Question

所以，我有一个嵌套列表（我们称它为A）。它的长度为2，应将它们视为单独的列表。我想分别遍历这2个，并删除长度不相等的子列表。例如，我希望我的输出与A相同，但要删除[['Table heading']]，因为它的长度与其他嵌套列表的长度不同。

A=[[[['1'], ['2'], ['3'], ['4']],
  [['x'],['y'],['z'],['w']],
  [['a'],['b'],['c'],['d']],
  [['11'], ['22'], ['33'], ['44']]],
  [[['Table heading']],
  [['A'], ['B'], ['C'], ['D']],
  [['X'], ['Y'], ['Z'], ['W']],
  [['1'], ['2'], ['3'], ['4']]]]

 output=[[[['1'], ['2'], ['3'], ['4']],
  [['x'],['y'],['z'],['w']],
  [['a'],['b'],['c'],['d']],
  [['11'], ['22'], ['33'], ['44']]],
  [[['A'], ['B'], ['C'], ['D']],
  [['X'], ['Y'], ['Z'], ['W']],
  [['1'], ['2'], ['3'], ['4']]]]

Answer 1

[ y for x in A for y in x if len(y) == 4]

或以可读格式。

out = []
for inner in A:
     for inner_inner in inner:
          if len(inner_inner) == 4:
               out.append(inner_inner)

如果您不知道内部列表的大小，但是知道只有一个列表的大小不匹配，您可以这样做。

out = []
for inner in A:
     for inner_inner in inner:
         inner_size = len(inner_inner) 
         try:
             if inner_size == previous_size:
                 out.append(inner_inner)
          except NameError:
                  previous_size = inner_size
                  out.append(inner_inner)

这是默认设置，如果要删除第一个inner_inner元素，则将删除整个列表元素，而不是第一个。

或者这样

from collections import Counter

size = Counter([ len(y) for x in A for y in x ]).most_common(1)[0][0]
[ y for x in A for y in x if len(y) == size]

此解决方案在整个列表上循环两次，具体取决于列表大小，这可能是一个限制。

Answer 2

检查一下。除非元素不再列出，否则它将删除该元素。


def  delete_odd(A):
    if isinstance(A[0],list)==False:
        return
    def delete_elem(a,if_not):
        print(a,'  ',if_not,'   ')

        i = 0
        while i<len(a):
            if len(a[i])!=if_not:
                del a[i]
            else:
                i+=1

    if len(A)<=2:
        pass
    else:
        n1 = len(A[0])
        n2 = len(A[1])
        n3 = len(A[2])
        if n1==n2:
            n = n1
        elif n2==n3:
            n = n2

        delete_elem(A,if_not=n)
    i = 0
    while i<len(A):
        delete_odd(A[i])
        i+=1




A = [[[['1'], ['2'], ['3'], ['4']],
  [['x'],['y'],['z'],['w']],
  [['a'],['b'],['c'],['d']],
  [['11'], ['22'], ['33'], ['44']]],
  [[['Table heading']],
  [['A'], ['B'], ['C'], ['D']],
  [['X'], ['Y'], ['Z'], ['W']],
  [['1'], ['2'], ['3'], ['4']]]]
delete_odd(A)

print(A)
#output
[[[['1'], ['2'], ['3'], ['4']], [['x'], ['y'], ['z'], ['w']], [['a'], ['b'], ['c'], ['d']], [['11'], ['22'], ['33'], ['44']]], [[['A'], ['B'], ['C'], ['D']], [['X'], ['Y'], ['Z'], ['W']], [['1'], ['2'], ['3'], ['4']]]]

根据条件从嵌套列表中删除某些子列表

2 个答案: