How To .remove All Matches In A Python List?

If I have a bit of Python like: n = [1, 3, 5, 1] n.remove(1) print n n will return [3, 5, 1] because .remove() quits after it finds its first match. What could I do to return just

Solution 1:

If creating a new list is fine:

n = [x for x in n if x != 1]

Of course you could also use slice assignment to modify the list in place:

n[:] = [x for x in n if x != 1]

Solution 2:

use filter function

n = [1, 3, 5, 1]
filter(lambda a:a isnot1,n)
[3,5]

edit

n = [1, 3, 5, 1]
filter(lambda a:a != 1,n)
[3,5]

Solution 3:

Simple List Comprehension -

>>>n = [1, 3, 5, 1]>>>n = [e for e in n if e != 1]>>>n
[3, 5]

If you wanted to remove more than one elements -

>>>g = [1,3]>>>n = [e for e in n if e notin g]>>>n
[5]

Solution 4:

This is not how I'd normally do this, but:

n = [1, 3, 5, 1]
REMOVECHAR = 1while REMOVECHAR in n:
    n.remove(REMOVECHAR)
# could also do:
# while1:
#     try: n.remove(REMOVECHAR)
#     except ValueError as e: break

If I were doing it, I would use a list comp to make a new list.

n = [1,3,5,1]
REMOVECHAR = 1nn = [element for element in n if element != REMOVECHAR]

Solution 5:

You could iterate backwards through the list and pop elements that match.

def remove_all_in_place(target, sequence):
    cursor= len(target) -1
    while cursor>=0:
        if sequence[cursor] == target:
            sequence.pop(cursor)
        cursor-=1

List comprehensions might be faster since they are optimized, but they aren't in-place operations. For native Python implementations, I think this would be fastest, as well as being true in-place.