Extracting from a list

[Cliff Wells]

On Wed, 10 Apr 2002 16:35:10 -0400
Joel Bender wrote:

> Hi,
> 
> 
> I would like to process the elements of a list that match a condition, 
> then remove them.  I'll use integers in my example, but my list items 
> are actually objects.
> 
> What's the "right" way to do this?  For example:
> 
>     >>> lst = [3, 2, 5, 1, 0, 7, 4, 8, 6, 9]
>     >>> dothese = [i for i in lst if i < 5]
>     >>> lst = [i for i in lst if i >= 5]
>     >>> for item in dothese:
>     ...     print item
> 
> I'd rather not scan the list twice, particularly if there's nothing to 
> do.  I'd rather not build another double-linked list.
> 
> Is there something like this?
> 
>     >>> for item in lst:
>     ...     if (item < 5):
>     ...         print item
>     ...         del lst[ xxxx ]
> 
> But the "trick" is figuring out what xxxx should be.  Using lst.remove() 
> would work if the internals used "is" and not "is equal to" and the 
> removal didn't modify the list where it would break the loop somehow.
> 
> I could rebuild the list for things that should not be processed:
> 
>     >>> newlst = []
>     >>> for item in lst:
>     ...     if (item < 5):
>     ...         print item
>     ...     else
>     ...         newlst.append(item)
>     >>> lst = newlst
> 
> But these seems like a lot of work when there's nothing in the list to 
> process.

How about:

lst = [3, 2, 5, 1, 0, 7, 4, 8, 6, 9]
for item in lst:
    if item < 5:
        print item
lst = [i for i in lst if i >= 5]

This way you only process the list twice and don't need a redundant copy.  If
you are really concerned about processing the list the second time when there's
nothing to be done, you could use a flag:

lst = [3, 2, 5, 1, 0, 7, 4, 8, 6, 9]
flag = 0
for item in lst:
    if item < 5:
        print item
        flag = 1
if flag:
    lst = [i for i in lst if i >= 5]

-- 
Cliff Wells, Software Engineer
Logiplex Corporation (www.logiplex.net)
(503) 978-6726 x308  (800) 735-0555 x308