Python列表从第二个元素

Question

扭转和排序我想改变这样的名单：

[[x, [1,2,3]], [y, [2,3,4]], [z, 4]] 

成为：

[[1, x], [2, [x,y]], [3, [x,y]], [4, z]] 

...基本上是将其“由内向外”，所以内部列表中的数字被分离出来，并且与它们相关的字母被收集到每个数字的列表中。

我该怎么做？

Answer 1

你可以做到这一点相当直接地使用defaultdict：

from collections import defaultdict 

data = [['x', [1, 2, 3]], ['y', [2, 3, 4]], ['z', [4]]] 

temp = defaultdict(list) 

for k, s in data: 
    for v in s: 
     temp[v].append(k) 

result = sorted(temp.items()) 
# [(1, ['x']), (2, ['x', 'y']), (3, ['x', 'y']), (4, ['y', 'z'])] 

Answer 2

您可以使用collections.defaultdict：

>>> d=defaultdict(list) 
>>> for k,v in l: 
...  if isinstance(v,list): 
...    for t in v: 
...     d[t].append(k) 
...  else: 
...     d[v].append(k) 
... 
>>> d 
defaultdict(<type 'list'>, {1: ['x'], 2: ['x', 'y'], 3: ['x', 'y'], 4: ['y', 'z']}) 

Answer 3

难以理解的列表理解的一点：

in_list = ['x', [1, 2, 3]], ['y', [2, 3, 4]], ['z', [4]] 
[[num, [sublist[0] for sublist in in_list if num in sublist[1]]] for num in sorted(list(set(num for sublist in in_list for num in sublist[1])))] 

得到：

Out[218]: [[1, ['x']], [2, ['x', 'y']], [3, ['x', 'y']], [4, ['y', 'z']]]