型编程

内存管理
- 1.new和delete
  - 1.1 new
  - 1.2 delete
- 2.智能指针

输入输出

变量

1.类型

算术类型
- 整形
  - 包括char和bool在内
- 浮点型
  - 单精度
  - 双精度
  - 扩展精度
空类型（void）

使用建议：

C++对象模型

Sun, 24 May 2026 00:00:00 +0000

第2章构造函数语意学

2.1 默认构造函数的构造操作

考虑如下代码；

 1class Foo {
 2public:
 3    int val;
 4    Foo *pnext;
 5};
 6
 7void foo_bar()
 8{
 9    //程序要求bar's members都被清为0
10    Foo bar;
11    if(bar.val || bar.pnext)
12        // ... do something
13    // ...
14}

上述代码是否会合成默认的拷贝构造函数？首先需要提到2个需要：

ConcurrentHashMap

Sun, 24 May 2026 00:00:00 +0000

我们对JDK1.8版本的ConcurrentHashMap进行说明，1.8版本的ConcurrentHashMap相比之前的版本主要做了两处改进:

使用CAS代替分段锁。
红黑树，这一点和HashMap是一致的。

put

最核心的便是put方法:

1public V put(K key, V value) {
2    return putVal(key, value, false);
3}

最后一个参数为onlyIfAbsent，表示只有在key对应的value不存在时才将value加入，所以putVal是put和putIfAbsent两个方法的真正实现。

 1final V putVal(K key, V value, boolean onlyIfAbsent) {
 2    if (key == null || value == null) throw new NullPointerException();
 3    int hash = spread(key.hashCode());
 4    int binCount = 0;
 5    //volatile读
 6    for (Node<K,V>[] tab = table;;) {
 7        Node<K,V> f; int n, i, fh;
 8        if (tab == null || (n = tab.length) == 0)
 9            //初始化
10            tab = initTable();
11        else if ((f = tabAt(tab, i = (n - 1) & hash)) == null) {
12            if (casTabAt(tab, i, null,
13                         new Node<K,V>(hash, key, value, null)))
14                break;                   // no lock when adding to empty bin
15        }
16        else if ((fh = f.hash) == MOVED)
17            tab = helpTransfer(tab, f);
18        else {
19            //节点添加
20            V oldVal = null;
21            synchronized (f) {
22                if (tabAt(tab, i) == f) {
23                    if (fh >= 0) {
24                        binCount = 1;
25                        for (Node<K,V> e = f;; ++binCount) {
26                            K ek;
27                            if (e.hash == hash &&
28                                ((ek = e.key) == key ||
29                                 (ek != null && key.equals(ek)))) {
30                                oldVal = e.val;
31                                if (!onlyIfAbsent)
32                                    e.val = value;
33                                break;
34                            }
35                            Node<K,V> pred = e;
36                            if ((e = e.next) == null) {
37                                pred.next = new Node<K,V>(hash, key, value, null);
38                                break;
39                            }
40                        }
41                    }
42                    else if (f instanceof TreeBin) {
43                        Node<K,V> p;
44                        binCount = 2;
45                        if ((p = ((TreeBin<K,V>)f).putTreeVal(hash, key, value)) != null) {
46                            oldVal = p.val;
47                            if (!onlyIfAbsent)
48                                p.val = value;
49                        }
50                    }
51                }
52            }
53            if (binCount != 0) {
54                if (binCount >= TREEIFY_THRESHOLD)
55                    treeifyBin(tab, i);
56                if (oldVal != null)
57                    return oldVal;
58                break;
59            }
60        }
61    }
62    addCount(1L, binCount);
63    return null;
64}

table便是其数据的存放载体:

ConcurrentLinkedQueue

Sun, 24 May 2026 00:00:00 +0000

内部类Node的类图如下:

算法

此类的实现基于算法Michael & Scott algorithm，论文:

Simple, Fast, and Practical Non-Blocking and Blocking Concurrent Queue Algorithms

offer

 1public boolean offer(E e) {
 2    //如果元素为null，那么抛出NPE
 3    checkNotNull(e);
 4    final Node<E> newNode = new Node<E>(e);
 5    for (Node<E> t = tail, p = t;;) {
 6        //volatile读
 7        Node<E> q = p.next;
 8        if (q == null) {
 9            // p is last node
10            if (p.casNext(null, newNode)) {
11                if (p != t) // hop two nodes at a time
12                    casTail(t, newNode);  // Failure is OK.
13                return true;
14            }
15            //casNext竞争失败，再次读取p.next，此时(即q)不为空
16        }
17        else if (p == q)
18            // We have fallen off list.  If tail is unchanged, it
19            // will also be off-list, in which case we need to
20            // jump to head, from which all live nodes are always
21            // reachable.  Else the new tail is a better bet.
22            p = (t != (t = tail)) ? t : head;
23        else
24            //每插入两个新节点时才会重新设置尾节点
25            p = (p != t && t != (t = tail)) ? t : q;
26    }
27}

注意，由于ConcurrentLinkedQueue为无界队列，所以offer方法永远返回true。为什么说插入两个节点之后才会尝试重新设置尾节点呢?

Condition

Sun, 24 May 2026 00:00:00 +0000

Condition

类图

ConditionObject是AbstractQueuedSynchronizer的内部类。

创建

创建是通过Lock.newCondition()来完成的。

ReentrantLock.newCondition:

1public Condition newCondition() {
2    return sync.newCondition();
3}

Sync.newCondition:

1final ConditionObject newCondition() {
2    return new ConditionObject();
3}

await

ConditionObject.await:

 1public final void await() throws InterruptedException {
 2    if (Thread.interrupted())
 3        throw new InterruptedException();
 4    Node node = addConditionWaiter();
 5    int savedState = fullyRelease(node);
 6    int interruptMode = 0;
 7    while (!isOnSyncQueue(node)) {
 8        LockSupport.park(this);
 9        if ((interruptMode = checkInterruptWhileWaiting(node)) != 0)
10            break;
11    }
12    if (acquireQueued(node, savedState) && interruptMode != THROW_IE)
13        interruptMode = REINTERRUPT;
14    if (node.nextWaiter != null) // clean up if cancelled
15        unlinkCancelledWaiters();
16    if (interruptMode != 0)
17        reportInterruptAfterWait(interruptMode);
18}

