Charles Sanders <[EMAIL PROTECTED]> writes: > Forgive any silly mistakes I have made (I've been teaching > myself python for about 1 week) but there is a moderately > well known algorithm for this that extends to arbitrary > lengths of both the list of alternatives and the length > of the required output, and avoids deeply nested loops.
s = "abcd"
def a(n):
if n==0:
yield ''
return
for c in s:
for r in a(n-1):
yield c+r
print list(a(3))
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