按第三个嵌套列表对Python中的深层嵌套列表进行排序

Question

If this is a duplicate, then I'd be glad to delete the question.

In python, I have a list that has the following structure, with none of the names necessarily being the same:

[[xy, ["Name 3", ["SubName 1", "SubName 2"]], 
      ["Name 1", ["NameSub 1", "NameSub 2"]], ...],
 [yz, ["Name 10", ["SubNamex 1", "SubNamex 2"]], 
      ["Name 6", ["NameSubx 1", "NameSubx 2"]], ...]]

I want to sort first by the first index, "xy" and "yz", and then sort the items with each nested list by "Name x". So my sorted list should look like this:

[[xy, ["Name 1", ["NameSub 1", "NameSub 2"]],
      ["Name 3", ["SubName 1", "SubName 2"]], ...],
 [yz, ["Name 6", ["NameSubx 1", "NameSubx 2"]], 
      ["Name 10", ["SubNamex 1", "SubNamex 2"]], ...]]

I have tried:

mylist.sort(key=lambda x: x[1]) 

and

mylist.sort(key=itemgetter(2,1))

and different variations of the indices, with no success.

Answer 1

First, iterate the array sorting the inner nested elements and finally sort the outer elements. The code is provided below:

srtli=[]
for ele in li:
    tmp=ele[1:]
    tmp.sort()
    for ele_tmp in tmp:
        ele_tmp[1].sort()
    srtli.append([ele[0]]+tmp)
srtli.sort()

Answer 2

try with this :-

a =     [['xy', ["Name 3", ["SubName 1", "SubName 2"]],
            ["Name 1", ["NameSub 1", "NameSub 2"]]],
        ['yz', ["Name 10", ["SubNamex 1", "SubNamex 2"]], 
                ["Name 6", ["NameSubx 1", "NameSubx 2"]]]]
# a.sort(key = lambda x: x[1])
for i in a:
    i.sort(key = lambda x : x[1])
    print i[-1],',',i[:-1] #or you can do print i only