如何通过检查python子列表中的值对列表进行排序？

Question

I have a list of lists in the following format:

[['a',[10]], ['b',[1]], ['c',[5,10]], ['d',[5,1,-10]], ['e',[5,1,-1]]]

I would like to sort if in an efficient way in python using the numeric list elements, matching the first element, and when it is the same, use the second, and so on. The result would be something like (I need reverse order this time):

['a',[10]]
['c',[5,10]]
['e',[5,1,-1]]
['d',[5,1,-10]
['b',[1]]

Thanks!

Answer 1

I think lists compare like you want, by default, if inverted:

>>> data = [['a',[10]], ['b',[1]], ['c',[5,10]], ['d',[5,1,-10]], ['e',[5,1,-1]]
>>> sorted(data, reverse = True, key = lambda pair: pair[1])
[['a', [10]], ['c', [5, 10]], ['e', [5, 1, -1]], ['d', [5, 1, -10]], ['b', [1]]]

You had a bracketing error in your input list, it's fixed in the code above.

Answer 2

>>> from operator import itemgetter
>>> L=[['a',[10]], ['b',[1]], ['c',[5,10]], ['d',[5,1,-10]], ['e',[5,1,-1]]]
>>> sorted(L, key=itemgetter(1), reverse=True)
[['a', [10]], ['c', [5, 10]], ['e', [5, 1, -1]], ['d', [5, 1, -10]], ['b', [1]]]
>>> 

I'd use itemgetter(1) here, which is roughly equivalent to the lambda function in the other answers. This effectively does the sort with the key being the sublists since they are item number 1. (item number 0 is the letters ae)

Answer 3

Use key to select the second element in the list, and reverse to change direction:

>>> l=[['a',[10]], ['b',[1]], ['c',[5,10]], ['d',[5,1,-10], ['e',[5,1,-1]]]
>>> sorted(l, key=lambda e:e[1], reverse=True)
[['a', [10]], ['c', [5, 10]], ['e', [5, 1, -1]], ['d', [5, 1, -10]], ['b', [1]]]

Lists are sorted by comparing their elements in order, just like a lexicon or regular dictionary. It's called 'lexographical comparison'.