# Parallel list comprehensions

Jon Fairbairn jon.fairbairn at cl.cam.ac.uk
Sun Feb 5 06:53:25 EST 2006

```On 2006-02-04 at 16:08EST Cale Gibbard wrote:
> cartesian xs ys = map (\[x,y] -> (x,y)) \$ sequence [xs,ys]

I'm lost. Isn't that just like
cartesian xs ys = [(x,y)|x<-xs, y<-ys]
?

Whereas...
> On 04/02/06, Jan-Willem Maessen <jmaessen at alum.mit.edu> wrote:
> >
> > On Feb 4, 2006, at 1:31 PM, Jon Fairbairn wrote:
> > > ...
> > > There ought to be a list_product somewhere (I mean [1..]
> > > `list_product` [4..] ==
> > > [(1,4),(2,4),(1,5),(3,4),(2,5),(1,6),...]). Is there?
> >
> > Not that I know of, but here's one which handles finite lists
> > correctly; it'd be a nice addition to Data.List:
> >
> > dzip :: [a] -> [b] -> [(a,b)]
> > dzip =  dzipWith (,)
> >
> > dzipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
> > dzipWith f [] ys = []
> > dzipWith f as [] = []
> > dzipWith f as (y:ys) = dzipK ys [y]
> >    where dzipK (b:bs) rbs =
> >              zipWith f as rbs ++ dzipK bs (b : rbs)
> >          dzipK []     rbs = dzipT as
> >            where dzipT ys@(_:yt) = zipWith f ys rbs ++ dzipT yt
> >                  dzipT []     = []
> >
> > -Jan-Willem Maessen

...does seem to work for infinite lists!
--
Jón Fairbairn                              Jon.Fairbairn at cl.cam.ac.uk

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