Arnaud Delobelle wrote:
>
> This is definitely the best way:
>
> from itertools import chain
>
> def overlaps(lst):
> bounds = chain(*(((x[1],i), (x[2], i)) for i,x in enumerate(lst)))
> inside = {}
> for x, i in sorted(bounds):
> if inside.pop(i, None) is None:
> for j, y in inside.iteritems():
> if y != x: yield i, j
> inside[i] = x
Why not just
def overlaps(lst):
for i,x in enumerate(lst):
for j,y in enumerate(lst):
if i != j and x[1] > y[2] > x[2]:
yield i,j
It's still n^2. Or am I missing something?
Cheers,
thomas
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