Why can't you assign to a list in a loop without enumerate?

[Danny Colligan]

I see.  Thanks for the helpful response.

Danny

Duncan Booth wrote:
> "Danny Colligan" <dannycolligan at gmail.com> wrote:
>
> > In the following code snippet, I attempt to assign 10 to every index in
> > the list a and fail because when I try to assign number to 10, number
> > is a deep copy of the ith index (is this statement correct?).
>
> No. There is no copying involved.
>
> Before the assignment, number is a reference to the object to which the ith
> element of the list also refers. After the assignment you have rebound the
> variable 'number' so it refers to the value 10. You won't affect the list
> that way.
>
> > My question is, what was the motivation for returning a deep copy of
> > the value at the ith index inside a for loop instead of the value
> > itself?
>
> There is no copying going on. It returns the value itself, or at least a
> reference to it.
>
> >  Also, is there any way to assign to a list in a for loop (with
> > as little code as used above) without using enumerate?
>
> a[:] = [10]*len(a)
>
> or more usually something like:
>
> a = [ fn(v) for v in a ]
>
> for some suitable expression involving the value. N.B. This last form
> leaves the original list unchanged: if you really need to mutate it in
> place assign to a[:] as in the first example, but if you are changing all
> elements in the list then you usually want a new list.