Remove all occurrences of a value from a list?

Question

In Python remove() will remove the first occurrence of value in a list.

How to remove all occurrences of a value from a list?

This is w

Answer 1

All of the answers above (apart from Martin Andersson's) create a new list without the desired items, rather than removing the items from the original list.

>>> import random, timeit
>>> a = list(range(5)) * 1000
>>> random.shuffle(a)

>>> b = a
>>> print(b is a)
True

>>> b = [x for x in b if x != 0]
>>> print(b is a)
False
>>> b.count(0)
0
>>> a.count(0)
1000

>>> b = a
>>> b = filter(lambda a: a != 2, x)
>>> print(b is a)
False

This can be important if you have other references to the list hanging around.

To modify the list in place, use a method like this

>>> def removeall_inplace(x, l):
...     for _ in xrange(l.count(x)):
...         l.remove(x)
...
>>> removeall_inplace(0, b)
>>> b is a
True
>>> a.count(0)
0

As far as speed is concerned, results on my laptop are (all on a 5000 entry list with 1000 entries removed)

• List comprehension - ~400us
• Filter - ~900us
• .remove() loop - 50ms

So the .remove loop is about 100x slower........ Hmmm, maybe a different approach is needed. The fastest I've found is using the list comprehension, but then replace the contents of the original list.

>>> def removeall_replace(x, l):
....    t = [y for y in l if y != x]
....    del l[:]
....    l.extend(t)

• removeall_replace() - 450us