How to create a set of random numbers with numpy

Question

My requirement is to create a set of random numbers. I know the set size and the random number range. Therefore:

import random def makeset(lo, hi, ss): s = set() while len(s) < ss: s.add(random.randint(lo, hi)) return s

I realise that certain combinations of lo, hi & ss could result in an impossible task but that's not in scope here.

I've started learning about numpy and wondered if that module offered something that could do this in a single step - i.e., no explicit loop.

But I can't find anything in the documentation. Maybe it's not there.

Any thoughts?

Do you have to do this with numpy? The same result can be achieved using random.sample see docs.python.org/3/library/random.html#random.sample — ChrisOram
– ChrisOram, Commented Jan 21, 2022 at 9:23
I do not need to use numpy. I was just wondering if there was something in that module. Having said that, random.sample is what I should have used in the first place. Thank you @ChrisOram — jackal
– jackal, Commented Jan 21, 2022 at 9:25
While not specifically about numpy, this and this are highly related and do have a few answers with numpy. Not sure if counts as a duplicate — Tomerikoo
– Tomerikoo, Commented Jan 21, 2022 at 11:42

hilberts_drinking_problem · Accepted Answer · 2022-01-21 10:01:33Z

2

You could use np.random.choice without replacement as follows:

import numpy as np gen = np.random.Generator(np.random.PCG64()) def with_numpy_gen(lo, hi, ss): return gen.choice(np.arange(lo, hi + 1), ss, replace=False)

edited Jan 21, 2022 at 10:01

answered Jan 21, 2022 at 9:29

hilberts_drinking_problem

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3 Comments

user2357112 Over a year ago

numpy.random.choice is extremely inefficient - it builds an entire permutation of the input array and then slices it to take the sample. They can't change it due to backward compatibility guarantees.

user2357112 Over a year ago

numpy.random.Generator.choice doesn't have the efficiency issue, but the old API still comes up first in all the search results, and the new API has the awkward requirement of forcing you to manage an explicit RNG object - using an explicit RNG is good practice, but it's hard to get anyone to care enough to do it.

hilberts_drinking_problem Over a year ago

@user2357112supportsMonica nevermind, I think my benchmarks were off. It still seems like using a generator is the way to go.

ansev · Accepted Answer · 2022-01-21 13:16:02Z

We could use np.random.default_rng

Construct a new Generator with the default BitGenerator (PCG64)

import numpy as np gen = np.random.default_rng() def makeset(lo, hi, ss): return set(gen.choice(np.arange(lo, hi + 1), ss, replace=False))

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