Sample from a 2d probability numpy array?

Question

Say that I have an 2d array ar like this:

0.9, 0.1, 0.3
0.4, 0.5, 0.1
0.5, 0.8, 0.5

And I want to sample from [1, 0] according to this probability array.

rdchoice = lambda x: numpy.random.choice([1, 0], p=[x, 1-x])

I have tried two methods:

1) reshape it into a 1d array first and use numpy.random.choice and then reshape it back to 2d:

np.array(list(map(rdchoice, ar.reshape((-1,))))).reshape(ar.shape)

2) use the vectorize function.

func = numpy.vectorize(rdchoice)
func(ar)

But these two ways are all too slow, and I learned that the nature of the vectorize is a for-loop and in my experiments, I found that map is no faster than vectorize.

I thought this can be done faster. If the 2d array is large it would be unbearably slow.

How fast is np.reshape([np.random.choice([0, 1], p=[x, 1-x]) for x in arr.flat], arr.shape)? — cs95
– cs95, Commented Feb 25, 2019 at 23:51

user545424 · Accepted Answer · 2019-02-26 00:00:28Z

4

You should be able to do this like so:

>>> p = np.array([[0.9, 0.1, 0.3], [0.4, 0.5, 0.1], [0.5, 0.8, 0.5]])
>>> (np.random.rand(*p.shape) < p).astype(int)

answered Feb 26, 2019 at 0:00

user545424

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2 Comments

Lerner Zhang Over a year ago

Great answer! Thanks very much. A follow-up. I wonder if any futher optimation would be possible if the samples are not [0, 1], say it is [-1, 1]. I know I can just add an additional step by first scaling it and then add a value: Y=2*X-1. But is it possible to do it in one step? What if the samples are arbitrary like [23, 15]?

Paul Panzer Over a year ago

@lerner np.where(np.random.random(p.shape) < p, 23, 15)

vidstige · Accepted Answer · 2023-02-14 07:43:44Z

-1

Actually I can use the np.random.binomial:

import numpy as np
p = [[0.9, 0.1, 0.3],
     [0.4, 0.5, 0.1],
     [0.5, 0.8, 0.5]]

np.random.binomial(1, p)

edited Feb 14, 2023 at 7:43

vidstige

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answered Apr 11, 2019 at 6:15

Lerner Zhang

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