How about:
list.sort(key=lambda x: x[3])
Does that work?
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Lasse Vågsæther Karlsen wrote:
> How about:
>
> list.sort(key=lambda x: x[3])
>
> Does that work?
>
Yes, on my linux-test-box it work, but I my developer pc I don't have
the 2.4 yet. I think that this is a good reason for update :)
Thanks,
Michele
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Michele Petrazzo wrote:
> Lasse Vågsæther Karlsen wrote:
>
>> How about:
>>
>> list.sort(key=lambda x: x[3])
Better to use key=operator.itemgetter(3)
> Yes, on my linux-test-box it work, but I my developer pc I don't have
> the 2.4 yet. I think that this is a good reason for update :)
or learn
Kent Johnson wrote:
> or learn about decorate-sort-undecorate:
>
> lst = [ ...whatever ] lst = [ x[3], i, x for i, x in enumerate(lst) ]
>
I think that here the code must be changed (for the future):
lst = [ (x[3], i, x) for i, x in enumerate(lst) ]
> lst.sort() lst = [ x for _, _, x in lst ]
Michele Petrazzo <[EMAIL PROTECTED]> wrote:
> Lasse Vågsæther Karlsen wrote:
> > How about:
> >
> > list.sort(key=lambda x: x[3])
> >
> > Does that work?
>
> Yes, on my linux-test-box it work, but I my developer pc I don't have
> the 2.4 yet. I think that this is a good reason for update :)
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