Is there a simple way to delete a list element by value?

Solution 1:

To remove an element's first occurrence in a list, simply use list.remove:

>>> a = ['a', 'b', 'c', 'd']
>>> a.remove('b')
>>> print(a)
['a', 'c', 'd']

Mind that it does not remove all occurrences of your element. Use a list comprehension for that.

>>> a = [10, 20, 30, 40, 20, 30, 40, 20, 70, 20]
>>> a = [x for x in a if x != 20]
>>> print(a)
[10, 30, 40, 30, 40, 70]

Solution 2:

Usually Python will throw an Exception if you tell it to do something it can't so you'll have to do either:

if c in a:
    a.remove(c)

or:

try:
    a.remove(c)
except ValueError:
    pass

An Exception isn't necessarily a bad thing as long as it's one you're expecting and handle properly.

Solution 3:

You can do

a=[1,2,3,4]
if 6 in a:
    a.remove(6)

but above need to search 6 in list a 2 times, so try except would be faster

try:
    a.remove(6)
except:
    pass

Solution 4:

Consider:

a = [1,2,2,3,4,5]

To take out all occurrences, you could use the filter function in python. For example, it would look like:

a = list(filter(lambda x: x!= 2, a))

So, it would keep all elements of a != 2.

To just take out one of the items use

a.remove(2)

Solution 5:

Here's how to do it inplace (without list comprehension):

def remove_all(seq, value):
    pos = 0
    for item in seq:
        if item != value:
           seq[pos] = item
           pos += 1
    del seq[pos:]