[Haskell-cafe] iota

Paul Keir pkeir at dcs.gla.ac.uk
Mon Jun 1 05:00:39 EDT 2009

Hi all,


I was looking for an APL-style "iota" function for array indices. I

"range" from Data.Ix which, with a zero for the lower bound (here

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

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 $

>>= genericReplicate r

                       lists = map buildRepList tups

                 in transpose lists


> length $ iota [2,3,4]

> 24




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