Is there such a beast as a "perfect" shuffle? :)

[Ian Bicking]

I'm late on the thread, but perhaps this is better:

def shuffle(l):
    l = [(random.random(), i) for i in l]
    l.sort()
    return [i for r, i in l]

If you use a real random number generator (e.g., /dev/urandom), instead
of random's pseudo-random generator, then you will be able to receive
the full len(l)! possibilities.  The precision of floats will lead to a
certain amount of non-randomness, with large enough l.  Perhaps you
could do something like:

  l.sort(lambda a, b: cmp(a[0], b[0]) or random.randrange(-1, 2))


I don't know what the distribution of this looks like:

def shuffle(l):
    l = l[:]
    l.sort(lambda a, b: cmp(random.randrange(-1, 2))
    return l

I guess it depends on sort's algorithm.  A nondeterministic sort is kind
of odd to think about.

  Ian