[Tutor] removing items from list in for loop

[Cameron Simpson]

On 13Feb2022 08:52, marcus.luetolf at bluewin.ch <marcus.luetolf at bluewin.ch> wrote:
>create from a list of n items, n copies of this list in a for loop with 
>one item successivly removed at each iteration.
>
>I should get n copies each with n-1 items.
>
>
>Beeing aware of problems arising when a list is alterated during an
>iteration I tried to get around using a copy oft he original list.
[...]

You code here:

    all_items = ['item1 ','item2 ', ......]
    copy_all_items = ... the same list definition ...
    for item in copy_all_items:
        copy_all_items.remove(item)
        copy_all_items = all_items
        print(item, copy_all_items)

does not do what you think. Specificly this step:

    copy_all_items = all_items

Does not make a copy of "all_items". Instead, it makes "copy_all_items" 
a reference to the original list referenced by "all_items". From then on 
you're modifying the "all_items" list. So your first pass for "item1 " 
removed from your original duplicate list, because that is what 
"copy_all_items" referred to. Every subsequent pass removed from the 
first list.

Only the fact that your for-loop iterates over the original 
"copy_all_items" brings even a semblance of normal operation.

The quick way to copy a list is either:

    list2 = list(list1)

which constructs a new list from the elements in "list1" or:

    list2 = list1[:]

which constructs a new list as a slice of from "list1", but the slice is 
the size of the entire list, thus a copy.

Both of these construct new lists. Compared with:

    list2 = list1

which just makes "list2" point at the same list as "list1". No second 
list is made/copied.

Note that _all_ Python variables are references, and an assigment 
statement never copies the contents of an object - it just takes an 
object reference (the right hand side) and places it in the variable on 
the left hand side.

Cheers,
Cameron Simpson <cs at cskk.id.au>