sorting a list of list

Matt Nordhoff mnordhoff at mattnordhoff.com
Wed Nov 21 12:44:35 CET 2007


Tim Chase wrote:
>> are there available library or pythonic algorithm for sorting a list
>> of list depending on the index of the list inside the list of my
>> choice?
> 
> The built-in sorted() function and the sort() method on various
> collections take an optional "key=function" keyword paramater
> with which you can pass a function (lambdas are convenient) to
> extract the bit on which you want to compare:
> 
>>>> d_list = [
> ...     ['a', 1, 9],
> ...     ['b', 2, 8],
> ...     ['c', 3, 7],
> ...     ['d', 4, 6],
> ...     ['e', 5, 5],
> ... ]
>>>> print sorted.__doc__
> sorted(iterable, cmp=None, key=None, reverse=False) --> new
> sorted list
>>>> sorted(d_list, key=lambda x: x[2]) # sort by the 3rd item
> [['e', 5, 5], ['d', 4, 6], ['c', 3, 7], ['b', 2, 8], ['a', 1, 9]]
>>>> sorted(d_list, key=lambda x: x[1]) # sort by the 2nd item
> [['a', 1, 9], ['b', 2, 8], ['c', 3, 7], ['d', 4, 6], ['e', 5, 5]]
>>>> sorted(d_list, key=lambda x: x[0]) # sort by the 1st item
> [['a', 1, 9], ['b', 2, 8], ['c', 3, 7], ['d', 4, 6], ['e', 5, 5]]

Not to be too complicated, but there are functions that return a
callable that is faster than a lambda.

>>> import operator

Then, instead of "lambda x: x[2]" use "operator.itemgetter(2)" and
instead of "lambda x: x.foo" use "operator.attrgetter('foo')".
-- 



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