条件队列

addConditionWaiter用于向Lock的锁队列的条件队列添加新的等待节点，什么是锁队列参见ReentrantLock。

CountDownLatch

Sun, 24 May 2026 00:00:00 +0000

CountDownLatch

类图

CountDownLatch本身没有任何父类或是实现任何接口，只含有一个内部类Sync.

await

CountDownLatch.await:

1public void await() throws InterruptedException {
2    sync.acquireSharedInterruptibly(1);
3}

AbstractQueuedSynchronizer.acquireSharedInterruptibly:

1public final void acquireSharedInterruptibly(int arg) throws InterruptedException {
2    if (Thread.interrupted()) throw new InterruptedException();
3    if (tryAcquireShared(arg) < 0)
4        doAcquireSharedInterruptibly(arg);
5}

这个套路已经见过很多次了，关键的逻辑便是Sync.tryAcquireShared:

1protected int tryAcquireShared(int acquires) {
2    return (getState() == 0) ? 1 : -1;
3}

可以看出，await成功的条件是当前读锁(共享锁)不再被其它线程持有。

countDown

1public void countDown() {
2    sync.releaseShared(1);
3}

AbstractQueuedSynchronizer.releaseShared:

1public final boolean releaseShared(int arg) {
2    if (tryReleaseShared(arg)) {
3        doReleaseShared();
4        return true;
5    }
6    return false;
7}

还是一样的套路，Sync.tryReleaseShared:

CyclicBarrier

Sun, 24 May 2026 00:00:00 +0000

CyclicBarrier

此类可管理一组线程，当所有线程都到达"检查点"时执行特定的动作(Runnable)，示例直接看其源码便好。

类图

和CountDownLatch类似，此类同样没有任何父类(除了Object)或是实现任何接口。

await

整个类的核心便是此方法:

1public int await() {
2    return dowait(false, 0L);
3}

doWait方法简略版源码:

 1private int dowait(boolean timed, long nanos) {
 2    final ReentrantLock lock = this.lock;
 3    lock.lock();
 4    try {
 5        final Generation g = generation;
 6        // 未到达(检查点)数减一
 7        int index = --count;
 8         // 已达到检查点，执行给定的任务
 9        if (index == 0) {
10            boolean ranAction = false;
11            try {
12                final Runnable command = barrierCommand;
13                if (command != null)
14                    command.run();
15                ranAction = true;
16                nextGeneration();
17                return 0;
18            } finally {
19                if (!ranAction)
20                    breakBarrier();
21            }
22        }
23
24        // 尚有线程未到达检查点
25        for (;;) {
26            try {
27                 // 等待
28                if (!timed)
29                    trip.await();
30                else if (nanos > 0L)
31                    nanos = trip.awaitNanos(nanos);
32            } catch (InterruptedException ie) {
33                if (g == generation && ! g.broken) {
34                    breakBarrier();
35                    throw ie;
36                } else {
37                    Thread.currentThread().interrupt();
38                }
39            }
40
41            if (g.broken)
42                throw new BrokenBarrierException();
43
44            if (g != generation)
45                return index;
46
47            if (timed && nanos <= 0L) {
48                breakBarrier();
49                throw new TimeoutException();
50            }
51        }
52    } finally {
53        lock.unlock();
54    }
55}

一目了然，但是有一个疑问，当未到达数减为0时，已经阻塞的线程是在哪里唤醒的呢?

EffectiveC++

Sun, 24 May 2026 00:00:00 +0000

一.让自己习惯C++

条款02：尽量以const,enum,inline替换#define

这个条款或许改为“宁可以编译器替换预处理器”比较好

1）以const替换#define

1#define ASPECT_RATIO 1.653
2替换为：
3const double AspectRatio = 1.653

调试的需要：#define的记号会被预处理器移走，记号名称可能没进入记号表内。因此当#define的宏名称获得一个编译错误时，会引起困惑，浪费大量时间追踪错误。而AspectRatio肯定会被编译器看到
更小的代码量：对浮点数而言，使用常量可能比使用#define导致较小量的代码，因为预处理器“盲目地将ASPECT_RATIO替换为1.653”可能导致目标码出现多份1.653

但是，以常量替换#define时要注意：

Enum

Sun, 24 May 2026 00:00:00 +0000

说明

对枚举这一看似平常的东西单开一节的目的是要探究一下为什么说枚举是最好的实现单例的方式。

从例子开始

假设有这么一个简单的枚举:

1public enum Test {
2
3    SOMEONE,
4
5    ANOTHER_ONE
6
7}

首先对其进行编译，然后使用命令javap -verbose Test.class对其反编译，得到如下部分:

1public final class Test extends java.lang.Enum<Test> {}

所以，对于Java来说，枚举是一个语法糖。

Why

Enum的类图:

finalize

Enum中将finalize方法直接定义为final，这就从根本上上避免了枚举类对此方法的实现:

1protected final void finalize() { }

序列化

以如下代码示例对枚举类型的序列化:

1File file = new File("C:/Users/xsdwe/Desktop/test");
2ObjectOutputStream out = new ObjectOutputStream(new FileOutputStream(file));
3out.writeObject(Person.EVIL);
4out.writeObject(Person.JACK);
5out.flush();
6out.close();

限制其实是在ObjectOutputStream的writeObject0方法中进行控制的，相关源码:

 1private void writeObject0(Object obj, boolean unshared) throws IOException {
 2    //...
 3    if (obj instanceof String) {
 4        writeString((String) obj, unshared);
 5    } else if (cl.isArray()) {
 6        writeArray(obj, desc, unshared);
 7    } else if (obj instanceof Enum) {
 8        writeEnum((Enum<?>) obj, desc, unshared);
 9    }
10    //...
11}

可以看出，对于枚举类型，这里使用了专门的方法writeEnum进行序列化:

FileChannel

Sun, 24 May 2026 00:00:00 +0000

我们以FileOutputStream的getChannel为例:

1public FileChannel getChannel() {
2    synchronized (this) {
3        if (channel == null) {
4            channel = FileChannelImpl.open(fd, path, false, true, append, this);
5        }
6        return channel;
7    }
8}

FileChannel类图:

有几点值得注意:

InputStream和OutputStream不是线程安全的，而通道(Channel)是线程安全的。
Channel根据注释的解释，是一组IO操作的联结。
GatheringByteChannel，顾名思义就是将一组ByteBuffer的数据收集/组合起来，所以它继承自WritableByteChannel。
ScatteringByteChannel，即将一个通道的数据分散到多个ByteBuffer之中。

