随机抽样的python实现

目录

随机抽样

numpy.random 模块对 Python 内置的 random 进行了补充，增加了一些用于高效生成多种概率分
布的样本值的函数，如正态分布、泊松分布等。

• numpy.random.seed(seed=None) Seed the generator.

seed() 用于指定随机数生成时所用算法开始的整数值，如果使用相同的 seed() 值，则每次生成的随
机数都相同，如果不设置这个值，则系统根据时间来自己选择这个值，此时每次生成的随机数因时间
差异而不同。

在对数据进行预处理时，经常加入新的操作或改变处理策略，此时如果伴随着随机操作，最好还是指
定唯一的随机种子，避免由于随机的差异对结果产生影响。

离散型随机变量

二项分布

二项分布可以用于一次实验只有两种结果，各结果对应的概率相等的多次实验的概率问题。比如
处理猜10次拳赢6次的概率等类似的问题。
二项分布概率函数的数学表示：

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• numpy.random.binomial(n, p, size=None) Draw samples from a binomial distribution

表示对一个二项分布进行采样， size 表示采样的次数， n 表示做了 n 重伯努利试验， p 表示成功的
概率，函数的返回值表示 n 中成功的次数。

【例】野外正在进行9（n=9）口石油勘探井的发掘工作，每一口井能够开发出油的概率是
0.1（p=0.1）。请问，最终所有的勘探井都勘探失败的概率？

import numpy as np
import matplotlib.pyplot as plt
from scipy import stats

np.random.seed(20201125)
n = 9# 做某件事情的次数
p = 0.1# 做某件事情成功的概率
size = 50000
x = np.random.binomial(n, p, size)#成功的次数

'''
或者使用binom.rvs
使用binom.rvs(n, p, size=1)函数模拟一个二项随机变量,可视化地表现概率
y = stats.binom.rvs(n, p, size=size)#返回一个numpy.ndarray
'''
print(np.sum(x == 0) / size) # 0.3897
plt.hist(x)
plt.xlabel('随机变量：成功次数')
plt.ylabel('样本中出现的次数')
plt.show()

0.3904


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plt.hist(x)
plt.xlabel('成功次数')
plt.ylabel('样本中次数')
plt.show()

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s = stats.binom.pmf(range(10), n, p)
print(np.around(s, 3))#它返回一个列表，列表中每个元素表示随机变量中对应值的概率

[0.387 0.387 0.172 0.045 0.007 0.001 0.    0.    0.    0.   ]

【例】模拟投硬币，投2次，请问两次都为正面的概率？

import numpy as np
from scipy import stats
import matplotlib.pyplot as plt
np.random.seed(20200605)
n = 2# 做某件事情的次数,这里是投两次硬币
p = 0.5#做某件事情成功的概率，在这里即投硬币为正面的概率
size = 50000
x = np.random.binomial(n, p, size)#成功的次数
'''或者使用binom.rvs
#使用binom.rvs(n, p, size=1)函数模拟一个二项随机变量,可视化地表现概率
y = stats.binom.rvs(n, p, size=size)#返回一个numpy.ndarray
'''
print(np.sum(x == 0) / size) # 0.25154
print(np.sum(x == 1) / size) # 0.49874
print(np.sum(x == 2) / size) # 0.24972
plt.hist(x, density=True)
plt.xlabel('随机变量：硬币为正面次数')
plt.ylabel('50000个样本中出现的次数')
plt.show()
#它返回一个列表，列表中每个元素表示随机变量中对应值的概率
s = stats.binom.pmf(range(n + 1), n, p)
print(np.around(s, 3))
# [0.25 0.5 0.25]

0.25154
0.49874
0.24972


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[0.25 0.5  0.25]

泊松分布

泊松分布主要用于估计某个时间段某事件发生的概率。泊松概率函数的数学表示：

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• numpy.random.poisson(lam=1.0, size=None) Draw samples from a Poisson distribution.

