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Is there a quick way to remove a list element based on its value?

Problem

If a value already exists in a list, I want to remove it (which it may not).

a = [1, 2, 3, 4]
b = a.index(6)

del a[b]
print(a)

The following error is shown in the above scenario (when it does not exist):

Traceback (most recent call last):
  File "D:\zjm_code\a.py", line 6, in <module>
    b = a.index(6)
ValueError: list.index(x): x not in list

So I have to do this:

a = [1, 2, 3, 4]

try:
    b = a.index(6)
    del a[b]
except:
    pass

print(a)

Isn’t there a more straightforward approach?

Asked by zjm1126

Solution #1

Simply use list to eliminate an element’s initial appearance in a list. remove:

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

Keep in mind that it does not eliminate all instances of your element. To do so, use a list comprehension exercise.

>>> 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]

Answered by Johannes Charra

Solution #2

If you ask Python to do something it can’t, it will usually throw an Exception, therefore you’ll have to perform one of two things:

if c in a:
    a.remove(c)

or:

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

An exception isn’t always a terrible thing, as long as you anticipate it and handle it properly.

Answered by Dave Webb

Solution #3

You can do

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

However, because searching 6 items in the list twice is inconvenient, trying unless would be a better option.

try:
    a.remove(6)
except:
    pass

Answered by YOU

Solution #4

Consider:

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

You may use Python’s filter function to remove all occurrences. For instance, consider the following:

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

As a result, all elements of a!= 2 would be preserved.

To simply remove one of the elements, use

a.remove(2)

Answered by mathwizurd

Solution #5

Here’s how to accomplish it in real time (without using a list):

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

Answered by jfs

Post is based on https://stackoverflow.com/questions/2793324/is-there-a-simple-way-to-delete-a-list-element-by-value