其实FileChannelImpl便是FileChannel的子类，位于包sun.nio.ch中，其源码可以从openjdk的jdk\src\share\classes\sun\nio\ch目录找到，open方法源码:

1public static FileChannel open(FileDescriptor fd, String path, boolean readable, boolean writable,
2    boolean append, Object parent) {
3    return new FileChannelImpl(fd, path, readable, writable, append, parent);
4}

可以看出，getChannel的原理就是构造了一个FileChannelImpl对象，此对象中保存有对应的流的是否可读、追加、文件描述符等属性。

HashMap

Sun, 24 May 2026 00:00:00 +0000

类图如下:

构造

我们以无参构造器为例:

1public HashMap() {
2    this.loadFactor = DEFAULT_LOAD_FACTOR; // all other fields defaulted
3}

其它的参数均取默认。

put

1public V put(K key, V value) {
2    return putVal(hash(key), key, value, false, true);
3}

putVal源码:

 1final V putVal(int hash, K key, V value, boolean onlyIfAbsent, boolean evict) {
 2    Node<K,V>[] tab; Node<K,V> p; int n, i;
 3    //1.初始化
 4    if ((tab = table) == null || (n = tab.length) == 0)
 5        n = (tab = resize()).length;
 6    //2.bin为null，初始化第一个节点
 7    if ((p = tab[i = (n - 1) & hash]) == null)
 8        tab[i] = newNode(hash, key, value, null);
 9    else {
10        Node<K,V> e; K k;
11        if (p.hash == hash && ((k = p.key) == key || (key != null && key.equals(k))))
12            //指定key的值已存在，那么记录下原先的值
13            e = p;
14        else if (p instanceof TreeNode)
15            //红黑树
16            e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
17        else {
18            //bin不为空，且与链表头不相同(==或equals)
19            //3.
20            for (int binCount = 0; ; ++binCount) {
21                if ((e = p.next) == null) {
22                    p.next = newNode(hash, key, value, null);
23                    //达到临界值转为红黑树
24                    if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
25                        treeifyBin(tab, hash);
26                    break;
27                }
28                if (e.hash == hash &&
29                    ((k = e.key) == key || (key != null && key.equals(k))))
30                    break;
31                p = e;
32            }
33        }
34        if (e != null) { // existing mapping for key
35            V oldValue = e.value;
36            if (!onlyIfAbsent || oldValue == null)
37                e.value = value;
38            //空实现，为LinkedHashMap预留
39            afterNodeAccess(e);
40            return oldValue;
41        }
42    }
43    ++modCount;
44    //4.
45    if (++size > threshold)
46        resize();
47    //空实现，为LinkedHashMap预留
48    afterNodeInsertion(evict);
49    return null;
50}

onlyIfAbsent参数如果为true，那么对于已经存在的key，将不替换其值。table即HashMap进行数据存储的核心变量:

IO

Sun, 24 May 2026 00:00:00 +0000

File

类图:

其实观察File里面多数方法的实现，其实是通过委托给属性fs完成的。

FileSystem

在 File中的初始化代码:

1private static final FileSystem fs = DefaultFileSystem.getFileSystem();

在Windows下的类图:

那DefaultFileSystem又是个什么东西呢?这东西是从JDK8加入的，全部的源码如下:

1class DefaultFileSystem {
2    public static FileSystem getFileSystem() {
3        return new WinNTFileSystem();
4    }
5}

可以想象，在Linux上一定是不同的实现。

文件创建

我们来看看创建文件时到底发生了什么，以代码:

1File file = new File("test");
2System.out.println(file.getAbsolutePath());

File构造器:

1public File(String pathname) {
2    this.path = fs.normalize(pathname);
3    this.prefixLength = fs.prefixLength(this.path);
4}

normalize

这一步是将文件分隔符、路径分隔符转换成系统特定的，比如对于windows平台，就将/转为\。转换的过程不再赘述，来看一下系统特定的分隔符是从哪来的，WinNTFileSystem构造器:

1public WinNTFileSystem() {
2    slash = AccessController.doPrivileged(new GetPropertyAction("file.separator")).charAt(0);
3    semicolon = AccessController.doPrivileged(new GetPropertyAction("path.separator")).charAt(0);
4    altSlash = (this.slash == '\\') ? '/' : '\\';
5}

其实就是System.getProperty。

LinkedHashMap

Sun, 24 May 2026 00:00:00 +0000

LinkedHashMap是HashMap的子类，内部使用双链表进行顺序的维护，内部类Entry为HashMap的Node的子类，类图:

构造器

三参数构造器:

1public LinkedHashMap(int initialCapacity, float loadFactor, boolean accessOrder) {
2    super(initialCapacity, loadFactor);
3    this.accessOrder = accessOrder;
4}

如果accessOrder为true，那么表示将顺序记录为访问顺序，否则为插入顺序，默认为false。而正是这一参数并对removeEldestEntry方法进行覆盖便可以快速实现一个简单的LRU缓存。

put

put操作其实和父类HashMap采用相同的实现，在HashMap部分也提到了afterNodeAccess和afterNodeInsertion方法其实是空实现，而在LinkedHashMap中对其进行了实现，Linked的特性也是在这里进行了体现。

newNode

LinkedhHashMap对此方法进行了覆盖:

1Node<K,V> newNode(int hash, K key, V value, Node<K,V> e) {
2    LinkedHashMap.Entry<K,V> p = new LinkedHashMap.Entry<K,V>(hash, key, value, e);
3    linkNodeLast(p);
4    return p;
5}

linkNodeLast方法负责双链表的连接:

 1private void linkNodeLast(LinkedHashMap.Entry<K,V> p) {
 2    LinkedHashMap.Entry<K,V> last = tail;
 3    tail = p;
 4    if (last == null)
 5        head = p;
 6    else {
 7        p.before = last;
 8        last.after = p;
 9    }
10}

afterNodeAccess

顾名思义，此方法应该在一次访问之后被调用，那么什么算是一次访问呢?LInkedHashMap对此做出了定义:

NIO

Sun, 24 May 2026 00:00:00 +0000

SelectionKey

SelectionKey的类图如下:

注意，SelectionKey使用AtomicReferenceFieldUpdater进行原子更新attachment，这货就像是Channel和Selector结合的结晶。

兴趣设置

SelectionKeyImpl.interestOps:

1public SelectionKey interestOps(int ops) {
2    ensureValid();
3    return nioInterestOps(ops);
4}

nioInterestOps利用Channel来实现:

