To remove an element and assign the list in one line

Question

So as the question says, is it possible to remove an element and return the list in one line?

So if lets say

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

This would set b to be NoneType, so I was wondering if there's a way to do this.

Answer 1

I suppose this would work:

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

This is assuming that you want to do both things:

1. Modify the original list by removing the element, and
2. Return the (now modified) list.

EDIT

Another alternative:

b = a.remove(1) or a

Both are fairly confusing; I'm not sure I would use either in code.

EDIT 2

Since you mentioned wanting to use this in a lambda... are you aware that you can use an actual named function any time you would otherwise use a lambda?

E.g. instead of something like this:

map(lambda x: x.remove(1) and x, foo)

You can do this:

def remove_and_return(x):
    x.remove(1)
    return x

map(remove_and_return, foo)

Answer 2

Since remove returns None, you could just or a to it:

>>> a = [1, 3, 2, 4]
>>> a.remove(1) or a
[3, 2, 4]

Answer 3

List a assign to b:

b = a.remove(1) or a  
a is b     # True

Using b is the same to use a, and it's dangerous if you get a final object like [b, c, d], because if you update one of this list element, the all elements will update too.

Compare to:

import copy
b = a.remove(1) or copy.copy(a)

I think this way should be safer.

Answer 4

pure functional solution that does not touch passed list

>>> a = [1,2,3]
>>> (lambda a,b: (lambda a0,a1: (a0.remove(b),a1)[-1] )( *((a.copy(),)*2) ))(a, 3)
[1,2]
>>> a
[1,2,3]