Re: Parallel list comprehensions

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 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