Python remove from list acting strange when used inline

Question

I want to loop through a list and remove a single element at a time:

l1 = ["a", "b"]
for l in l1:
    l2 = l1[:]
    l2.remove(l)
    print(l2)

# ['b']
# ['a']

The behavior of the code changes when I use the remove call inline:

l1 = ["a", "b"]
for l in l1:
    l2 = l1[:].remove(l)
    print(l2)

# None
# None

Any idea what is happening? Is this an artefact of remove() or a more general issue with functions that use mutability of objects?

Thanks

Asked By: Carsten

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Source

Answer 1

Note that remove() does not return anything. That is why l2 is always None in your second example. The change is only applied to l1.
In your first example example the list is duplicated into l2 and then an element is removed from l2.

Answered By: Andi R

Answer 2

The problem is that when you excecute the line:

l2 = l1[:].remove(l)

you are doing this:

x = l1[:]
l2 = x.remove(l)

As you can see you are assinging to l2 the result of apply the method remove on x. The element will be removed, but remove always return None, that’s why l2 is None.

Answered By: alejandroklever

Answer 3

The below code will do the job.

l1 = ["a", "b", "c", "d"]
for a,b in enumerate(l1):
    l2 = l1[:].pop(a)
    print(l2)

and that’s because .pop() function actually removes and returns it’s value, in the other hand .remove() returns None (and that’s what you get).

Happy Coding.!!!

Answered By: Ioannis E. Kommas

Answer 4

Actually, the remove() function returns None, rather than the list with element removed.

Moreover, remove() delete the element in O(N) complexity, which is cumbersome for your demand. I prefer to adopt the following approach

for idx in range(len(l1)):
    l2 = l1[:idx] + l1[idx+1:]
    print(l2)

Answered By: Yangtian Sun

Python remove from list acting strange when used inline

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