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