[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,
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 at 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
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