Boris Borcic wrote: > does > > x.sort(cmp = lambda x,y : cmp(random.random(),0.5)) > > pick a random shuffle of x with uniform distribution ? > > Intuitively, assuming list.sort() does a minimal number of comparisons to > achieve the sort, I'd say the answer is yes. But I don't feel quite > confortable > with the intuition... can anyone think of a more solid argumentation ?
Try this:
x.sort(key=lambda x: random.random())
Raymond
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