表示对一个泊松分布进行采样， size 表示采样的次数， lam 表示一个单位内发生事件的平均值，函
数的返回值表示一个单位内事件发生的次数。

【例】假定某航空公司预定票处平均每小时接到42次订票电话，那么10分钟内恰好接到6次电话的概
率是多少？

import numpy as np
from scipy import stats
import matplotlib.pyplot as plt

np.random.seed(20200605)
lam = 42 / 6# 平均值：平均每十分钟接到42/6次订票电话
size = 50000
x = np.random.poisson(lam, size)
'''或者
#模拟服从泊松分布的50000个随机变量
x = stats.poisson.rvs(lam,size=size)
'''

print(np.sum(x == 6) / size) # 0.14988
plt.hist(x)
plt.xlabel('随机变量：每十分钟接到订票电话的次数')
plt.ylabel('50000个样本中出现的次数')
plt.show()

#用poisson.pmf(k, mu)求对应分布的概率:概率质量函数 (PMF)
x = stats.poisson.pmf(6, lam)
print(x) # 0.14900277967433773

0.14988

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0.14900277967433773

超几何分布

在超几何分布中，各次实验不是独立的，各次实验成功的概率也不等。 超几何分布概率函数的数学表
示：

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• numpy.random.hypergeometric(ngood, nbad, nsample, size=None) Draw samples from a
  Hypergeometric distribution.

表示对一个超几何分布进行采样， size 表示采样的次数， ngood 表示总体中具有成功标志的元素个
数， nbad 表示总体中不具有成功标志的元素个数， ngood+nbad 表示总体样本容量， nsample 表示抽
取元素的次数（小于或等于总体样本容量），函数的返回值表示抽取 nsample 个元素中具有成功标识
的元素个数

【例】一共20只动物里有7只是狗，抽取12只有3只狗的概率（无放回抽样）。

import numpy as np
from scipy import stats
import matplotlib.pyplot as plt
np.random.seed(20200605)
size = 500000
x = np.random.hypergeometric(ngood=7, nbad=13, nsample=12, size=size)
'''或者
#用rvs(M, n, N, loc=0, size=1, random_state=None)模拟
x = stats.hypergeom.rvs(M=20,n=7,N=12,size=size)
'''

print(np.sum(x == 3) / size) # 0.198664
plt.hist(x, bins=8) #条数为8
plt.xlabel('狗的数量')
plt.ylabel('50000个样本中出现的次数')
plt.title('超几何分布',fontsize=20)
plt.show()


"""
M 为总体容量
n 为总体中具有成功标志的元素的个数
N,k 表示抽取N个元素有k个是成功元素
"""
x = range(8)
#用hypergeom.pmf(k, M, n, N, loc)来计算k次成功的概率
s = stats.hypergeom.pmf(k=x, M=20, n=7, N=12)
print(np.round(s, 3))  #保留三位小数
# [0. 0.004 0.048 0.199 0.358 0.286 0.095 0.01 ]

0.198664


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[0.    0.004 0.048 0.199 0.358 0.286 0.095 0.01 ]

超几何分布的均值与方差
均值E(x) = N(n/M)
方差Var(x) = N(n/M)(1‐n/M)((M‐N)/(M‐1))
注释：考虑n次实验的超几何分布，令p=n/M,当总体容量足够大时((M‐N)/(M‐1))近似于1，此时数学期望为
Np，方差为Np(1‐p).

#用stats(M, n, N, loc=0, moments='mv')计算均值和方差
stats.hypergeom.stats(20,7,12,moments='mv')  #Mean(‘m’), variance(‘v’), skew(‘s’), and/or kurtosis(‘k’).

(array(4.2), array(1.14947368))

连续型随机变量

均匀分布

• numpy.random.uniform(low=0.0, high=1.0, size=None) Draw samples from a uniform
  distribution.

【例】在low到high范围内，创建大小为size的均匀分布的随机数

import numpy as np
import matplotlib.pyplot as plt
from scipy import stats

np.random.seed(20200614)
a = 0
b = 100
size = 50000
x = np.random.uniform(a, b, size=size)

print(np.all(x >= 0)) # True
print(np.all(x < 100)) # True
y = (np.sum(x < 50)- np.sum(x < 10)) / size
print(y) # 0.40144

plt.hist(x, bins=20)
plt.show()
a = stats.uniform.cdf(10, 0, 100)
b = stats.uniform.cdf(50, 0, 100)
print(b- a) # 0.4

True
True
0.40144

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0.4

作为 uniform() 的特列，可以得到 [0,1) 之间的均匀分布的随机数。

• numpy.random.rand(d0, d1, …, dn) Random values in a given shape.

