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
On 6/1/09, Paul Keir
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 _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe