Loop over list of pairs

sismex01 at hebmex.com sismex01 at hebmex.com
Fri Jun 6 15:51:41 CEST 2003

```> From: SISMEX01
> Sent: Thursday, June 05, 2003 3:59 PM
>
> > From: Alexander Schmolck [mailto:a.schmolck at gmx.net]
> > Sent: Thursday, June 05, 2003 3:58 PM
> >
> > Thomas Güttler <guettler at thomas-guettler.de> writes:
> >
> > > Hi!
> > >
> > > What is the prefered way of loop over
> > > a list like this?
> > >  mylist=[1, "one", 2, "two", 3, "three"]
> >
> > you could use this:
> >
> > def xgroup(iter,n=2):
> >     """
> >     >>> list(xgroup(range(9), 3))
> >     [(0, 1, 2), (3, 4, 5), (6, 7, 8)]
> >     """
> >     last = []
> >     for elt in iter:
> >         last.append(elt)
> >         if len(last) == n: yield tuple(last); last = []
> >
>
> Since 'n' is known from the start you don't need to
> test against it, nor build a list piecemeal.  Something
> like this seems a bit more solid:
>
> def xgroup(Iterable, group=2):
>    """Return a groupwise iterator for Iterable."""
>    Iterator = iter(Iterable)
>    Length = range(group)
>    while 1:
>       yield [ Iterator.next() for i in xrange(group) ]
>
> hth!
>
> -gca
>

Yuck!  Sorry for replying to my own post, darn untested code.
Please remove "Length = ..." from the previous, as it's
superfluous.

-gca

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