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POHaskell. Keeping track of indices in order to generate a new list
 

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  1. POHaskell. Keeping track of indices in order to generate a new list
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    <p>I decided to learn Haskell and also learn to think in a more functional way, so I'm trying to solve very simple exercises trying to use a good approach in this paradigm.</p> <p>I'm trying to achieve this simple exercise in Haskell:</p> <pre><code>Input: [2, 4, 1, 1, 2] Output: [True, True, False, False, False, False, True, False, True, True] </code></pre> <p>So, the elements in the <code>Input</code> list will fall to be <code>False</code> in the <code>Output</code> list, and the odd elements will be <code>True</code>; for each one repeated as much times as the value on the <code>Input</code> list indicates.</p> <p>Traverse the <code>Input</code> list, if the <em>i</em>&thinsp;ᵗʰ item is in a pair position, append to output <code>True</code> <em>i</em> times to <code>Output</code>; if the <em>i</em>&thinsp;ᵗʰ item is in an odd position, append <code>False</code> <em>i</em> times to the <code>Output</code> list.</p> <p>It seems to be a very simple problem, and it is. But for me, without any functional programming background, I don't know how to express that in Haskell.</p> <p>I tried to keep track of the current index by using a λ-function within the list comprehension.</p> <pre><code> row :: [Integer] -&gt; [Bool] row xs = [ (last $ zipWith (\i x -&gt; x) [1..] [0..i]) `mod` 2 == 0 | j &lt;- xs, i &lt;- [0..j-1] ] </code></pre> <p>But I'm not understanding its behavior, so I ended using <code>findIndices</code> as a quick alternative:</p> <pre><code> row :: [Integer] -&gt; [Bool] row xs = [ (head $ findIndices (==j) (xs)) `mod` 2 == 0 | j &lt;- xs, i &lt;- [0..j-1] ] </code></pre> <p>Using this last approach it seems OK:</p> <pre><code> &gt; let xs = [ 1, 4, 3, 2 ] &gt; print $ row xs [True,False,False,False,False,True,True,True,False,False] </code></pre> <p>but the problem is not solved, for the items are not necessarily unique:</p> <pre><code> &gt; let xs = [ 2, 2, 4, 3] &gt; print $ row xs [True,True,True,True,True,True,True,True,False,False,False] </code></pre> <p>because <code>head findIndices</code> only gives the first of the occurrences. (Though I think, event that if worked, that's not a very efficient way of solving this problem.)</p> <p>How can I achieve the result I'm looking for in a <em>Haskellian</em> way?</p>
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    1. USJ. A. Corbal
    1. USJ. A. Corbal
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    1. COI think your example output should be changed. You are missing one false for the 4.
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      1. USDiegoNolan
    2. CO@DiegoNolan Thanks
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    3. COYou're not appending true/false i times, right? you're appending it list[i] times.
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      1. USJustin L.
 

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