How Can I Remove All Instances of an Element from a List in Python? [Duplicate]

Question

Lets say I have a list a:

a = [[1, 1], [2, 2], [1, 1], [3, 3], [1, 1]]

Is there a function that removes all instances of [1, 1]?

Accepted Answer

If you want to modify the list in-place,

a[:] = [x for x in a if x != [1, 1]]

Answer 2

Use a list comprehension:

[x for x in a if x != [1, 1]]

Answer 3

Google finds Delete all items in the list, which includes gems such as

from functools import partial
from operator import ne
a = filter(partial(ne, [1, 1]), a)

Answer 4

def remAll(L, item):
    answer = []
    for i in L:
        if i!=item:
            answer.append(i)
    return answer

Answer 5

Here is an easier alternative to Alex Martelli's answer:

a = [x for x in a if x != [1,1]]

Answer 6

new_list = filter(lambda x: x != [1,1], a)

Or as a function:

def remove_all(element, list):
    return filter(lambda x: x != element, list)

a = remove_all([1,1],a)

Or more general:

def remove_all(elements, list):
    return filter(lambda x: x not in elements, list)

a = remove_all(([1,1],),a)

Answer 7

filter([1,1].__ne__,a)

Answer 8

pure python no modules version, or no list comp version ( simpler to understand ?)

>>> x = [1, 1, 1, 1, 1, 1, 2, 3, 2]
>>> for item in xrange(x.count(1)):
...     x.remove(1)
...
>>>
>>> x
[2, 3, 2]

can be made into a def pretty easily also

def removeThis(li,this):
    for item in xrange(li.count(this)):
           li.remove(this)
    return li

8 Answers