1public SelectionKey nioInterestOps(int ops) {
2    channel.translateAndSetInterestOps(ops, this);
3    interestOps = ops;
4    return this;
5}

这里服务器和客户端有不同的实现。

客户端

SocketChannelImpl.translateAndSetInterestOps:

 1public void translateAndSetInterestOps(int ops, SelectionKeyImpl sk) {
 2    int newOps = 0;
 3    if ((ops & SelectionKey.OP_READ) != 0)
 4        newOps |= Net.POLLIN;
 5    if ((ops & SelectionKey.OP_WRITE) != 0)
 6        newOps |= Net.POLLOUT;
 7    if ((ops & SelectionKey.OP_CONNECT) != 0)
 8        newOps |= Net.POLLCONN;
 9    sk.selector.putEventOps(sk, newOps);
10}

服务器

ServerSocketChannelImpl.translateAndSetInterestOps:

Pandas Beginner

Sun, 24 May 2026 00:00:00 +0000

title: "2天学会Pandas" date: 2026-05-24T14:09:31Z

2天学会Pandas

0.导语

1.Series

2.DataFrame

2.1 DataFrame的简单运用

3.pandas选择数据

3.1 实战筛选

3.2 筛选总结

4.Pandas设置值

4.1 创建数据

4.2 根据位置设置loc和iloc

4.3 根据条件设置

4.4 按行或列设置

4.5 添加Series序列(长度必须对齐)

4.6 设定某行某列为特定值

4.7 修改一整行数据

5.Pandas处理丢失数据

5.1 创建含NaN的矩阵

5.2 删除掉有NaN的行或列

5.3 替换NaN值为0或者其他

5.4 是否有缺失数据NaN

6.Pandas导入导出

6.1 导入数据

6.2 导出数据

7.Pandas合并操作

7.1 Pandas合并concat

7.2.Pandas 合并 merge

*  7.2.1 定义资料集并打印出
*  7.2.2 依据key column合并,并打印
*  7.2.3 两列合并
*  7.2.4 Indicator设置合并列名称
*  7.2.5 依据index合并
*  7.2.6 解决overlapping的问题

8.Pandas plot出图

9.学习来源

0.导语

Pandas是基于Numpy构建的，让Numpy为中心的应用变得更加简单。

Pandas50

Sun, 24 May 2026 00:00:00 +0000

title: "Pandas练习题50题" date: 2026-05-24T14:09:49Z

Pandas练习题50题

作者：王大毛，和鲸社区

出处：https://www.kesci.com/home/project/5ddc974ef41512002cec1dca

修改：黄海广

Pandas 是基于 NumPy 的一种数据处理工具，该工具为了解决数据分析任务而创建。Pandas 纳入了大量库和一些标准的数据模型，提供了高效地操作大型数据集所需的函数和方法。这些练习着重DataFrame和Series对象的基本操作，包括数据的索引、分组、统计和清洗。

基本操作

1.导入 Pandas 库并简写为 pd，并输出版本号

1import pandas as pd
2pd.__version__

'0.22.0'

2. 从列表创建 Series

1arr = [0, 1, 2, 3, 4]
2df = pd.Series(arr) # 如果不指定索引，则默认从 0 开始
3df

0    0
1    1
2    2
3    3
4    4
dtype: int64

3. 从字典创建 Series

1d = {'a':1,'b':2,'c':3,'d':4,'e':5}
2df = pd.Series(d)
3df

a    1
b    2
c    3
d    4
e    5
dtype: int64

4. 从 NumPy 数组创建 DataFrame

1import numpy as np
2dates = pd.date_range('today', periods=6)  # 定义时间序列作为 index
3num_arr = np.random.randn(6, 4)  # 传入 numpy 随机数组
4columns = ['A', 'B', 'C', 'D']  # 将列表作为列名
5df = pd.DataFrame(num_arr, index=dates, columns=columns)
6df

Process

Sun, 24 May 2026 00:00:00 +0000

命令执行

我们以以下代码为例:

1Runtime runtime = Runtime.getRuntime();
2String cmd = "ls -l";
3Process process = runtime.exec(cmd);

实际调用了下面的exec方法:

1public Process exec(String command, String[] envp, File dir) {
2    StringTokenizer st = new StringTokenizer(command);
3    String[] cmdarray = new String[st.countTokens()];
4    for (int i = 0; st.hasMoreTokens(); i++)
5        cmdarray[i] = st.nextToken();
6    return exec(cmdarray, envp, dir);
7}

第二个参数为环境变量数组，其格式为key=value，第三个参数为命令的执行路径，如果为null，那么便为当前Java进程的工作路径，即user.dir。

StringTokenizer默认以空格、换行符、制表符，换页符(\f)为分隔单位，所以这里将命令ls -l分割为了ls和-l两部分。

1public Process exec(String[] cmdarray, String[] envp, File dir) {
2    return new ProcessBuilder(cmdarray)
3        .environment(envp)
4        .directory(dir)
5        .start();
6}

其实是用ProcessBuilder实现，此类在JDK1.5时加入。

Python打卡Day1

Sun, 24 May 2026 00:00:00 +0000

Question 1

Question:

Write a program which will find all such numbers which are divisible by 7 but are not a multiple of 5, between 2000 and 3200 (both included).The numbers obtained should be printed in a comma-separated sequence on a single line.

Hints:

Consider use range(#begin, #end) method.

Main author's Solution: Python 2

1l=[]
2for i in range(2000, 3201):
3    if (i%7==0) and (i%5!=0):
4        l.append(str(i))
5
6print ','.join(l)

My Solution: Python 3

Using for loops

1for i in range(2000,3201):
2    if i%7 == 0 and i%5!=0:
3        print(i,end=',')
4print("\b")

Using generators and list comprehension

1print(*(i for i in range(2000, 3201) if i%7 == 0 and i%5 != 0), sep=",")

Question 2

Question:

Write a program which can compute the factorial of a given numbers.The results should be printed in a comma-separated sequence on a single line.Suppose the following input is supplied to the program: 8 Then, the output should be:40320

Python打卡Day10

Sun, 24 May 2026 00:00:00 +0000

Question 31

Question:

Define a function which can print a dictionary where the keys are numbers between 1 and 20 (both included) and the values are square of keys.

Hints:

1Use dict[key]=value pattern to put entry into a dictionary.Use ** operator to get power of a number.Use range() for loops.

