On Wed, 2006-12-06 at 17:33 +0100, Schüle Daniel wrote: > def permute3gen(lst): > lenlst = len(lst) > def permute(perm, level): > if level == 1: > yield perm > return # not sure return without a value is allowed, > theoretically it could be replaces with if/else block > for i in lst: > if i not in perm: > permute(perm + (i,), level - 1) ##1## > for item in lst: > yield permute((item,), lenlst) # makes generator from the outer > function too ##2##
Your only problem is in knowing what to do with recursively invoked sub-generators. You have two such invocations, which I marked above with ##1## and ##2##, and neither one is handled correctly. In ##1##, you construct a sub-generator and simply discard it. In ##2##, you construct a sub-generator, and yield it (instead of yielding its elements). In neither case do you actually consume any of the elements that the sub-generators are prepared to produce. To usefully invoke a sub-generator, you need to consume it (i.e. iterate over it) and yield whatever it produces. Hope this helps, Carsten -- http://mail.python.org/mailman/listinfo/python-list
