使用python生成随机单词

Question

I have a list of words

count=100    
list = ['apple','orange','mango']

for the count above using random function is it possible to select 40% of the time apple, 30% of the time orange and 30% of the time mango?

for ex:

for the count=100, 40 times apple, 30 times orange and 30 times mango.

this select has to happen randomly

Answer 1

Based on an answer to the question about generating discrete random variables with specified weights , you can use numpy.random.choice to get 20 times faster code than with random.choice :

from numpy.random import choice

sample = choice(['apple','orange','mango'], p=[0.4, 0.3, 0.3], size=1000000)

from collections import Counter
print(Counter(sample))

Outputs:

Counter({'apple': 399778, 'orange': 300317, 'mango': 299905})

Not to mention that it is actually easier than "to build a list in the required proportions and then shuffle it".

Also, shuffle would always produce exactly 40% apples, 30% orange and 30% mango, which is not the same as saying "produce a sample of million fruits according to a discrete probability distribution". The latter is what both choice solutions do (and the bisect too). As can be seen above, there is about 40% apples, etc., when using numpy .

Answer 2

The easiest way is to build a list in the required proportions and then shuffle it.

>>> import random
>>> result = ['apple'] * 40 + ['orange'] * 30 + ['mango'] * 30
>>> random.shuffle(result)

Edit for the new requirement that the count is really 1,000,000:

>>> count = 1000000
>>> pool = ['apple'] * 4 + ['orange'] * 3 + ['mango'] * 3
>>> for i in xrange(count):
        print random.choice(pool)

A slower but more general alternative approach is to bisect a cumulative probability distribution :

>>> import bisect
>>> choices = ['apple', 'orange', 'mango']
>>> cum_prob_dist = [0.4, 0.7]
>>> for i in xrange(count):
        print choices[bisect.bisect(cum_prob_dist, random.random())]