I was going through the docs for module-random<http://docs.python.org/lib/module-random.html> And I came through this, *shuffle*( x[, random]) Shuffle the sequence x in place. The optional argument random is a 0-argument function returning a random float in [0.0, 1.0); by default, this is the function random().
Note that for even rather small len(x), the total number of permutations of x is larger than the period of most random number generators; this implies that most permutations of a long sequence can never be generated. Why would we need to generate the total number of permutations for the list while trying to shuffle it? Should not we just use a knuth shuffle<http://en.wikipedia.org/wiki/Knuth_shuffle#Shuffling_algorithms>which does not require us to generate the total number of permutations. As long as len(x) is smaller than period of random number generator, a permutation can be generated. A quick first draft of implemetation might be, import random def shuffle(list_): for i in range(len(list_)): j = random.randint(i, len(list_)-1) list_[i], list_[j] = list_[j], list_[i] if __name__ == '__main__': a = range(100) shuffle(a) print a This is my first post to the list, I am not sure if this is the correct place to ask these questions. Please advise if this is not the correct place. --Shabda
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