Main Author's Solution: Python 2

1def printDict():
2	d=dict()
3	for i in range(1,21):
4		d[i]=i**2
5	print d
6
7printDict()

My Solution: Python 3

1def printDict():
2    dict={i:i**2 for i in range(1,21)}   # Using comprehension method and
3    print(dict)
4
5printDict()

Question 32

Question:

Define a function which can generate a dictionary where the keys are numbers between 1 and 20 (both included) and the values are square of keys. The function should just print the keys only.

Python打卡Day11

Sun, 24 May 2026 00:00:00 +0000

Question 38

Question:

With a given tuple (1,2,3,4,5,6,7,8,9,10), write a program to print the first half values in one line and the last half values in one line.

Hints:

Use [n1:n2] notation to get a slice from a tuple.

Main Author's Solution: Python 2

1tp = (1,2,3,4,5,6,7,8,9,10)
2tp1 = tp[:5]
3tp2 = tp[5:]
4print tp1
5print tp2

My Solution: Python 3

1tpl = (1,2,3,4,5,6,7,8,9,10)
2
3for i in range(0,5):
4    print(tpl[i],end = ' ')
5print()
6for i in range(5,10):
7    print(tpl[i],end = ' ')

 1tpl = (1,2,3,4,5,6,7,8,9,10)
 2lst1,lst2 = [],[]
 3
 4for i in range(0,5):
 5    lst1.append(tpl[i])
 6
 7for i in range(5,10):
 8    lst2.append(tpl[i])
 9
10print(lst1)
11print(lst2)

Python打卡Day12

Sun, 24 May 2026 00:00:00 +0000

Question 44

Question:

Write a program which can map() to make a list whose elements are square of numbers between 1 and 20 (both included).

Hints:

Use map() to generate a list. Use lambda to define anonymous functions.

Main Author's Solution: Python 2

1squaredNumbers = map(lambda x: x**2, range(1,21))
2print squaredNumbers

My Solution: Python 3

1def sqr(x):
2    return x*x
3
4squaredNumbers = list(map(sqr, range(1,21)))
5print (squaredNumbers)

Question 45

Question:

Define a class named American which has a static method called printNationality.

Python打卡Day13

Sun, 24 May 2026 00:00:00 +0000

Question 47

Question

Define a class named Circle which can be constructed by a radius. The Circle class has a method which can compute the area.

Hints

Use def methodName(self) to define a method.

Main author's Solution: Python 2

1class Circle(object):
2    def __init__(self, r):
3        self.radius = r
4
5    def area(self):
6        return self.radius**2*3.14
7
8aCircle = Circle(2)
9print aCircle.area()

My Solution: Python 3

 1class Circle():
 2    def __init__(self,r):
 3        self.radius = r
 4
 5    def area(self):
 6        return 3.1416*(self.radius**2)
 7
 8
 9circle = Circle(5)
10print(circle.area())

Question 48

Question

Define a class named Rectangle which can be constructed by a length and width. The Rectangle class has a method which can compute the area.

Python打卡Day14

Sun, 24 May 2026 00:00:00 +0000

Question 51

Question

Write a function to compute 5/0 and use try/except to catch the exceptions.

Hints

Use try/except to catch exceptions.

Main author's Solution: Python 2

 1def throws():
 2    return 5/0
 3
 4try:
 5    throws()
 6except ZeroDivisionError:
 7    print "division by zero!"
 8except Exception, err:
 9    print 'Caught an exception'
10finally:
11    print 'In finally block for cleanup'

My Solution: Python 3

1def divide():
2    return 5/0
3
4try:
5    divide()
6except ZeroDivisionError as ze:
7    print("Why on earth you are dividing a number by ZERO!!")
8except:
9    print("Any other exception")

Question 52

Question

Define a custom exception class which takes a string message as attribute.

Python打卡Day15

Sun, 24 May 2026 00:00:00 +0000

Question 54

Question

Assuming that we have some email addresses in the "username@companyname.com" format, please write program to print the company name of a given email address. Both user names and company names are composed of letters only.

Example: If the following email address is given as input to the program:

1john@google.com

Then, the output of the program should be:

1google

In case of input data being supplied to the question, it should be assumed to be a console input.

Python打卡Day16

Sun, 24 May 2026 00:00:00 +0000

Question 60

Question

Write a program to compute:

1f(n)=f(n-1)+100 when n>0
2and f(0)=0

with a given n input by console (n>0).

Example: If the following n is given as input to the program:

Then, the output of the program should be:

In case of input data being supplied to the question, it should be assumed to be a console input.

Hints

We can define recursive function in Python.

Python打卡Day17

Sun, 24 May 2026 00:00:00 +0000

Question 65

Question

Please write assert statements to verify that every number in the list [2,4,6,8] is even.

Hints

Use "assert expression" to make assertion.

Main author's Solution: Python 2

1li = [2,4,6,8]
2for i in li:
3    assert i%2==0

My Solution: Python 3

1data = [2,4,5,6]
2for i in data:
3    assert i%2 == 0, "{} is not an even number".format(i)

Question 66

Question

Please write a program which accepts basic mathematic expression from console and print the evaluation result.

Example: If the following n is given as input to the program:

Python打卡Day18

Sun, 24 May 2026 00:00:00 +0000

Question 70

Question

Please write a program to output a random even number between 0 and 10 inclusive using random module and list comprehension.

Hints

Use random.choice() to a random element from a list.

Main author's Solution: Python 2

1li = [2,4,6,8]
2import random
3print random.choice([i for i in range(11) if i%2==0])

My Solution: Python 3

1import random
2resp = [i for i in range(0,11,2)]
3print(random.choice(resp))

Question 71

Question

Please write a program to output a random number, which is divisible by 5 and 7, between 10 and 150 inclusive using random module and list comprehension.

Python打卡Day19

Sun, 24 May 2026 00:00:00 +0000

Question 75

Question

Please write a program to randomly print a integer number between 7 and 15 inclusive.

Hints

Use random.randrange() to a random integer in a given range.

Solution:

1import random
2print random.randrange(7,16)

Question 76

Question

Please write a program to compress and decompress the string "hello world!hello world!hello world!hello world!".

Hints

Use zlib.compress() and zlib.decompress() to compress and decompress a string.

Solution:

1import zlib
2s = 'hello world!hello world!hello world!hello world!'
3t = zlib.compress(s)
4print t
5print zlib.decompress(t)

1'''Solution by: anas1434 
2'''
3s = 'hello world!hello world!hello world!hello world!'
4# In Python 3 zlib.compress() accepts only DataType <bytes>
5y = bytes(s, 'utf-8')
6x = zlib.compress(y)
7print(x)
8print(zlib.decompress(x))

Question 77

Question

Please write a program to print the running time of execution of "1+1" for 100 times.