Create an array of the given shape and populate it with random samples from a uniform
distribution over [0, 1) .

【例】根据指定大小产生[0,1)之间均匀分布的随机数。

import numpy as np
np.random.seed(20200614)
print(np.random.rand())
# 0.7594819171852776

print(np.random.rand(5))
# [0.75165827 0.16552651 0.0538581 0.46671446 0.89076925]

print(np.random.rand(4, 3))
# [[0.10073292 0.14624784 0.40273923]
# [0.21844459 0.22226682 0.37246217]
# [0.50334257 0.01714939 0.47780388]
# [0.08755349 0.86500477 0.70566398]]

np.random.seed(20200614)
print(np.random.uniform()) 
# 0.7594819171852776

print(np.random.uniform(size=5))
# [0.75165827 0.16552651 0.0538581 0.46671446 0.89076925]

print(np.random.uniform(size=(4, 3)))
# [[0.10073292 0.14624784 0.40273923]
# [0.21844459 0.22226682 0.37246217]
# [0.50334257 0.01714939 0.47780388]
# [0.08755349 0.86500477 0.70566398]]

0.7594819171852776
[0.75165827 0.16552651 0.0538581  0.46671446 0.89076925]
[[0.10073292 0.14624784 0.40273923]
 [0.21844459 0.22226682 0.37246217]
 [0.50334257 0.01714939 0.47780388]
 [0.08755349 0.86500477 0.70566398]]
0.7594819171852776
[0.75165827 0.16552651 0.0538581  0.46671446 0.89076925]
[[0.10073292 0.14624784 0.40273923]
 [0.21844459 0.22226682 0.37246217]
 [0.50334257 0.01714939 0.47780388]
 [0.08755349 0.86500477 0.70566398]]

作为 uniform 的另一特例，可以得到 [low,high) 之间均匀分布的随机整数

• numpy.random.randint(low, high=None, size=None, dtype=‘l’) Return random integers from
  low (inclusive) to high (exclusive)

Return random integers from the “discrete uniform” distribution of the specified dtype in
the “half-open” interval [low, high). If high is None (the default), then results are from [0,
low).

【例】若 high 不为 None 时，取[low,high)之间随机整数，否则取值[0,low)之间随机整数。

import numpy as np
np.random.seed(20200614)
x = np.random.randint(2, size=10)
print(x)
# [0 0 0 1 0 1 0 0 0 0]

x = np.random.randint(1, size=10)
print(x)
# [0 0 0 0 0 0 0 0 0 0]

x = np.random.randint(5, size=(2, 4))
print(x)
# [[3 3 0 1]
# [1 1 0 1]]

x = np.random.randint(1, 10, [3, 4])
print(x)
# [[2 1 7 7]
# [7 2 4 6]
# [8 7 2 8]]

[0 0 0 1 0 1 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0]
[[3 3 0 1]
 [1 1 0 1]]
[[2 1 7 7]
 [7 2 4 6]
 [8 7 2 8]]

正态分布

标准正态分布数学表示：

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• numpy.random.randn(d0, d1, …, dn) Return a sample (or samples) from the “standard
  normal” distribution

【例】根据指定大小产生满足标准正态分布的数组（均值为0，标准差为1）。

import numpy as np
import matplotlib.pyplot as plt
from scipy import stats

np.random.seed(20200614)
size = 50000
x = np.random.randn(size)
y1 = (np.sum(x < 1) - np.sum(x < -1)) / size
y2 = (np.sum(x < 2) - np.sum(x < -2)) / size
y3 = (np.sum(x < 3) - np.sum(x < -3)) / size
print(y1) # 0.68596
print(y2) # 0.95456
print(y3) # 0.99744

plt.hist(x, bins=20)
plt.show()
y1 = stats.norm.cdf(1) - stats.norm.cdf(-1)
y2 = stats.norm.cdf(2) - stats.norm.cdf(-2)
y3 = stats.norm.cdf(3) - stats.norm.cdf(-3)
print(y1) # 0.6826894921370859
print(y2) # 0.9544997361036416
print(y3) # 0.9973002039367398

