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
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.
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())] 
