How can I remove all instances of an element from a list in Python? [duplicate]

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]?

0

8 Answers

If you want to modify the list in-place,

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

3

Use a list comprehension:

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

2

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)

2

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

1

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

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

1

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)

1

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

1

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