Python编程产生非均匀随机数的几种方法代码分享

1.反变换法

设需产生分布函数为F(x)的连续随机数X。若已有[0,1]区间均匀分布随机数R,则产生X的反变换公式为：

F(x)=r, 即x=F^-1(r)

反函数存在条件：如果函数y=f(x)是定义域D上的单调函数,那么f(x)一定有反函数存在,且反函数一定是单调的。分布函数F(x)为是一个单调递增函数，所以其反函数存在。从直观意义上理解，因为r一一对应着x,而在[0,1]均匀分布随机数R≤r的概率P(R≤r)=r。 因此,连续随机数X≤x的概率P(X≤x)=P(R≤r)=r=F(x)

即X的分布函数为F(x)。

例子:下面的代码使用反变换法在区间[0, 6]上生成随机数，其概率密度近似为P(x)=e^-x

import numpy as np
import matplotlib.pyplot as plt
# probability distribution we're trying to calculate
p = lambda x: np.exp(-x)
# CDF of p
CDF = lambda x: 1-np.exp(-x)
# invert the CDF
invCDF = lambda x: -np.log(1-x)
# domain limits
xmin = 0 # the lower limit of our domain
xmax = 6 # the upper limit of our domain
# range limits
rmin = CDF(xmin)
rmax = CDF(xmax)
N = 10000 # the total of samples we wish to generate
# generate uniform samples in our range then invert the CDF
# to get samples of our target distribution
R = np.random.uniform(rmin, rmax, N)
X = invCDF(R)
# get the histogram info
hinfo = np.histogram(X,100)
# plot the histogram
plt.hist(X,bins=100, label=u'Samples');
# plot our (normalized) function
xvals=np.linspace(xmin, xmax, 1000)
plt.plot(xvals, hinfo[0][0]*p(xvals), 'r', label=u'p(x)')
# turn on the legend
plt.legend()
plt.show()

一般来说，直方图的外廓曲线接近于总体X的概率密度曲线。

2.舍选抽样法(Rejection Methold)

用反变换法生成随机数时，如果求不出F-1(x)的解析形式或者F(x)就没有解析形式，则可以用F-1(x)的近似公式代替。但是由于反函数计算量较大，有时也是很不适宜的。另一种方法是由Von Neumann提出的舍选抽样法。下图中曲线w(x)为概率密度函数，按该密度函数产生随机数的方法如下：

基本的rejection methold步骤如下：

1. Draw x uniformly from [xmin xmax]

2. Draw x uniformly from [0, ymax]

3. if y < w(x)，accept the sample, otherwise reject it

4. repeat

即落在曲线w(x)和X轴所围成区域内的点接受，落在该区域外的点舍弃。

例子:下面的代码使用basic rejection sampling methold在区间[0, 10]上生成随机数，其概率密度近似为P(x)=e^-x

# -*- coding: utf-8 -*-
'''
The following code produces samples that follow the distribution P(x)=e^−x 
for x=[0, 10] and generates a histogram of the sampled distribution.
'''
import numpy as np
import matplotlib.pyplot as plt
P = lambda x: np.exp(-x)
# domain limits
xmin = 0 # the lower limit of our domain
xmax = 10 # the upper limit of our domain
# range limit (supremum) for y
ymax = 1
N = 10000  # the total of samples we wish to generate
accepted = 0 # the number of accepted samples
samples = np.zeros(N)
count = 0  # the total count of proposals
# generation loop
while (accepted < N):
    # pick a uniform number on [xmin, xmax) (e.g. 0...10)
  x = np.random.uniform(xmin, xmax)
    # pick a uniform number on [0, ymax)
  y = np.random.uniform(0,ymax)
    # Do the accept/reject comparison
  if y < P(x):
    samples[accepted] = x
    accepted += 1
    count +=1
  print count, accepted
# get the histogram info
# If bins is an int, it defines the number of equal-width bins in the given range 
(n, bins)= np.histogram(samples, bins=30) # Returns: n-The values of the histogram,n是直方图中柱子的高度
# plot the histogram
plt.hist(samples,bins=30,label=u'Samples')  # bins=30即直方图中有30根柱子
# plot our (normalized) function
xvals=np.linspace(xmin, xmax, 1000)
plt.plot(xvals, n[0]*P(xvals), 'r', label=u'P(x)')
# turn on the legend
plt.legend()
plt.show()

>>>

99552 10000

3.推广的舍取抽样法

从上图中可以看出，基本的rejection methold法抽样效率很低，因为随机数x和y是在区间[xmin xmax]和区间[0 ymax]上均匀分布的，产生的大部分点不会落在w(x)曲线之下(曲线e-x的形状一边高一边低，其曲线下的面积占矩形面积的比例很小，则舍选抽样效率很低)。为了改进简单舍选抽样法的效率，可以构造一个新的密度函数q(x)(called a proposal distribution from which we can readily draw samples)，使它的形状接近p(x)，并选择一个常数k使得kq(x)≥w(x)对于x定义域内的值都成立。对应下图，首先从分布q(z)中生成随机数z0，然后按均匀分布从区间[0 kq(z0)]生成一个随机数u0。 if u0 > p(z0) then the sample is rejected,otherwise u0 is retained. 即下图中灰色区域内的点都要舍弃。可见，由于随机点u0只出现在曲线kq(x)之下，且在q(x)较大处出现次数较多，从而大大提高了采样效率。显然q(x)形状越接近p(x)，则采样效率越高。

根据上述思想，也可以表达采样规则如下：

1. Draw x from your proposal distribution q(x)

2. Draw y uniformly from [0 1]

3. if y < p(x)/kq(x) , accept the sample, otherwise reject it

4. repeat

下面例子中选择函数p(x)=1/(x+1)作为proposal distribution，k=1。曲线1/(x+1)的形状与e-x相近。

import numpy as np
import matplotlib.pyplot as plt
p = lambda x: np.exp(-x)     # our distribution
g = lambda x: 1/(x+1)      # our proposal pdf (we're choosing k to be 1)
CDFg = lambda x: np.log(x +1)  # generates our proposal using inverse sampling
# domain limits
xmin = 0 # the lower limit of our domain
xmax = 10 # the upper limit of our domain
# range limits for inverse sampling
umin = CDFg(xmin)
umax = CDFg(xmax)
N = 10000 # the total of samples we wish to generate
accepted = 0 # the number of accepted samples
samples = np.zeros(N)
count = 0 # the total count of proposals
# generation loop
while (accepted < N):
    # Sample from g using inverse sampling
  u = np.random.uniform(umin, umax)
  xproposal = np.exp(u) - 1
  # pick a uniform number on [0, 1)
  y = np.random.uniform(0, 1)
  # Do the accept/reject comparison
  if y < p(xproposal)/g(xproposal):
    samples[accepted] = xproposal
    accepted += 1
    count +=1
  print count, accepted
# get the histogram info
hinfo = np.histogram(samples,50)
# plot the histogram
plt.hist(samples,bins=50, label=u'Samples');
# plot our (normalized) function
xvals=np.linspace(xmin, xmax, 1000)
plt.plot(xvals, hinfo[0][0]*p(xvals), 'r', label=u'p(x)')
# turn on the legend
plt.legend()
plt.show()

 >>>

24051 10000

可以对比基本的舍取法和改进的舍取法的结果，前者产生符合要求分布的10000个随机数运算了99552步，后者运算了24051步，可以看到效率明显提高。

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