如何概率性地在python中填充列表?

时间:2019-01-30 20:53:03

标签: python list numpy for-loop probability

我想使用一个基本的for循环来填充Python中的值列表,但我希望以概率方式计算这些值,以便在(玩具)公式1和100-p中计算值的时间为p%在公式2中计算值的时间的百分比。

这是到目前为止我得到的:

    # generate list of random probabilities 
    p_list = np.random.uniform(low=0.0, high=1.0, size=(500,))
    my_list = []

    # loop through but where to put 'p'? append() should probably only appear once
    for p in p_list:
        calc1 = x*y # equation 1
        calc2 = (x-y) # equation 2
        my_list.append(calc1)
        my_list.append(calc2)

4 个答案:

答案 0 :(得分:2)

您已经生成了一个概率列表-p_list-与您要生成的my_list中的每个值相对应。这样做的pythonic方法是通过三元运算符和列表理解:

import random
my_list = [(x*y if random() < p else x-y) for p in p_list]

如果我们将其扩展为适当的for循环:

my_list = []
for p in p_list:
    if random() < p:
        my_list.append(x*y)
    else:
        my_list.append(x-y)

如果我们想对calc1calc2更具Python风格,可以将它们变成lambda:

calc1 = lambda x,y: x*y
calc2 = lambda x,y: x-y
...
my_list = [calc1(x,y) if random() < p else calc2(x,y) for p in p_list]

或者,根据xy的功能变化(假设它们不是静态的),您甚至可以分两个步骤进行理解:

calc_list = [calc1 if random() < p else calc2 for p in p_list]
my_list = [calc(x,y) for calc in calc_list]

答案 1 :(得分:1)

我采用了对原始代码进行最少更改并易于理解的语法的方法:

import numpy as np

p_list = np.random.uniform(low=0.0, high=1.0, size=(500,))

my_list = []

# uncomment below 2 lines to make this code syntactially correct
#x = 1
#y = 2

for p in p_list:
        # randoms are uniformly distributed over the half-open interval [low, high)
        # so check if p is in [0, 0.5) for equation 1 or [0.5, 1) for equation 2
        if p < 0.5:
                calc1 = x*y # equation 1
                my_list.append(calc1)
        else:
                calc2 = (x-y) # equation 2
                my_list.append(calc2)

答案 2 :(得分:0)

其他答案似乎假设您想保持计算出的机会不变。如果您所需要的只是列表的结果,那么使用等式1的时间为p%,使用等式2的时间为100-p%,这就是您所需要的:

from random import random, seed

inputs = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

# change the seed to see different 'random' outcomes
seed(1)
results = [x * x if random() > 0.5 else 2 * x for x in inputs]

print(results)

答案 3 :(得分:0)