Create random list of integers in Python

问题:

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?

Some results from the answer

Andrew's: 0.000290962934494

gnibbler's: 0.0058455221653

KennyTM's: 0.00219276118279

NumPy came, saw, and conquered.

回答1:

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...

回答2:

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)]

For example,

• random.randint calls random.randrange.
• random.randrange has a bunch of overhead to check the range before returning istart + istep*int(self.random() * n).

NumPy is much faster still of course.

回答3:

Your question about performance is moot—both functions are very fast. The speed of your code will be determined by what you do with the random numbers.

However it's important you understand the difference in behaviour of those two functions. One does random sampling with replacement, the other does random sampling without replacement.

回答4:

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

---

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

range(1000)*(10000/1000)

?