[Tutor] Sorting Nested Lists

[Sarma Tangirala]

On 9 January 2012 18:26, Steven D'Aprano <steve at pearwood.info> wrote:

> Sarma Tangirala wrote:
>
>> Hi list,
>>
>> I was banging my head about a pythonic way of doing the following,
>>
>> Given a nested list, how do I sort the uppermost list based on one key and
>> when a special condition occurs a sort on another key should be performed?
>> For example, [[1,2], [2, 2], [3, 2], [4, 0]] would be sorted, in my
>> example
>> as, [[4, 0], [3, 2], [2, 2], [1, 2]]. That is, sort on the second value
>> and
>> in case they are equal, reverse sort on the first value.
>>
>
> That is not exactly a good example. There are at least two other ways to
> get the result you show, both much more obvious:
>
> py> L = [[1,2], [2, 2], [3, 2], [4, 0]]
> py> list(reversed(L))
>
> [[4, 0], [3, 2], [2, 2], [1, 2]]
> py> sorted(L, reverse=True)
>
> [[4, 0], [3, 2], [2, 2], [1, 2]]
>
>
> If I ignore your example, and just use the description:
>
> "sort on the second value, and in case they are equal, reverse sort on the
> first value"
>
> the way to do this is with a double sort. Note that this works because
> Python's sort is stable: in the event of ties, the first item remains
> first. In earlier versions of Python, this was not always the case.
>
> So, given this list:
>
> L = [[1,2], [4,0], [3,2], [2,2], [5,1], [1,1]]
>
> first sort in reverse by the first item, then by the second:
>
>
> py> L.sort(key=lambda sublist: sublist[0], reverse=True)
> py> L.sort(key=lambda sublist: sublist[1])
> py> print L
> [[4, 0], [5, 1], [1, 1], [3, 2], [2, 2], [1, 2]]
>
>
>
> Note that using a key function is MUCH more efficient than a cmp function.
> Comparison functions end up doing much more work, and hence are very much
> slower, than a key function.
>
> Also note that in recent versions of Python, you can do without the lambda
> function and use the special "itemgetter" function:
>
>
> py> from operator import itemgetter
> py> L = [[1,2], [4,0], [3,2], [2,2], [5,1], [1,1]]
> py> L.sort(key=itemgetter(0), reverse=True)
> py> L.sort(key=itemgetter(1))
> py> print L
> [[4, 0], [5, 1], [1, 1], [3, 2], [2, 2], [1, 2]]
>
>
I tried this a lot yesterday but seemed to get a wrong answer but now I
realize its because of a bad test case. Thank you.


>
> Last but not least, I will show how to do it using a custom cmp function.
> But I recommend that you don't use this!
>
> def my_cmp(list1, list2):
>    x = cmp(list1[1], list2[1])
>    if x == 0:  # a tie
>        x = cmp(list2[0], list1[0])  # swap the order for reverse sort
>        # or if you prefer, x = -cmp(list1[0], list2[0])
>    return x
>
> sorted(L, my_cmp)
>
>
>
>
> --
> Steven
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-- 
Sarma Tangirala,
Class of 2012,
Department of Information Science and Technology,
College of Engineering Guindy - Anna University
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