Python打卡Day2

Sun, 24 May 2026 00:00:00 +0000

Question 4

Question:

Write a program which accepts a sequence of comma-separated numbers from console and generate a list and a tuple which contains every number.Suppose the following input is supplied to the program:

134,67,55,33,12,98

Then, the output should be:

1['34', '67', '55', '33', '12', '98']
2('34', '67', '55', '33', '12', '98')

Hints:

In case of input data being supplied to the question, it should be assumed to be a console input.tuple() method can convert list to tuple

Python打卡Day20

Sun, 24 May 2026 00:00:00 +0000

Question 80

Question

Please write a program to print the list after removing even numbers in [5,6,77,45,22,12,24].

Hints

Use list comprehension to delete a bunch of element from a list.

Main author's Solution: Python 2

1li = [5,6,77,45,22,12,24]
2li = [x for x in li if x%2!=0]
3print li

My Solution: Python 3

1def isEven(n):
2    return n%2!=0
3
4li = [5,6,77,45,22,12,24]
5lst = list(filter(isEven,li))
6print(lst)

1li = [5,6,77,45,22,12,24]
2lst = list(filter(lambda n:n%2!=0,li))
3print(lst)

Question 81

Question

By using list comprehension, please write a program to print the list after removing numbers which are divisible by 5 and 7 in [12,24,35,70,88,120,155].

Python打卡Day21

Sun, 24 May 2026 00:00:00 +0000

Question 85

Question

By using list comprehension, please write a program to print the list after removing the 0th,4th,5th numbers in [12,24,35,70,88,120,155].

Hints

Use list comprehension to delete a bunch of element from a list.Use enumerate() to get (index, value) tuple.

Main author's Solution: Python 2

1li = [12,24,35,70,88,120,155]
2li = [x for (i,x) in enumerate(li) if i not in (0,4,5)]
3print li

My Solution: Python 3

1li = [12,24,35,70,88,120,155]
2li = [li[i] for i in range(len(li)) if i not in (0,4,5)]
3print(li)

1'''Solution by: pratikb0501
2'''
3li = [12, 24, 35, 70, 88, 120, 155]
4print(list(j for i, j in enumerate(li) if i != 0 and i != 4 and i != 5))

Question 86

Question

By using list comprehension, please write a program to print the list after removing the value 24 in [12,24,35,24,88,120,155].

Python打卡Day22

Sun, 24 May 2026 00:00:00 +0000

Question 90

Question

Please write a program which count and print the numbers of each character in a string input by console.

Example: If the following string is given as input to the program:

1abcdefgabc

Then, the output of the program should be:

1a,2
2c,2
3b,2
4e,1
5d,1
6g,1
7f,1

Hints

Use dict to store key/value pairs. Use dict.get() method to lookup a key with default value.

Main author's Solution: Python 2

Python打卡Day23

Sun, 24 May 2026 00:00:00 +0000

The extended part of the repository starts from this page. Previous 94 problems were collected from the repository mentioned in intro. The following problems are collected from Hackerrank and other resources from internet.All the given solutions are in python 3.

Question 95

Question

Given the participants' score sheet for your University Sports Day, you are required to find the runner-up score. You are given scores. Store them in a list and find the score of the runner-up.

Python打卡Day24

Sun, 24 May 2026 00:00:00 +0000

Question 100

Question

You are given words. Some words may repeat. For each word, output its number of occurrences. The output order should correspond with the input order of appearance of the word. See the sample input/output for clarification.

If the following string is given as input to the program:
14
2bcdef
3abcdefg
4bcde
5bcdef
Then, the output of the program should be:
13
22 1 1

Hints

Make a list to get the input order and a dictionary to count the word frequency

Python打卡Day3

Sun, 24 May 2026 00:00:00 +0000

Question 10

Question

Write a program that accepts a sequence of whitespace separated words as input and prints the words after removing all duplicate words and sorting them alphanumerically.

Suppose the following input is supplied to the program:

1hello world and practice makes perfect and hello world again

Then, the output should be:

1again and hello makes perfect practice world

Hints:

In case of input data being supplied to the question, it should be assumed to be a console input.We use set container to remove duplicated data automatically and then use sorted() to sort the data.

ReadWriteLock

Sun, 24 May 2026 00:00:00 +0000

类图

ReadWriteLock

Sync

ReentrantReadWriteLock内部的Sync的类图如下:

构造

默认构造器:

1public ReentrantReadWriteLock() {
2	this(false);
3}

含参构造器:

1public ReentrantReadWriteLock(boolean fair) {
2	sync = fair ? new FairSync() : new NonfairSync();
3	readerLock = new ReadLock(this);
4	writerLock = new WriteLock(this);
5}

写锁和读锁的构造器是一个套路，以读锁为例:

1protected ReadLock(ReentrantReadWriteLock lock) {
2	sync = lock.sync;
3}

可以看出，读写锁内部的sync其实就是ReentrantReadWriteLock的sync。

读锁

lock

ReadLock.lock:

1public void lock() {
2	sync.acquireShared(1);
3}

AbstractQueuedSynchronizer.acquireShared:

1public final void acquireShared(int arg) {
2	if (tryAcquireShared(arg) < 0)
3		doAcquireShared(arg);
4}

tryAcquireShared方法的实现位于Sync:

ReentrantLock

Sun, 24 May 2026 00:00:00 +0000

ReentrantLock

类图

如下:

Sync

观察ReentrantLock关键方法的源码可以发现，其实现是委托给内部类Sync实现的，所以逻辑的关键便在于此类。例子:

1public void lock() {
2    sync.lock();
3}

类图

很明显，锁的公平性便是由NonFairSync和FairSync实现的，从ReentrantLock的构造器可以印证这一点:

1public ReentrantLock(boolean fair) {
2    sync = fair ? new FairSync() : new NonfairSync();
3}

锁的公平性指的是等待时间最长的线程最有机会获得锁，但是这样会导致性能的下降，因为此时对于线程的调度和操作系统的调度是矛盾的。参考:

ReentrantLock(重入锁)以及公平性

非公平锁

lock

NonfairSync.lock:

1final void lock() {
2    if (compareAndSetState(0, 1))
3        setExclusiveOwnerThread(Thread.currentThread());
4    else
5        acquire(1);
6}

