all pairs of items in a list without indexing?

[Michael Sparks]

On Tue, 28 Sep 2004, Steven Bethard wrote:

> So I need to do something like:
>
> for i in range(len(l)):
>     for j in range(i+1, len(l)):
>         # do something with (l[i], l[j])
>
> where I get all pairs of items in a list (where I'm thinking of pairs
> as sets, not tuples, so order doesn't matter).  There isn't really
> anything wrong with the solution here, but since Python's for-each
> construction is so nice, I usually try to avoid range(len(..)) type
> calls and list indexing when I can...  Is there any way to do this
> without indexing, e.g.:

You want to do this: (eg based on what it sounds like you want to do)

>>> taglist=["WINDOW", [10,10,40,40], "TITLE", "Hello World"]
>>> for i in pairs(taglist):
...    print i
...
('WINDOW', [10, 10, 40, 40])
('TITLE', 'Hello World')
>>>

So let's just write pairs:

def pairs(l):
   x=iter(l)
   while 1:
      yield x.next(),x.next()

There might be a more trivial way of doing this, but this is the first
approach that sprang to mind. This doesn't strictly speaking require a
list as source, and obviously doesn't require an intermediate list.

Regards,


Michael.