0.68596
0.95456
0.99744

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0.6826894921370859
0.9544997361036416
0.9973002039367398

还可以指定分布以及所需参数来进行随机，例如高斯分布中的mu和sigma。

• numpy.random.normal(loc=0.0, scale=1.0, size=None) Draw random samples from a
  normal (Gaussian) distribution.
  
  normal() 为创建均值为 loc（mu），标准差为 scale（sigma），大小为 size 的数组。

• sigma * np.random.randn(…) + mu

【例】

import numpy as np
import matplotlib.pyplot as plt

np.random.seed(20200614)
x = 0.5 * np.random.randn(2, 4) + 5

'''或者
#模拟10000个随机变量
x = 0.5*stats.norm.rvs(size=(2,4))+5
'''
print(x)
# [[5.39654234 5.4088702 5.49104652 4.95817289]
# [4.31977933 4.76502391 4.70720327 4.36239023]]

np.random.seed(20200614)
mu = 5#平均值
sigma = 0.5#标准差
x = np.random.normal(mu, sigma, (2, 4))
print(x)
# [[5.39654234 5.4088702 5.49104652 4.95817289]
# [4.31977933 4.76502391 4.70720327 4.36239023]]size = 50000

x = np.random.normal(mu, sigma, size)
print(np.mean(x)) # 4.996403463175092
print(np.std(x, ddof=1)) # 0.4986846716715106（#样本标准差）
'''
ddof：int, optional
Means Delta Degrees of Freedom. The divisor used in calculations is N ‐ ddof, where N
represents the number of elements. By default ddof is zero.
'''

[[5.39654234 5.4088702  5.49104652 4.95817289]
 [4.31977933 4.76502391 4.70720327 4.36239023]]
[[5.39654234 5.4088702  5.49104652 4.95817289]
 [4.31977933 4.76502391 4.70720327 4.36239023]]
4.996403463175092
0.4986846716715106





'\nddof：int, optional\nMeans Delta Degrees of Freedom. The divisor used in calculations is N ‐ ddof, where N\nrepresents the number of elements. By default ddof is zero.\n'

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plt.hist(x, bins=20)
plt.show()

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指数分布

指数分布描述时间发生的时间长度间隔。指数分布的数学表示：

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• numpy.random.exponential(scale=1.0, size=None) Draw samples from an exponential
  distribution.

【例】 scale = 1/lambda

import numpy as np
import matplotlib.pyplot as plt
from scipy import stats

np.random.seed(20200614)
lam = 7
size = 50000
x = np.random.exponential(1 / lam, size)
'''或者
#rvs(loc=0, scale=1/lam, size=size, random_state=None)模拟
'''
y1 = (np.sum(x < 1 / 7)) / size
y2 = (np.sum(x < 2 / 7)) / size
y3 = (np.sum(x < 3 / 7)) / size
print(y1) # 0.63218
print(y2) # 0.86518
print(y3) # 0.95056

plt.hist(x, bins=20)
plt.show()
y1 = stats.expon.cdf(1 / 7, scale=1 / lam)
y2 = stats.expon.cdf(2 / 7, scale=1 / lam)
y3 = stats.expon.cdf(3 / 7, scale=1 / lam)
print(y1) # 0.6321205588285577
print(y2) # 0.8646647167633873
print(y3) # 0.950212931632136

0.63218
0.86518
0.95056

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0.6321205588285577
0.8646647167633873
0.950212931632136

其它随机函数

随机从序列中获取元素

• numpy.random.choice(a, size=None, replace=True, p=None) Generates a random sample
  from a given 1-D array.