快速尝试

AbstractQueuedSynchronizer.compareAndSetState:

1protected final boolean compareAndSetState(int expect, int update) {
2    return unsafe.compareAndSwapInt(this, stateOffset, expect, update);
3}

AbstractQueuedSynchronizer内部有一个state变量:

Reflection

Sun, 24 May 2026 00:00:00 +0000

核心便是Class类，类图如下:

方法获取

我们以getDeclaredMethods为例:

1@CallerSensitive
2public Method[] getDeclaredMethods() throws SecurityException {
3    return copyMethods(privateGetDeclaredMethods(false));
4}

privateGetDeclaredMethods:

 1private Method[] privateGetDeclaredMethods(boolean publicOnly) {
 2    Method[] res;
 3    ReflectionData<T> rd = reflectionData();
 4    if (rd != null) {
 5        res = publicOnly ? rd.declaredPublicMethods : rd.declaredMethods;
 6        if (res != null) return res;
 7    }
 8    // No cached value available; request value from VM
 9    res = Reflection.filterMethods(this, getDeclaredMethods0(publicOnly));
10    if (rd != null) {
11        if (publicOnly) {
12            rd.declaredPublicMethods = res;
13        } else {
14            rd.declaredMethods = res;
15        }
16    }
17    return res;
18}

publicOnly为false便说明需要获取各种访问级别的方法。

ScheduledThreadPool

Sun, 24 May 2026 00:00:00 +0000

创建

我们一般使用下面的方式进行创建:

1public static ScheduledExecutorService newSingleThreadScheduledExecutor() {
2    return new DelegatedScheduledExecutorService(new ScheduledThreadPoolExecutor(1));
3}

DelegatedScheduledExecutorService实际上是对ScheduledExecutorService接口方法的转发，目的是只将ScheduledExecutorService接口的public方法暴露出来，这其实就是门面模式。

显然这里的核心便是ScheduledThreadPoolExecutor了:

ScheduledThreadPoolExecutor构造器:

1public ScheduledThreadPoolExecutor(int corePoolSize) {
2    super(corePoolSize, Integer.MAX_VALUE, 0, NANOSECONDS, new DelayedWorkQueue());
3}

所以，默认情况下创建的是corePoolSize为1的线程池，而maximumPoolSize却为int最大值!

其工作队列DelayedWorkQueue是ScheduledThreadPoolExecutor的嵌套类:

单次调度

1public <V> ScheduledFuture<V> schedule(Callable<V> callable, long delay, TimeUnit unit) {
2    RunnableScheduledFuture<V> t = decorateTask(callable,
3        new ScheduledFutureTask<V>(callable, triggerTime(delay, unit)));
4    delayedExecute(t);
5    return t;
6}

触发时间计算

1private long triggerTime(long delay, TimeUnit unit) {
2    return triggerTime(unit.toNanos((delay < 0) ? 0 : delay));
3}
4long triggerTime(long delay) {
5     return now() + ((delay < (Long.MAX_VALUE >> 1)) ? delay : overflowFree(delay));
6}

now方法即返回当前纳秒表示的时间，所以触发时间就是当前时间加延时。

Socket

Sun, 24 May 2026 00:00:00 +0000

客户端连接

先来看一下使用socket作为客户端的使用。假设有代码如下:

1public static void main(String[] args) {
2    Socket socket = new Socket();
3    socket.connect(new InetSocketAddress(8080));
4}

构造器

Socket的默认构造器源码如下:

1 public Socket() {
2    setImpl();
3}

setImpl()方法的目的在于设置系统采用的Socket实现，采用SocksSocketImpl。其类图:

SocksSocketImpl.jpg的父类会在静态代码块中进行初始化工作，下面分部分说明。

网络库加载

AbstractPlainSocketImpl静态代码块源码:

1static {
2    java.security.AccessController.doPrivileged(
3        new java.security.PrivilegedAction<Void>() {
4            public Void run() {
5                System.loadLibrary("net");
6                return null;
7            }
8        });
9    }

通过debug可以发现，对于Windows来说，这加载的实际上是位于jre下的bin/net.dll。

启动参数获取

PlainSocketImpl静态代码块源码:

 1static {
 2    java.security.AccessController.doPrivileged( new PrivilegedAction<Object>() {
 3            public Object run() {
 4                version = 0;
 5                try {
 6                    version = Float.parseFloat(System.getProperties().getProperty("os.version"));
 7                    preferIPv4Stack = Boolean.parseBoolean(
 8                                      System.getProperties().getProperty("java.net.preferIPv4Stack"));
 9                    exclBindProp = System.getProperty("sun.net.useExclusiveBind");
10                } catch (NumberFormatException e ) {
11                    assert false : e;
12                }
13                return null; // nothing to return
14            } });
15
16    // (version >= 6.0) implies Vista or greater.
17    if (version >= 6.0 && !preferIPv4Stack) {
18            useDualStackImpl = true;
19    }
20    if (exclBindProp != null) {
21        // sun.net.useExclusiveBind is true
22        exclusiveBind = exclBindProp.length() == 0 ? true
23                : Boolean.parseBoolean(exclBindProp);
24    } else if (version < 6.0) {
25        exclusiveBind = false;
26    }
27}

java.net.preferIPv4Stack参数默认为false，表示如果系统支持IPV6，那么即使用IPV6 socket，如果设置为true，那么将只使用V4.

STL源码剖析

Sun, 24 May 2026 00:00:00 +0000

一.简介

1.GNU源代码开放精神

全世界所有的STL实现版本，都源于Alexander Stepanov和Meng Lee完成的原始版本，这份原始版本有Hewlett-Packard Compant(惠普公司)拥有。每一个头文件都有一份声明，允许任何人任意运用、拷贝、修改、传播、贩卖这些代码，无需付费，唯一的条件是必须将声明置于使用者新开发的文件内

Thread

Sun, 24 May 2026 00:00:00 +0000

构造器

我们以喜闻乐见的为例:

1public Thread(Runnable target) {
2    init(null, target, "Thread-" + nextThreadNum(), 0);
3}

第一个参数为线程组，最后一个为栈大小。init方法就是一些内部属性的赋值操作。

启动

1public synchronized void start() {
2    if (threadStatus != 0)
3        throw new IllegalThreadStateException();
4    boolean started = false;
5    start0();
6    started = true;
7}

核心便在于native方法start0.

