[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
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
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