Comparison of np.random.choice vs np.random.shuffle for samples without replacement

Question

My use case is a bit specific. I want to sample 2 items without replacement from a list/array (of 50, or 100 elements). So I don't have to worry about arrays of sizes of 10^4 or 10^5 or multidimensional data.

I want to know

1. Which one, numpy.random.choice() or numpy.random.shuffle() is faster for this purpose, and why?
2. If they both produce random samples of "good quality"? That is, are both generating good random samples for my purpose, or does one produce less random samples? (Just a sanity check to make sure I am not overlooking something regarding the source codes of these functions).

For Question 1, I tried timing both functions (code below), and the shuffle method seems to about 5-6 times faster. Any insight you can give about this is most welcome. If there are faster ways to achieve my purpose, I will be glad to hear about them (I had looked at the options of python random module, but the fastest method from my testing was using np.random.shuffle()).

def shuffler(size, num_samples):
    items = list(range(size))
    np.random.shuffle(items)
    return items[:num_samples]
    
def chooser(size, num_samples):
    return np.random.choice(size, num_samples, replace=False)

%timeit shuffler(50, 2)
#> 1.84 µs ± 17.2 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
%timeit chooser(50, 2)
#> 13 µs ± 1.09 µs per loop (mean ± std. dev. of 7 runs, 100000 loops each)

You may think it's already optimized and I am wasting time trying to save pennies. But np.random.choice() is called 5000000 times in my code and takes about 8% of my runtime. It is being used in a loop to obtain 2 random samples from the population for each iteration. Pseudocode:

for t in range(5000000):
    # Random sample of 2 from the population without replacement.

If there is a smarter implementations for my requirement, I am open to suggestions.

PS: I am aware that shuffle performs in place operation, but as I just require the indices of the two random elements I do not essentially have to perform it on my original array. There are other questions that compares the two functions from python random module. But I require 2 samples without replacement.

Answer 1

To answer your questions:

1. shuffle seems to be the fastest implementation
2. It should give the same answers (in fact, it seems to be the same thing under the hood)

Let's start @SvenMarnach's answer here. This is not a dupe of that question, but the answer is useful. Unfortunately that answer doesn't bring it in line with shuffler timewise:

%timeit shuffler(50, 2)
2.47 µs ± 180 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

%timeit chooser(50, 2)
52.5 µs ± 3.58 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
    
rng = np.random.default_rng()
def chooser2(size, num_samples):
    return rng.choice(size, num_samples, replace=False)

%timeit chooser2(50, 2)
15.9 µs ± 1.41 µs per loop (mean ± std. dev. of 7 runs, 100000 loops each)

the random.sample answer is better:

import random 
def sampler(size, num_samples):
    return np.array(random.sample(range(size), num_samples))

%timeit sampler(50, 2)
4.6 µs ± 140 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)    

Still slower, though.

Since I'm not able to parse c code, I'll take sven's word for it that random.choice is doing shuffle-and-split in the background, so the methods should be equivalent. Why it's so much faster here is baffling to me though.

EDIT: A sample_indices based on @DaniMesejo's answer (slightly slower for num_samples = 2):

def sample_indices(pop, size, num_samples):
    arr = np.random.rand(pop, size)
    return np.argpartition(arr, num_samples, axis = 1)[:, :num_samples] 

Answer 2

See the source code for numpy.random.choice; with replace=False it creates a 50-item temporary list, shuffles that list, and takes two items from that list.

Since version 1.17, the implementation decisions for numpy.random.choice and numpy.random.shuffle, as with other numpy.random functions, can't be changed without affecting backward compatibility (see the recent RNG policy for NumPy). See also these questions:

• Why is random.sample faster than numpy's random.choice?
• Why does numpy.random.choice not use arithmetic coding?
• Does numpy.random.seed() always give the same random number every time?

Compare numpy.random.choice with numpy.random.Generator.choice, which is the newer way to sample items in NumPy 1.17 and later. The advantage is that numpy.random.Generator.choice is not subject to the same compatibility guarantee as numpy.random.choice or numpy.random.shuffle. If you care about performance of numpy.random.Generator you can file an issue in NumPy's GitHub repository.

Answer 3

You could use an alternative solution, the idea is to generate a random array and then find the position of the min and max:

import numpy as np


def sample_indices(ran, size):
    arr = np.random.rand(ran, size)
    mi = np.argmin(arr, axis=1).reshape((-1, 1))
    ma = np.argmax(arr, axis=1).reshape((-1, 1))
    return np.hstack((mi, ma))


def shuffler(size, num_samples):
    items = list(range(size))
    np.random.shuffle(items)
    return items[:num_samples]


def chooser(size, num_samples):
    return np.random.choice(size, num_samples, replace=False)


def sample_indices_shuffler(ran, size):
    return np.array([shuffler(size, 2) for _ in range(ran)])


def sample_indices_chooser(ran, size):
    return np.array([chooser(size, 2) for _ in range(ran)])

Here are the timings:

%timeit sample_indices_chooser(1000, 50)
17.3 ms ± 1.74 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
%timeit sample_indices_shuffler(1000, 50)
2.69 ms ± 215 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit sample_indices(1000, 50)
553 µs ± 22.3 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

For:

res = sample_indices(10, 50)
print(res)

Output

[[ 9  6]
 [31 42]
 [17 42]
 [24 45]
 [ 2 49]
 [27 31]
 [21 19]
 [ 7 16]
 [20 28]
 [32 36]]