On 5/17/18 11:57 AM, Abdur-Rahmaan Janhangeer wrote:
x = [0,1]
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?


Abdur-Rahmaan Janhangeer

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
of making a copy of the 1-less x.  Can you elaborate on the problem you
trying to solve?


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[:]
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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