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

Duncan Booth duncan.booth at invalid.invalid
Tue Oct 31 20:04:44 CET 2006


"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.






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