python 中的嵌套列表和排序

Question

不知道可不可以。我试图找到一种在以下条件下对嵌套列表进行排序的方法

我想将表格 1 指向另一个（不是整个列表，只是其中的一部分）
应该根据子列表的第 3 个元素进行排序

我想要的想法：

    PAE=[['a',0,8],
         ['b',2,1],
         ['c',4,3],
         ['d',7,2],
         ['e',8,4]]

    #PAE[1:4].sort(key=itemgetter(2))  (something like this)    
    or    
    #sorted(PAE[1:4],key=itemgetter(2))  (something like this)  
`   # ^ i know both are wrong but just for an idea
`
    #output should be like this  
    ['a', 0, 8]
    ['b', 2, 1]
    ['d', 7, 2]
    ['c', 4, 3]
    ['e', 8, 4]

我是python的新手，但我尽我所能找到解决方案但失败了。

Answer 1

应该这样做：

from operator import itemgetter
PAE=[['a',0,8],
    ['b',2,1],
    ['c',4,3],
    ['d',7,2],
    ['e',8,4]]

split_index = 1

print PAE[:split_index]+sorted(PAE[split_index:],key=itemgetter(2))
#=> [['a', 0, 8], ['b', 2, 1], ['d', 7, 2], ['c', 4, 3], ['e', 8, 4]]

Answer 2

对切片进行排序并写回:

>>> PAE[1:4] = sorted(PAE[1:4], key=itemgetter(2))
>>> PAE
[['a', 0, 8], ['b', 2, 1], ['d', 7, 2], ['c', 4, 3], ['e', 8, 4]]

Answer 3

这里不拆分，问题是什么是最好的可读性

PAE=[['a',0,8],
    ['b',2,1],
    ['c',4,3],
    ['d',7,2],
    ['e',8,4]]

print (sorted(PAE, key=lambda PAE: PAE[1] if not PAE[1] else PAE[2]))
>>> [['a', 0, 8], ['b', 2, 1], ['d', 7, 2], ['c', 4, 3], ['e', 8, 4]]

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