2d array in Haskell

Question

I'm learning how to use arrays in Haskell, for example, generating a times table:

Prelude Data.Array> array ((0,0),(10,12)) [((x,y),x*y) | x<-[0..10], y<-[0..12]] 
array ((0,0),(10,12)) [((0,0),0),((0,1),0),((0,2),0),((0,3),0),((0,4),0),((0,5),0),((0,6),0),((0,7),0),((0,8),0),((0,9),0),((0,10),0),((0,11),0),((0,12),0),((1,0),0),((1,1),1),((1,2),2),((1,3),3),((1,4),4),((1,5),5),((1,6),6),((1,7),7),((1,8),8),((1,9),9),((1,10),10),((1,11),11),((1,12),12),((2,0),0),((2,1),2),((2,2),4),((2,3),6),((2,4),8),((2,5),10),((2,6),12),((2,7),14),((2,8),16),((2,9),18),((2,10),20),((2,11),22),((2,12),24),((3,0),0),((3,1),3),((3,2),6),((3,3),9),((3,4),12),((3,5),15),((3,6),18),((3,7),21),((3,8),24),((3,9),27),((3,10),30),((3,11),33),((3,12),36),((4,0),0),((4,1),4),((4,2),8),((4,3),12),((4,4),16),((4,5),20),((4,6),24),((4,7),28),((4,8),32),((4,9),36),((4,10),40),((4,11),44),((4,12),48),((5,0),0),((5,1),5),((5,2),10),((5,3),15),((5,4),20),((5,5),25),((5,6),30),((5,7),35),((5,8),40),((5,9),45),((5,10),50),((5,11),55),((5,12),60),((6,0),0),((6,1),6),((6,2),12),((6,3),18),((6,4),24),((6,5),30),((6,6),36),((6,7),42),((6,8),48),((6,9),54),((6,10),60),((6,11),66),((6,12),72),((7,0),0),((7,1),7),((7,2),14),((7,3),21),((7,4),28),((7,5),35),((7,6),42),((7,7),49),((7,8),56),((7,9),63),((7,10),70),((7,11),77),((7,12),84),((8,0),0),((8,1),8),((8,2),16),((8,3),24),((8,4),32),((8,5),40),((8,6),48),((8,7),56),((8,8),64),((8,9),72),((8,10),80),((8,11),88),((8,12),96),((9,0),0),((9,1),9),((9,2),18),((9,3),27),((9,4),36),((9,5),45),((9,6),54),((9,7),63),((9,8),72),((9,9),81),((9,10),90),((9,11),99),((9,12),108),((10,0),0),((10,1),10),((10,2),20),((10,3),30),((10,4),40),((10,5),50),((10,6),60),((10,7),70),((10,8),80),((10,9),90),((10,10),100),((10,11),110),((10,12),120)]

I'm wondering if this is the correct way to hold a matrix or 2d-array of values? Why does it give a list of ((x,y),value) instead of giving a table of values? Is there a way to change how it prints the array?

Tikhon Jelvis · Accepted Answer · 2012-01-17 21:40:13Z

11

Using a tuple as the index is the correct way of getting a multidimensional array. If you want to print it out differently, you'll have to write your own function to convert it to a string.

For example, you could have something like this:

showTable arr = 
  unlines $ map (unwords . map (show . (arr !))) indices
  where indices = [[(x, y) | x <- [startX..endX]] | y <- [startY..endY]]
        ((startX, startY), (endX, endY)) = bounds arr

edited Jan 17, 2012 at 21:40

answered Jan 17, 2012 at 20:29

Tikhon Jelvis

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4 Comments

ehird Over a year ago

indices = range $ bounds arr would work too, by the way (and avoid the shadowing of length, plus it'd also work on arrays of every index type).

NoBugs Over a year ago

unfortunately it is missing the last row and last column.

Tikhon Jelvis Over a year ago

@ehird: Wait, never mind. The way I did it, I need the indices in a nested list by row.

hammar Over a year ago

A more robust version would take the lower bounds into account as well, instead of hard-coding them at zero.

Daniel Fischer · Accepted Answer · 2012-01-17 20:31:06Z

3

That's just the Show instance. The Array constructor is not exported from Data.Array, so you can't directly construct an array. The Show instance produces valid Haskell code that can be used to construct the array from the list of its associations.

answered Jan 17, 2012 at 20:31

Daniel Fischer

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