[Haskell-cafe] iota
Raynor Vliegendhart
shinnonoir at gmail.com
Mon Jun 1 06:27:37 EDT 2009
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 <pkeir at dcs.gla.ac.uk> wrote:
>
>
>
> 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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