从序列中获取元素，若 a 为整数，元素取值从 np.range(a) 中随机获取；若 a 为数组，取值从 a 数组
元素中随机获取。该函数还可以控制生成数组中的元素是否重复 replace ，以及选取元素的概率 p 。

import numpy as np
np.random.seed(20200614)
x = np.random.choice(10, 3)
print(x) # [2 0 1]
x = np.random.choice(10, 3, p=[0.05, 0, 0.05, 0.9, 0, 0, 0, 0, 0, 0])
print(x) # [3 2 3]
x = np.random.choice(10, 3, replace=False, p=[0.05, 0, 0.05, 0.9, 0, 0, 0, 0, 0, 0])
print(x) # [3 0 2]
aa_milne_arr = ['pooh', 'rabbit', 'piglet', 'Christopher']
x = np.random.choice(aa_milne_arr, 5, p=[0.5, 0.1, 0.1, 0.3])
print(x) # ['pooh' 'rabbit' 'pooh' 'pooh' 'pooh']
np.random.seed(20200614)
x = np.random.randint(0, 10, 3)
print(x) # [2 0 1]

[2 0 1]
[3 2 3]
[3 0 2]
['pooh' 'rabbit' 'pooh' 'pooh' 'pooh']
[2 0 1]

对数据集进行洗牌操作

数据一般都是按照采集顺序排列的，但是在机器学习中很多算法都要求数据之间相互独立，所以需要
先对数据集进行洗牌操作。

• numpy.random.shuffle(x) Modify a sequence in-place by shuffling its contents.

This function only shuffles the array along the first axis of a multi-dimensional array. The
order of sub-arrays is changed but their contents remains the same.

对 x 进行重排序，如果 x 为多维数组，只沿第 0 轴洗牌，改变原来的数组，输出为None。

【例】洗牌，改变自身内容，打乱顺序。

import numpy as np
np.random.seed(20200614)
x = np.arange(10)
np.random.shuffle(x)
print(x)
# [6 8 7 5 3 9 1 4 0 2]
print(np.random.shuffle([1, 4, 9, 12, 15]))
# None
x = np.arange(20).reshape((5, 4))
print(x)# [[ 0 1 2 3]
# [ 4 5 6 7]
# [ 8 9 10 11]
# [12 13 14 15]
# [16 17 18 19]]
np.random.shuffle(x)
print(x)
# [[ 4 5 6 7]
# [ 0 1 2 3]
# [ 8 9 10 11]
# [16 17 18 19]
# [12 13 14 15]]

[6 8 7 5 3 9 1 4 0 2]
None
[[ 0  1  2  3]
 [ 4  5  6  7]
 [ 8  9 10 11]
 [12 13 14 15]
 [16 17 18 19]]
[[ 8  9 10 11]
 [ 0  1  2  3]
 [12 13 14 15]
 [16 17 18 19]
 [ 4  5  6  7]]

numpy.random.permutation(x) Randomly permute a sequence, or return a permuted
range.

If x is a multi-dimensional array, it is only shuffled along its first index.

permutation() 函数的作用与 shuffle() 函数相同，可以打乱第0轴的数据，但是它不会改变原来的数
组。

【例】

import numpy as np
np.random.seed(20200614)
x = np.arange(10)
y = np.random.permutation(x)
print(y)
# [6 8 7 5 3 9 1 4 0 2]
print(np.random.permutation([1, 4, 9, 12, 15]))
# [ 4 1 9 15 12]
x = np.arange(20).reshape((5, 4))
print(x)
# [[ 0 1 2 3]
# [ 4 5 6 7]
# [ 8 9 10 11]
# [12 13 14 15]
# [16 17 18 19]]
y = np.random.permutation(x)
print(y)
# [[ 8 9 10 11]
# [ 0 1 2 3]
# [12 13 14 15]
# [16 17 18 19]
# [ 4 5 6 7]]

[6 8 7 5 3 9 1 4 0 2]
[ 4  1  9 15 12]
[[ 0  1  2  3]
 [ 4  5  6  7]
 [ 8  9 10 11]
 [12 13 14 15]
 [16 17 18 19]]
[[ 8  9 10 11]
 [ 0  1  2  3]
 [12 13 14 15]
 [16 17 18 19]
 [ 4  5  6  7]]