如果此子列表中包含特定元素，Python將刪除列表中的子列表

Question

例如

list_of_specific_element = [2,13]
list = [[1, 0], [2, 1], [2, 3], [13, 12], [13, 14], [15, 13]]

我希望從列表中刪除子列表，包括特定元素列表中的任何值。
因此[2,1],[2,3],[13,12],[13,14]應從列表中刪除元素[2,1],[2,3],[13,12],[13,14] 。
最終的輸出列表應為[[1,0],[15,13]]

Answer 1

listy=[elem for elem in listy if (elem[0] not in list_of_specific_element) and (elem[1] not in list_of_specific_element)]

使用列表理解單行

Answer 2

你可以使用set intersection：

>>> exclude = {2, 13}
>>> lst = [[1, 0], [2, 1], [2, 3], [13, 12], [13, 14], [15, 13]]
>>> [sublist for sublist in lst if not exclude.intersection(sublist)]
[[1, 0]]

Answer 3

你可以寫：

list_of_specific_element = [2,13]
set_of_specific_element = set(list_of_specific_element)
mylist = [[1, 0], [2, 1], [2, 3], [13, 12], [13, 14], [15, 13]]
output = [
    item
    for item in mylist
    if not set_of_specific_element.intersection(set(item))
]

這使：

>>> output
[[1, 0]]

這使用集合，集合交集和列表理解。

Answer 4

一個簡單的列表解決方案：

list = [x for x in list if all(e not in list_of_specific_element for e in x)]

你真的不應該把它list ！

Answer 5

過濾器和lambda版本

list_of_specific_element = [2,13]
list = [[1, 0], [2, 1], [2, 3], [13, 12], [13, 14], [15, 13]]

filtered = filter(lambda item: item[0] not in list_of_specific_element, list)

print(filtered)

>>> [[1, 0], [15, 13]]

Answer 6

你可以使用list comprehension和any()來做這個例子：

list_of_specific_element = [2,13]
# PS: Avoid calling your list a 'list' variable
# You can call it somthing else.
my_list = [[1, 0], [2, 1], [2, 3], [13, 12], [13, 14], [15, 13]]
final = [k for k in my_list if not any(j in list_of_specific_element for j in k)]
print(final)

輸出：

[[1, 0]]

Answer 7

我猜你可以用：

match = [2,13]
lst = [[1, 0], [2, 1], [2, 3], [13, 12], [13, 14], [15, 13]]

[[lst.remove(subl) for m in match if m in subl]for subl in lst[:]]

演示