Generate random numbers according to a specific distribution

Question

How can I generate random numbers say between 0 and 1, according to the distribution 20%, 80% i.e.

for i in range(0, 10):
    print random.randint(0, 1)

>>> 1 1 0 1 0 1 1 1 1 1

Martijn Pieters · Accepted Answer · 2014-12-12T20:25:42.647

3

Make use of the fact that booleans are also integers; True is 1 and False is 0; the following produces 1 80% of the time, 0 otherwise:

int(random.random() < 0.8)

Demo:

>>> import random
>>> from collections import Counter
>>> Counter(int(random.random() < 0.8) for _ in range(1000))
Counter({1: 798, 0: 202})

edited Dec 12 '14 at 20:25

answered Dec 12 '14 at 20:12

Martijn Pieters

You don't think this is a duplicate question @MartijnPieters? – simonzack Dec 12 '14 at 20:55
1

@simonzack: I think the specific problem is better solved this way than the approach for more numbers as used in the duplicate. – Martijn Pieters Dec 12 '14 at 21:47

score 2 · Answer 2 · answered Dec 12 '14 at 20:08

2

One way to handle it:

>>>import random
>>>random.choice(2*[0]+8*[1])

To elaborate, this will select at random from a list containing 2 zeros and 8 ones.

answered Dec 12 '14 at 20:08

SmearingMap

2 Answers2