"János Juhász" <[EMAIL PROTECTED]> wrote
> So I can modify any item in it.
> >>> for index in range(len(array)): array[index] *= 2
> ...
> >>> array
> [2, 4, 6, 8, 10]
>
> So I typed this:
> >>> for item in array: item *= 2
This is equivalent to
index = 0
while index < len(array):
item = array[index] # references the content of the array
item = item * 2 # assigns a new value to item, no change
to the array content
index +=1
> It confused me a little, so made another test.
> >>> item1 = array[0]
> >>> item1
> 1
> >>> item1 = 'changed'
> >>> array
> [2, 4, 6, 8, 10]
> >>> item1
> 'changed'
This shows that you can assign item1 to the array content and you
can change the assignment to something else, all without affecting
the list itself
> But how can I iterate the iterate the items as mutable object,
> like the pointers in C ?
> Is the only way to manage the iteration with indexes ?
If you want to change the contents of the array then you need to
access the array, so yes you need the index.
> Or is it any trick like
> >>> for item in array[:]: item *= 2
That only creates a new temporary list referencing the same
objects as the original it doesn't change anything in the existing
one.
The normal way to change a single item, in a lst is to use
the index. If you are doing bulk changes use a list
comprehension to build a new list:
>>> array = [1,2,3,4,5]
>>> print array
[1, 2, 3, 4, 5]
>>> array[2] = 9
>>> print array
[1, 2, 9, 4, 5]
>>> array = [n*2 for n in array]
>>> print array
[2, 4, 18, 8, 10]
>>>
HTH,
--
Alan Gauld
Author of the Learn to Program web site
http://www.freenetpages.co.uk/hp/alan.gauld
_______________________________________________
Tutor maillist - Tutor@python.org
http://mail.python.org/mailman/listinfo/tutor
