如果嵌套列表在索引处包含数字，则删除嵌套列表

Question

如果嵌套列表在特定索引处包含特定数字，我想删除该嵌套列表。

列表示例：

permutations_list = [[9, 7, 14, 4, 2, 10], [9, 7, 2, 10, 14, 4], [9, 7, 2, 10, 4, 14], [9, 7, 2, 14, 10, 4], [9, 7, 2, 14, 4, 10], [9, 7, 2, 4, 10, 14], [9, 7, 2, 4, 14, 10], [9, 7, 4, 10, 14, 2], [9, 7, 4, 10, 2, 14], [9, 7, 4, 14, 10, 2], [9, 7, 4, 14, 2, 10], [9, 7, 4, 2, 10, 14], [9, 7, 4, 2, 14, 10]]

我想要检查每个嵌套列表是否在索引 4 处包含数字 14。如果发生这种情况，请删除任何符合这些规范的嵌套列表，从而生成以下列表列表：

permutations_list = [[9, 7, 14, 4, 2, 10], [9, 7, 2, 10, 4, 14], [9, 7, 2, 14, 10, 4], [9, 7, 2, 14, 4, 10], [9, 7, 2, 4, 10, 14], [9, 7, 4, 10, 2, 14], [9, 7, 4, 14, 10, 2], [9, 7, 4, 14, 2, 10], [9, 7, 4, 2, 10, 14]]

这是我尝试过的：

for i in permutations_list:
    for c in i:
        if c =='10' and c[4]:
            permutations_list.remove(i)

所有这些都会导致：

TypeError: 'int' object is not subscriptable

Answer 1

您可以只循环一次主列表并检查此列表中索引 4 处的元素是否为 14。如果是14，删除它，如果不是，什么也不做。就像我在下面做的那样

for i in permutations_list :
   if i[4] == 14 :
      permutations_list.remove(i)

Answer 2

使用列表推导

例如：

permutations_list = [[9, 7, 14, 4, 2, 10], [9, 7, 2, 10, 14, 4], [9, 7, 2, 10, 4, 14], [9, 7, 2, 14, 10, 4], [9, 7, 2, 14, 4, 10], [9, 7, 2, 4, 10, 14], [9, 7, 2, 4, 14, 10], [9, 7, 4, 10, 14, 2], [9, 7, 4, 10, 2, 14], [9, 7, 4, 14, 10, 2], [9, 7, 4, 14, 2, 10], [9, 7, 4, 2, 10, 14], [9, 7, 4, 2, 14, 10]]
permutations_list = [i for i in permutations_list if not i[4] == 14]
print(permutations_list)

---

或使用filter

permutations_list = list(filter(lambda x: x[4] != 14, permutations_list))

输出：

[[9, 7, 14, 4, 2, 10],
 [9, 7, 2, 10, 4, 14],
 [9, 7, 2, 14, 10, 4],
 [9, 7, 2, 14, 4, 10],
 [9, 7, 2, 4, 10, 14],
 [9, 7, 4, 10, 2, 14],
 [9, 7, 4, 14, 10, 2],
 [9, 7, 4, 14, 2, 10],
 [9, 7, 4, 2, 10, 14]]