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    <p><strong>A Solution Using a Parsing Library</strong></p> <p>Since you'll probably have a number of people responding with code that parses strings of <code>Int</code>s into <code>[[Int]]</code> (<code>map (map read . words) . lines $ contents</code>), I'll skip that and introduce one of the parsing libraries. If you were to do this task for real work you'd probably use such a library that parses <code>ByteString</code> (instead of <code>String</code>, which means your IO reads everything into a linked list of individual characters).</p> <pre><code>import System.Environment import Control.Monad import Data.Attoparsec.ByteString.Char8 import qualified Data.ByteString as B </code></pre> <p>First, I imported the Attoparsec and bytestring libraries. You can see these libraries and their documentation on <a href="http://hackage.haskell.org/package/attoparsec" rel="nofollow noreferrer">hackage</a> and install them using the <code>cabal</code> tool.</p> <pre><code>main = do (fname:_) &lt;- getArgs putStrLn fname parsed &lt;- parseX fname print parsed </code></pre> <p><code>main</code> is basically unchanged.</p> <pre><code>parseX :: FilePath -&gt; IO (Int, Int, [Int], [[Int]]) parseX fname = do bs &lt;- B.readFile fname let res = parseOnly parseDrozzy bs -- We spew the error messages right here either (error . show) return res </code></pre> <p><code>parseX</code> (renamed from parse to avoid name collision) uses the bytestring library's readfile, which reads in the file packed, in contiguous bytes, instead of into cells of a linked list. After parsing I use a little shorthand to return the result if the parser returned <code>Right result</code> or print an error if the parser returned a value of <code>Left someErrorMessage</code>.</p> <pre><code>-- Helper functions, more basic than you might think, but lets ignore it sint = skipSpace &gt;&gt; int int = liftM floor number parseDrozzy :: Parser (Int, Int, [Int], [[Int]]) parseDrozzy = do m &lt;- sint n &lt;- sint skipSpace ks &lt;- manyTill sint endOfLine arr &lt;- count m (count n sint) return (m,n,ks,arr) </code></pre> <p>The real work then happens in <code>parseDrozzy</code>. We get our <code>m</code> and <code>n</code> <code>Int</code> values using the above helper. In most Haskell parsing libraries we must explicitly handle whitespace - so I skip the newline after <code>n</code> to get to our <code>ks</code>. <code>ks</code> is just all the int values before the next newline. Now we can actually use the previously specified number of rows and columns to get our array.</p> <p>Technically speaking, that final bit <code>arr &lt;- count m (count n sint)</code> doesn't follow your format. It will grab <code>n</code> ints even if it means going to the next line. We could copy Python's behavior (not verifying the number of values in a row) using <code>count m (manyTill sint endOfLine)</code> or we could check for each end of line more explicitly and return an error if we are short on elements.</p> <p><strong>From Lists to a Matrix</strong></p> <p>Lists of lists are not 2 dimensional arrays - the space and performance characteristics are completely different. Let's pack our list into a real matrix using Data.Array.Repa (<code>import Data.Array.Repa</code>). This will allow us to access the elements of the array efficiently as well as perform operations on the entire matrix, optionally spreading the work among all the available CPUs.</p> <p>Repa defines the dimensions of your array using a slightly odd syntax. If your row and column lengths are in variables <code>m</code> and <code>n</code> then <code>Z :. n :. m</code> is much like the C declaration <code>int arr[m][n]</code>. For the one dimensional example, <code>ks</code>, we have:</p> <pre><code>fromList (Z :. (length ks)) ks </code></pre> <p>Which changes our type from <code>[Int]</code> to <code>Array DIM1 Int</code>.</p> <p>For the two dimensional array we have:</p> <pre><code>let matrix = fromList (Z :. m :. n) (concat arr) </code></pre> <p>And change our type from <code>[[Int]]</code> to <code>Array DIM2 Int</code>.</p> <p>So there you have it. A parsing of your file format into an efficient Haskell data structure using production-oriented libraries.</p>
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    1. POHow do I parse a matrix of integers in Haskell?
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    1. USThomas M. DuBuisson
    1. USThomas M. DuBuisson
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    1. COCool! Will this Array type allow me to query for things like 1st column or 2nd row?
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    2. COIt sure does. One question about that operation is [here](http://stackoverflow.com/questions/6170008/how-to-take-an-array-slice-with-repa-over-a-range). The general form, assuming you want column `i`, is `slice arr (Z :. All :. i)`.
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