On Jan 5, 11:36 pm, [EMAIL PROTECTED] wrote: > > This one is good. Someone commented that you destroy the list, but > that can be fixed: > > 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]= seq[L - i - 1],seq[r] > > just pushing elements without the property to the end of the list > (list is mutated, true, but shuffle will do that too). In each > iteration of the for loop, there is only one random element, one check > of the property, and rebinding of elements without altering the lenght > of the list. This is optimal and has all the functionality. > > Two more comments: > for buzcor: deleting an element in the middle of a list is costly > for martin: that is certainly enough for me > > Thanks everybody!
How about some benchmarks as a token of gratitude? ;) Or at least a nice hint on your expected number of lists (and their sizes) :) -- http://mail.python.org/mailman/listinfo/python-list
