On Jan 29, 8:55 pm, pataphor <[EMAIL PROTECTED]> wrote:
> On Tue, 29 Jan 2008 11:51:04 -0800 (PST)
>
> [EMAIL PROTECTED] wrote:
> > Any elegant way of breaking out of the outer for loop than below, I
> > seem to have come across something, but it escapes me
>
> > for i in outerLoop:
> > for j in innerLoop:
> > if condition:
> > break
> > else:
> > continue
> > break
>
> Ha! Think outside the box to begin with ...
>
> P.
>
> def cross(args):
> ans = [[]]
> for arg in args:
> ans = [x+[y] for x in ans for y in arg]
> return ans
While we're at it, a generator version:
def iproduct(head=None, *tail):
if head is None:
return ((),)
else:
return ((x,)+y for x in head for y in iproduct(*tail))
for a, b, c in iproduct('124', 'ab', 'AB'):
print a, b, c
;-)
--
Arnaud
--
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