Kent Johnson, 11.02.2010 14:16:
> On Thu, Feb 11, 2010 at 4:44 AM, Stefan Behnel wrote:
>
>> 2) given that you have lists as items in the 'data' list, it's enough to
>> call sort() once, as the comparison of lists is defined as the comparison
>> of each item to the corresponding item of the other list. If you want to
>> sort based on the second item before the first item, it's best to exchange
>> both items before sorting and swap them back afterwards.
>
> No, that is the old decorate-sort-undecorate idiom which has been
> replaced by the key= parameter to sort.
Nothing keeps you from writing
data.sort(key=lambda x:x[::-1])
or
data.sort(key=operator.itemgetter(slice(None,None,-1))
(which is equivalent)
Stefan
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