[Haskell-cafe] RE: iota

Paul Keir pkeir at dcs.gla.ac.uk
Mon Jun 1 07:24:05 EDT 2009

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,



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 at haskell.org
Subject: iota


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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