Create random list of integers in Python

Question

I\'d like to create a random list of integers for testing purposes. The distribution of the numbers is not important. The only thing that is counting is time

Answer 1

Firstly, you should use randrange(0,1000) or randint(0,999), not randint(0,1000). The upper limit of randint is inclusive.

For efficiently, randint is simply a wrapper of randrange which calls random, so you should just use random. Also, use xrange as the argument to sample, not range.

You could use

[a for a in sample(xrange(1000),1000) for _ in range(10000/1000)]

to generate 10,000 numbers in the range using sample 10 times.

(Of course this won't beat NumPy.)

$ python2.7 -m timeit -s 'from random import randrange' '[randrange(1000) for _ in xrange(10000)]'
10 loops, best of 3: 26.1 msec per loop

$ python2.7 -m timeit -s 'from random import sample' '[a%1000 for a in sample(xrange(10000),10000)]'
100 loops, best of 3: 18.4 msec per loop

$ python2.7 -m timeit -s 'from random import random' '[int(1000*random()) for _ in xrange(10000)]' 
100 loops, best of 3: 9.24 msec per loop

$ python2.7 -m timeit -s 'from random import sample' '[a for a in sample(xrange(1000),1000) for _ in range(10000/1000)]'
100 loops, best of 3: 3.79 msec per loop

$ python2.7 -m timeit -s 'from random import shuffle
> def samplefull(x):
>   a = range(x)
>   shuffle(a)
>   return a' '[a for a in samplefull(1000) for _ in xrange(10000/1000)]'
100 loops, best of 3: 3.16 msec per loop

$ python2.7 -m timeit -s 'from numpy.random import randint' 'randint(1000, size=10000)'
1000 loops, best of 3: 363 usec per loop

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But since you don't care about the distribution of numbers, why not just use:

range(1000)*(10000/1000)

?