On Thu, Feb 11, 2010 at 8:37 AM, Stefan Behnel <stefan...@behnel.de> wrote: > 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))
Ok, I thought you meant make a new list. These examples don't swap anything back... You can also use data.sort(key=operator.itemgetter(1, 0)) Kent _______________________________________________ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
