On 5/17/18 11:57 AM, Abdur-Rahmaan Janhangeer wrote:
x = [0,1]
x.remove(0)
new_list = x
instead i want in one go
x = [0,1]
new_list = x.remove(0) # here a way for it to return the modified list by
adding a .return() maybe ?
There isn't a way to do that in one line. I often find myself splitting
long statements into more, shorter, statements to express myself more
clearly.
I don't know if you have a real piece of code in mind, so I don't know
if you can tell us: why is it useful to have another variable referring
to the same list?
--Ned.
Abdur-Rahmaan Janhangeer
https://github.com/Abdur-rahmaanJ
On Thu, 17 May 2018, 19:54 Alexandre Brault, <abra...@mapgears.com> wrote:
On 2018-05-17 11:26 AM, Abdur-Rahmaan Janhangeer wrote:
I don't understand what this would return? x? You already have x. Is it
meant to make a copy? x has been mutated, so I don't understand the
benefit
of making a copy of the 1-less x. Can you elaborate on the problem you
are
trying to solve?
--Ned.
assignment to another var
You already have access to the list before removal, the list after
removal and the element to be removed.
Do need a copy of the list before removing x?
old_list = list[:]
list.remove(x)
Do you need the list after removing x?
list.remove(x) # list is the modified list
Do you need x?
list.remove(x) # x is x
What else would need to be assigned to another var?
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