Hello,
i think it could be done by using itertools functions even if i can not
see the trick. i would like to have all available "n-uples" from each
list of lists.
example for a list of 3 lists, but i should also be able to handle any
numbers of items (any len(lol))
lol = (['a0', 'a1', 'a2'], ['b0', 'b1'], ['c0', 'c1', 'c2', 'c3'])
=>
[('a0', 'b0', 'c0'), ('a0', 'b0', 'c1'), ('a0', 'b0', 'c2'), ('a0',
'b0', 'c3'), ('a0', 'b1', 'c0'), ('a0', 'b1', 'c1'), ('a0', 'b1',
'c2'), ('a0', 'b1', 'c3'), ('a1', 'b0', 'c0'), ('a1', 'b0', 'c1'),
('a1', 'b0', 'c2'), ('a1', 'b0', 'c3'), ('a1', 'b1', 'c0'), ('a1',
'b1', 'c1'), ('a1', 'b1', 'c2'), ('a1', 'b1', 'c3'), ('a2', 'b0',
'c0'), ('a2', 'b0', 'c1'), ('a2', 'b0', 'c2'), ('a2', 'b0', 'c3'),
('a2', 'b1', 'c0'), ('a2', 'b1', 'c1'), ('a2', 'b1', 'c2'), ('a2',
'b1', 'c3')]
maybe tee(lol, len(lol)) can help ?
it could be done by a recursive call, but i am interested in using and
understanding generators.
i also have found a convenient function, here :
http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/65285 (paste
below)
but i am curious of how you will do it or refactorize this one with
generators...
def permuteflat(*args):
outs = []
olen = 1
tlen = len(args)
for seq in args:
olen = olen * len(seq)
for i in range(olen):
outs.append([None] * tlen)
plq = olen
for i in range(len(args)):
seq = args[i]
plq = plq / len(seq)
for j in range(olen):
si = (j / plq) % len(seq)
outs[j][i] = seq[si]
for i in range(olen):
outs[i] = tuple(outs[i])
return outs
many thanx
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