> itertools only looks for changes to the key value (the one returned by
> operator.itemgetter(0) in your case); it doesn't sort the list for you.
>
> this should work:
>
>for k, g in itertools.groupby(sorted(vals), operator.itemgetter(0)):
> print k, [i for i in g]
footnote: to turn
[EMAIL PROTECTED] wrote:
> What am I doing wrong here?
>
import operator
import itertools
vals = [(1, 11), (2, 12), (3, 13), (4, 14), (5, 15),
> ... (1, 16), (2, 17), (3, 18), (4, 19), (5, 20)]
for k, g in itertools.groupby(iter(vals), operator.itemgetter(0)):
> ... p
[EMAIL PROTECTED] wrote:
> What am I doing wrong here?
>
import operator
import itertools
vals = [(1, 11), (2, 12), (3, 13), (4, 14), (5, 15),
> ... (1, 16), (2, 17), (3, 18), (4, 19), (5, 20)]
for k, g in itertools.groupby(iter(vals), operator.itemgetter(0)):
> ... pr
