comple list slices

[William Meyer]

 <johnzenger <at> gmail.com> writes:

> Although I don't know if this is faster or more efficient than your
> current solution, it does look cooler:
> 
> def grouprows(inrows):
>     rows = []
>     rows[:] = inrows  #  makes a copy because we're going to be
> deleting
>     while len(rows) > 0:
>         rowspan = rows[0]["rowspan"]
>         yield rows[0:rowspan]  # "returns" this value, but control flow
> unaffected
>         del rows[0:rowspan]  # remove what we just returned from the
> list, and loop
> 
> grouper = grouprows(copyrows)
> print [x for x in grouper]

wow, i think this is much better then my solution. And you can easily call it
for subgroups:

grouper = grouprows(rows)
for x in grouper
    grouperTwo = grouprows(x)
    for y in grouperTwo

Do i need to copy the list in the iterator? (if I am not planning on using rows
again) The reference count for the list members will get bumped on yield right?