如何在Python中排列嵌套列表的外部和内部子列表？ [英] How to sort the outer and inner sublist of a nested list in Python?

问题描述

首先，如果这太天真了，我很抱歉（我是初学者）。
我有以下类型的列表，我想先按内容列表的最后一个成员按升序排序：

  data = [[1，.45，0]，[2，.49，2]，[3，.98，0]，[4，.82，1]，[5， 77，1]，[6，.98，2]] 
  

code> sorted（data，key = operator.itemgetter（2），reverse = True），这给了我：

  [[1，.45，0]，[3，.98，0]，[4，.82，1]，[5，.77，1]，[2，.49 ，2]，[6，.98，2]] 
  

现在，我想在子列表，首先按照降序将中间的成员作为关键，将列表的最后一个成员作为0排序。然后用1作为其最后一个成员排列子列表，依此类推。请注意，每个子列表中的元素数量不同，不知道。最终列表应如下所示：

  [[3，.98，0]，[1，.45，0] ，[4，.82，1]，[5，.77，1]，[6，.98，2]，[2，.49，2]] 
  pre> 
 
 
列表实际上相当庞大，因此我正在寻找一个高效的实现。 
任何指导将不胜感激！
解决方案
 
 >>> data = [[1，.45,0]，[2，.49,2]，[3，.98,0]，[4，.82，1]，[5，.77，1]，[6 ，.98，2]] 
>>>排序（data，key = lambda x：（x [2]，-x [1]））
 [[3，0.98,0]，[1，0.45,0]，[4，0.82,1] ，[5，0.77，1]，[6，0.98,2]，[2，0.49，2]] 
  
您可以根据需要向lambda添加额外的术语
 
First of all, apologies if this too naive (I am a beginner). 
I have the following type of list of lists, which I would like to first sort by the last member of the inner list in ascending order:
data =  [[1, .45, 0], [2, .49, 2], [3, .98, 0], [4, .82, 1], [5, .77, 1], [6, .98, 2] ]
I accomplish this by using : sorted(data,key=operator.itemgetter(2),reverse = True), which gives me:
[[1, .45, 0], [3, .98, 0],[4, .82, 1], [5, .77, 1], [2, .49, 2], [6, .98, 2]]
Now, I would like to sort within the sub-lists i.e. first sort the list with its last member as '0' using the middle member as the key, in descending order. Then sort the sub-list with '1' as its last member and so on. Note that number of elements in each sub-list are different and are not know. The final list should look like this:
[[3, .98, 0],[1, .45, 0], [4, .82, 1], [5, .77, 1], [6, .98, 2],[2, .49, 2] ]
The list is actually quite huge, therefore I am looking for an efficient implementation.
Any guidance would be appreciated !
解决方案
>>> data =  [[1, .45, 0], [2, .49, 2], [3, .98, 0], [4, .82, 1], [5, .77, 1], [6, .98, 2] ]
>>> sorted(data, key=lambda x:(x[2], -x[1]))
[[3, 0.98, 0], [1, 0.45, 0], [4, 0.82, 1], [5, 0.77, 1], [6, 0.98, 2], [2, 0.49, 2]]
You can add extra terms to the lambda as required

                        
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