That is quite spectacular. I revised my knowledge of sequence with a little function, akin to "sequence [xs1,xs2]": seq2 xs1 xs2 = do x1 <- xs1 x2 <- xs2 return [x1,x2]

seq2 [0,1] [0,1,2]

[[0,0],[0,1],[0,2],[1,0],[1,1],[1,2]]

I like your point-free style too; and that's a nice use of pred. Many thanks, Paul The iota function you're looking for can be a whole lot simpler if you know about monads (list monad in particular) and sequence. For lists, sequence has the following behaviour: sequence [xs1,xs2, ... xsn] = [[x1,x2, ... , xn] | x1 <- xs1, x2 <- xs2, ... , xn <- xsn] Using this, you can reduce your iota function to a powerful one-liner: iota = sequence . map (enumFromTo 0 . pred) Kind regards, Raynor Vliegendhart From: Paul Keir Sent: 01 June 2009 10:01 To: haskell-cafe@haskell.org Subject: iota Hi all, I was looking for an APL-style "iota" function for array indices. I noticed "range" from Data.Ix which, with a zero for the lower bound (here (0,0)), gives the values I need:

let (a,b) = (2,3)

index ((0,0),(a-1,b-1))

[(0,0),(0,1),(0,2),(1,0),(1,1),(1,2)]

However, I need the results as a list of lists rather than a list of tuples; and my input is a list of integral values. I ended up writing the following function instead. The function isn't long, but longer than I first expected. Did I miss a simpler approach? iota :: (Integral a) => [a] -> [[a]] iota is = let count = product is tups = zip (tail $ scanr (*) 1 is) is buildRepList (r,i) = genericTake count $ cycle $ [0..i-1]

...

= genericReplicate r

lists = map buildRepList tups in transpose lists

length $ iota [2,3,4]

24

Thanks, Paul