ThreadLocal

Sun, 24 May 2026 00:00:00 +0000

get

 1public T get() {
 2    Thread t = Thread.currentThread();
 3    ThreadLocalMap map = getMap(t);
 4    if (map != null) {
 5        ThreadLocalMap.Entry e = map.getEntry(this);
 6        if (e != null) {
 7            @SuppressWarnings("unchecked")
 8            T result = (T)e.value;
 9            return result;
10        }
11    }
12    return setInitialValue();
13}

getMap实现:

1ThreadLocalMap getMap(Thread t) {
2    return t.threadLocals;
3}

可以看出，其实ThreadLocalMap作为线程Thread的属性而存在:

ThreadPool

Sun, 24 May 2026 00:00:00 +0000

创建

我们以下列代码为例:

1public static ExecutorService newFixedThreadPool(int nThreads) {
2    return new ThreadPoolExecutor(nThreads, nThreads,
3        0L, TimeUnit.MILLISECONDS,
4        new LinkedBlockingQueue<Runnable>());
5}

可见默认使用LinkedBlockingQueue作为工作队列，其构造器:

1public LinkedBlockingQueue() {
2    this(Integer.MAX_VALUE);
3}

可见，这其实是一个有界队列，虽然大小为int最大值。

ThreadPoolExecutor便是JDK线程池的核心了，类图:

ThreadPoolExecutor构造器:

1public ThreadPoolExecutor(int corePoolSize,
2                          int maximumPoolSize,
3                          long keepAliveTime,
4                          TimeUnit unit,
5                          BlockingQueue<Runnable> workQueue) {
6    this(corePoolSize, maximumPoolSize, keepAliveTime, unit, workQueue,
7         Executors.defaultThreadFactory(), defaultHandler);
8}

线程工厂

默认的线程工厂是Executors的内部类，核心的newThread方法:

1public Thread newThread(Runnable r) {
2    Thread t = new Thread(group, r, namePrefix + threadNumber.getAndIncrement(), 0);
3    if (t.isDaemon())
4        t.setDaemon(false);
5    if (t.getPriority() != Thread.NORM_PRIORITY)
6        t.setPriority(Thread.NORM_PRIORITY);
7    return t;
8}

namePrefix定义:

TreeMap

Sun, 24 May 2026 00:00:00 +0000

类图:

TreeMap的构造器可以接受一个Comparator参数用以决定key的顺序，如果没有指定，那么尝试将key强转为Comparable，如果转换失败那就抛出异常了。

put

 1public V put(K key, V value) {
 2    Entry<K,V> t = root;
 3    if (t == null) {
 4        compare(key, key); // type (and possibly null) check
 5        root = new Entry<>(key, value, null);
 6        size = 1;
 7        modCount++;
 8        return null;
 9    }
10    int cmp;
11    Entry<K,V> parent;
12    // split comparator and comparable paths
13    Comparator<? super K> cpr = comparator;
14    //按Comparator查找
15    if (cpr != null) {
16        do {
17            parent = t;
18            cmp = cpr.compare(key, t.key);
19            if (cmp < 0)
20                t = t.left;
21            else if (cmp > 0)
22                t = t.right;
23            else
24                return t.setValue(value);
25        } while (t != null);
26    }
27    else {
28        if (key == null)
29            throw new NullPointerException();
30            //强转为Comparable
31            Comparable<? super K> k = (Comparable<? super K>) key;
32        do {
33            parent = t;
34            cmp = k.compareTo(t.key);
35            if (cmp < 0)
36                t = t.left;
37            else if (cmp > 0)
38                t = t.right;
39            else
40                return t.setValue(value);
41        } while (t != null);
42    }
43    Entry<K,V> e = new Entry<>(key, value, parent);
44    if (cmp < 0)
45        parent.left = e;
46    else
47        parent.right = e;
48    //修复红黑树
49    fixAfterInsertion(e);
50    size++;
51    modCount++;
52    return null;
53}

红黑树首先是一棵二叉查找树，put方法首先便是在树中进行搜索寻找合适的插入位置，然后修复红黑树。

UDP

Sun, 24 May 2026 00:00:00 +0000

一个简单的Server/Client交互过程参见udp包。从中可以看出其实对于UDP来说，服务器和客户端代码其实是一样的。在此我们将"服务器"简单的理解为数据的接收方。

类图

DatagramSocket

DatagramSocketImpl

数据接收

示例代码:

1DatagramSocket server = new DatagramSocket(8080);
2byte[] buffer = new byte[1024];
3DatagramPacket packet = new DatagramPacket(buffer, 1024);
4server.receive(packet);
5System.out.println("Server接收到: " + new String(buffer, 0, packet.getLength()));

Socket创建

创建和TCP相似，真正的实现位于DualStackPlainDatagramSocketImpl.datagramSocketCreate，简略版源码:

1protected void datagramSocketCreate() {
2    int newfd = socketCreate(false /* v6Only */);
3    fdAccess.set(fd, newfd);
4}

socketCreate为native方法，Windows实现位于DualStackPlainDatagramSocketImpl.c的Java_java_net_DualStackPlainDatagramSocketImpl_socketCreate方法，简略版源码:

1fd = (int) socket(AF_INET6, SOCK_DGRAM, 0);
2rv = setsockopt(fd, IPPROTO_IPV6, IPV6_V6ONLY, (char *) &opt, sizeof(opt));

结合TCP的创建过程可以发现，两者的创建都是通过Windows API socket方法完成，区别在于第二个参数，TCP是SOCK_STREAM而UDP是SOCK_DGRAM。

URLConnection

Sun, 24 May 2026 00:00:00 +0000

类图:

我们以urlconnection包的简单对百度首页的读取来开启我们的源码阅读旅程。

连接建立

URL构建

URL负责协议的解析以及相应的协议处理器的创建，协议解析的过程其实就是一个字符串的处理过程。协议处理器即URLStreamHandler，类图:

其子类如下图所示:

每种协议的处理器便在sun.net.www.protocol.xxx中，URL获取处理器就是一个手动拼接类名，用反射生成实例的过程。

创建连接对象

URL.openConnection:

1public URLConnection openConnection() {
2    return handler.openConnection(this);
3}

注意此处并未执行真正的连接操作，从上一节可以看出，这里的Handler应该是sun.net.www.protocol.http.Handler:

1protected java.net.URLConnection openConnection(URL u, Proxy p) {
2    //p为空
3    return new HttpURLConnection(u, p, this);
4}

这里的HttpURLConnection同样位于sun.net.www.protocol.http包下，构造器源码: