On Sep 28, 11:25 pm, Gabriel Zachmann <[EMAIL PROTECTED]
clausthal.de> wrote:
> could some kind soul please explain to me why the following trivial code is
> misbehaving?
>
> lst = [ 0, 1, 2 ]
> s = []
> l = [ lst[0] ]
> r = lst[1:]
> while r:
> x = (l,r)
> print x
> s.append( x )
> l.append( r.pop(0) )
> print s
What TeroV said - you need to copy the lists into s otherwise they'll
still mutate when you pop or append to them.
Try putting an extra "print s" after the s.append(x) line, and you'll
find that s changes between there and the final print statement
outside the loop.
Here's code that works:
s, l, r = [], [0], [1, 2]
while r:
x = (list(l), list(r))
print x
s.append(x)
l.append(r.pop(0))
print s
I'm using list(l) to copy the list, Tero uses l[:], but the idea is
the same.
But do you just want all proper partitions of lst? Then this is much
simpler:
lst = [0, 1, 2]
s = [(lst[:i], lst[i:]) for i in range(1, len(lst))]
--
Paul Hankin
--
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