For example
list_of_specific_element = [2,13]
list = [[1, 0], [2, 1], [2, 3], [13, 12], [13, 14], [15, 13]]
I want the sublist including any value inside the list of specific element be removed from the list.
So the element [2,1],[2,3],[13,12],[13,14] should be removed from the list.
the final output list should be[[1,0],[15,13]]
listy=[elem for elem in listy if (elem[0] not in list_of_specific_element) and (elem[1] not in list_of_specific_element)]
Using list comprehension one-liner
You could use set intersections:
>>> 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]]
You could write:
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))
]
which gives:
>>> output
[[1, 0]]
This uses sets, set intersection and list comprehension.
A simple list-only solution:
list = [x for x in list if all(e not in list_of_specific_element for e in x)]
And you really shouldn't call it list!
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]]
You can use list comprehension and any() to do the trick like this example:
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)
Output:
[[1, 0]]
I guess you can use:
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[:]]
[15, 13]remain? It does contain a13.