Hi Jeff,
On 27/02/07, Smith, Jeff <[EMAIL PROTECTED]> wrote:
> I'm probably missing something simple here but is there anyway to
> accomplish the following with a list comprehension?
>
> def get_clists():
> return [1, 2, 3]
>
> def get_clist(num):
> if num == 1:
> return ['a', 'b', 'c']
> if num == 2:
> return ['x', 'y', 'z']
> if num == 3:
> return ['p', 'q']
This would be better represented as a dictionary:
>>> clists = { 1:['a', 'b', 'c'],
... 2:['x', 'y', 'z'],
... 3:['p', 'q'] }
You may also be able to replace get_clists() with a call to
clists.keys() (or just simple iteration), depending on what you are
doing.
>>> for k in clists:
... print clists[k]
...
['a', 'b', 'c']
['x', 'y', 'z']
['p', 'q']
> files = list()
> for clist in get_clists():
> files += get_clist(clist)
Just a comment -- you could write this as
"files.extend(get_clist(clist))", which would be slightly more
efficient.
> My first attempt was to try
> [get_clist(c) for c in get_clists()]
>
> but this returns a list of lists rather than the flat list from the
> original.
This will do it:
>>> [x for k in clists for x in clists[k]]
['a', 'b', 'c', 'x', 'y', 'z', 'p', 'q']
Or [x for k in get_clists() for x in get_clist(k)] using your original
structure.
HTH!
--
John.
_______________________________________________
Tutor maillist - Tutor@python.org
http://mail.python.org/mailman/listinfo/tutor
