How can I ignore ValueError when I try to remove an element from a list?

Question

How can I ignore the \"not in list\" error message if I call a.remove(x) when x is not present in list a?

This is my situation:

Answer 1

When I only care to ensure the entry is not in a list, dict, or set I use contextlib like so:

import contextlib

some_list = []
with contextlib.suppress(ValueError):
    some_list.remove(10)

some_set = set()
some_dict = dict()
with contextlib.suppress(KeyError):
    some_set.remove('some_value')
    del some_dict['some_key']

Answer 2

As an alternative to ignoring the ValueError

try:
    a.remove(10)
except ValueError:
    pass  # do nothing!

I think the following is a little more straightforward and readable:

if 10 in a:
    a.remove(10)

Answer 3

you have typed wrong input. syntax: list.remove(x)

and x is the element of your list. in remove parenthesis ENTER what already have in your list. ex: a.remove(2)

i have entered 2 because it has in list. I hape this data help you.

Answer 4

I'd personally consider using a set instead of a list as long as the order of your elements isn't necessarily important. Then you can use the discard method:

>>> S = set(range(10))
>>> S
set([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])
>>> S.remove(10)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
KeyError: 10
>>> S.discard(10)
>>> S
set([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])

Answer 5

How about list comprehension?

a = [x for x in a if x != 10]

Answer 6

A better way to do this would be

source_list = list(filter(lambda x: x != element_to_remove,source_list))

Because in a more complex program, the exception of ValueError could also be raised for something else and a few answers here just pass it, thus discarding it while creating more possible problems down the line.