Numpy随机选择分配错误

Question

I have a list of numbers and another list of probabilities which corresponds to these numbers. I use numpy.random.choice to create a random 2d array:

choice = numpy.random.choice([10, 22, 30], [10, 10], p=[0.45, 0.45, 0.10])

In choice should be 45 ones, 45 twos and 10 zeros but after several runs i never get the correct distribution.

unique, counts = numpy.unique(choice, return_counts=True)
print(dict(zip(unique, counts)))

{10: 49, 22: 37, 30: 14}
{10: 47, 22: 42, 30: 11}
{10: 40, 22: 51, 30: 9}

What did i miss?

Answer 1

You are completely missing how sampling from a distribution works in practice. You never "get" the correct distribution, you always get an approximation to it, because you are sampling.

Only in the case where the number of samples is very large you should eventually converge to the target distribution. But since sampling is a stochastic process, there is always randomness on the results of the process.

And this of course applies to generating numbers with a (pseudo-)random number generator.

Answer 2

So if you flipped a coin a thousand times, you'd expect to always get exactly 500 heads?

If you want to control the exact count of each result, you can't rely on probabilities - instead, chose (without replacement) from a list in which each result is present with the desired multiplicity:

numpy.random.choice([10] * 45 + [22] * 45 + [30] * 10, [10, 10], replace=False)

Answer 3

What Matias said is true.

If you do want to create an array with exactly 45 zeros, 45 ones, and 10 twos, with a shape of (10, 10) but in a random order, you can do something like this:

import numpy as np
zeros = np.array([0]*45)
ones = np.array([1]*45)
twos = np.array([2]*10)
myarr = np.concatenate([zeros, ones, twos])

# Random permutation, followed by reshaping in (10, 10) form
choice = np.random.permutation(myarr).reshape(10,10)
unique, counts = np.unique(choice, return_counts=True)
print(dict(zip(unique, counts)))
{0: 45, 1: 45, 2: 10}

Answer 4

The sampling won't be exact, you could force all numbers to be in the output array by making a list of all numbers you want and then randomly shuffling it:

import numpy
import numpy.random

numbers = numpy.asarray(45*[10]+45*[22]+10*[30])
print (numbers)
numpy.random.shuffle(numbers) # numbers is changed in place
choice = numbers.reshape((10,10))

print (choice)
unique, counts = numpy.unique(choice, return_counts=True)
print(dict(zip(unique, counts)))