fastest method to choose a random element

Paul Hankin paul.hankin at gmail.com
Sat Jan 5 21:50:43 CET 2008


On Jan 5, 5:12 pm, Paul Hankin <paul.han... at gmail.com> wrote:
> On Jan 5, 4:14 pm, c... at mailinator.com wrote:
>
> > On Jan 5, 5:07 pm, c... at mailinator.com wrote:
>
> > > Hello, Paul and Arnaud.
> > > While I think about your answers: do you think there is any way to
> > > avoid shuffle?
> > > It may take unnecessary long on a long list most of whose elements
> > > have the property.
>
> You could generate a random shuffle of range(len(seq)) lazily, and use
> that to iterate over your sequence.
>
> import random
> import itertools
>
> def randxrange(n):
>     "Generate the numbers 0 to n-1 in a random order"
>     x = range(n)
>     for i in xrange(n):
>         r = random.randrange(n - i)
>         yield x[r]
>         x[r] = x[n - i - 1]
>
> def shuffled(seq):
>     "Generate the elements of seq in a random order"
>     return (seq[i] for i in randxrange(len(seq)))
>
> def pick_random(seq, prop):
>     return itertools.ifilter(prop, shuffled(seq)).next()

At the risk of filling this thread with my posts, here's a much
simplified and faster version of this code that uses no extra storage.

import random

def pick_random(seq, prop):
    L = len(seq)
    for i in xrange(L):
        r = random.randrange(L - i)
        if prop(seq[r]):
            return seq[r]
        seq[r] = seq[L - i - 1]

--
Paul Hankin



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