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. I know generating random numbers is a time-consuming task, but there must be a better way.
Here's my current solution:
import random import timeit # Random lists from [0-999] interval print [random.randint(0, 1000) for r in xrange(10)] # v1 print [random.choice([i for i in xrange(1000)]) for r in xrange(10)] # v2 # Measurement: t1 = timeit.Timer('[random.randint(0, 1000) for r in xrange(10000)]', 'import random') # v1 t2 = timeit.Timer('random.sample(range(1000), 10000)', 'import random') # v2 print t1.timeit(1000)/1000 print t2.timeit(1000)/1000
v2 is faster than v1, but it is not working on such a large scale. It gives the following error:
ValueError: sample larger than population
Is there a fast, efficient solution that works at that scale?
NumPy came, saw, and conquered.
It is not entirely clear what you want, but I would use numpy.random.randint:
import numpy.random as nprnd import timeit t1 = timeit.Timer('[random.randint(0, 1000) for r in xrange(10000)]', 'import random') # v1 ### Change v2 so that it picks numbers in (0, 10000) and thus runs... t2 = timeit.Timer('random.sample(range(10000), 10000)', 'import random') # v2 t3 = timeit.Timer('nprnd.randint(1000, size=10000)', 'import numpy.random as nprnd') # v3 print t1.timeit(1000)/1000 print t2.timeit(1000)/1000 print t3.timeit(1000)/1000
which gives on my machine:
0.0233682730198 0.00781716918945 0.000147947072983
Note that randint is very different from random.sample (in order for it to work in your case I had to change the 1,000 to 10,000 as one of the commentators pointed out -- if you really want them from 0 to 1,000 you could divide by 10).
And if you really don't care what distribution you are getting then it is possible that you either don't understand your problem very well, or random numbers -- with apologies if that sounds rude...
All the random methods end up calling
random.random() so the best way is to call it directly:
[int(1000*random.random()) for i in xrange(10000)]
random.randrangehas a bunch of overhead to check the range before returning
istart + istep*int(self.random() * n).
NumPy is much faster still of course.