Python numpy: iterate random.rand() without using loop

Question

I'm trying to have randomize number between 0-1 until the sum of those are >=1. This process iterate for nTime. Code below:

from random import random

def funct(n): #10_000
    media = 0
    for _ in range(n):
        result = 0
        count = 0
        while result < 1:
            x = random()
            result += x
            count += 1
            if result >= 1:
                break
        media += count
    return media / n

I need to optimize the above code, utilize only numpy library, utilize numpy array, no loop python, no if python(numpy.where for example). How should I do it?

I need to optimize my code, because I'm try to reduce the time used. and my english is not the best

score 0 · Answer 1 · answered Apr 04 '22 at 21:14

0

I don't think you'll find any significant improvements by trying to avoid a loop here. If you for some reason need a significantly faster version of this simple system, would probably recommend simply using a faster language than Python.

answered Apr 04 '22 at 21:14

Otto Hanski

382
1
6

Ali_Sh · Answer 2 · 2022-04-06T15:57:06.357

Inspired by one of the SO answers, np.random.dirichlet or np.random.gamma could be used, cautiously, in this regard. In one of my tests:

spec_len = 15
result = np.random.dirichlet(np.ones(spec_len), size=1)

# [[0.04798813 0.04001169 0.03344301 0.03314941 0.02594012 0.00738061
#   0.05712809 0.20175982 0.0605601  0.17495591 0.01114262 0.12018851
#   0.08900246 0.05989448 0.03745501]]
# SUM --> 0.9999999999999999

shape, scale = 0.5, 0.25
result = np.random.gamma(shape, scale, spec_len)

# [5.44188272e-03 2.81687195e-01 1.68385119e-01 1.77502131e-03
#  1.21338381e-03 1.12168743e-02 5.98830384e-02 5.69830641e-02
#  3.13285820e-02 3.92879720e-01 1.69169125e-02 3.99294001e-07
#  1.24306290e-01 1.06084121e-02 4.63536093e-02]
# SUM --> 1.208979502912514

Both will get results near 1 (np.random.gamma is more flexible which can be used for some other magnitudes; It can be controlled by shape and scale).
These methods can be used in loop to ensure the sum >= 1; The proposed ways will be satisfying in terms of time. E.g.:

sum = 0
while sum <= 1:  # sum < 1
  result = np.random.dirichlet(np.ones(spec_len), size=1)
  sum = np.sum(result)

# SUM --> 1.0000000000000002

The fastest way that I know, if iteration is not matter, is using Numba decorators on your function. its decorator can be used as:

@nb.njit('float64(int64)', parallel=True)  # without parallelization: @nb.njit() --> with signatures: @nb.njit('float64(int64)')
def funct(n): #10_000
    media = 0
    for _ in nb.prange(n):  # change to 'for _ in range(n):' if not parallelized
        result = 0
        count = 0
        while result < 1:
            x = random()
            result += x
            count += 1
            if result >= 1:
                break
        media += count
    return media / n

Numba is the fastest way --> ~ 16 times faster

Benchmarks (Colab CPU)

The results are for n=10000, n=100000, n=1000000, n=10000000 respectively:

OP code:
100 loops, best of 5: 6.45 ms per loop
100 loops, best of 5: 64.9 ms per loop
100 loops, best of 5: 655  ms per loop
n=10000000 --> not tested estimated ~ 6 s per loop

NumPy dirichlet without while: --> SUM must be checked to ensure >= 1    (~ x12)
100 loops, best of 5: 528  µs per loop
100 loops, best of 5: 5.59 ms per loop
100 loops, best of 5: 55.7 ms per loop
100 loops, best of 5: 550  ms per loop

NumPy dirichlet wit while:                                               (~ x6)
100 loops, best of 5: 1.1  ms per loop
100 loops, best of 5: 11.4 ms per loop
100 loops, best of 5: 107  ms per loop
100 loops, best of 5: 1.07  s per loop

Numba nopython(njit) ~ nopython(njit) + signatures:                      (~ x7)
100 loops, best of 5: 899  µs per loop
100 loops, best of 5: 9.13 ms per loop
100 loops, best of 5: 91.5 ms per loop
100 loops, best of 5: 914  ms per loop

Numba nopython(njit) + signatures + parallel: <-- THE FASTEST            (~ x16)
100 loops, best of 5: 404  µs per loop
100 loops, best of 5: 3.96 ms per loop
100 loops, best of 5: 40   ms per loop
100 loops, best of 5: 401  ms per loop

Python numpy: iterate random.rand() without using loop

2 Answers2

Benchmarks (Colab CPU)