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Rather than starting by trying to fit <code>map</code> in somehow, consider how you might simplify and generalize your current function. Starting from this: <pre><code>dotProduct :: [(Float, Integer)] -> Float -> [(Float, Integer)] dotProduct [] _ = [] dotProduct [(x,y)] z = [(x*z,y)] dotProduct ((x,y):xys) z = (x*z,y):dotProduct (xys) z </code></pre> First, we'll rewrite the second case using the <code>(:)</code> constructor: <pre><code>dotProduct ((x,y):[]) z = (x*z,y):[] </code></pre> Expanding the <code>[]</code> in the result using the first case: <pre><code>dotProduct ((x,y):[]) z = (x*z,y):dotProduct [] z </code></pre> Comparing this to the third case, we can see that they're identical except for this being specialized for when <code>xys</code> is <code>[]</code>. So, we can simply eliminate the second case entirely: <pre><code>dotProduct :: [(Float, Integer)] -> Float -> [(Float, Integer)] dotProduct [] _ = [] dotProduct ((x,y):xys) z = (x*z,y):dotProduct (xys) z </code></pre> Next, generalizing the function. First, we rename it, and let <code>dotProduct</code> call it: <pre><code>generalized :: [(Float, Integer)] -> Float -> [(Float, Integer)] generalized [] _ = [] generalized ((x,y):xys) z = (x*z,y):generalized (xys) z dotProduct :: [(Float, Integer)] -> Float -> [(Float, Integer)] dotProduct xs z = generalized xs z </code></pre> First, we parameterize it by the operation, specializing to multiplication for <code>dotProduct</code>: <pre><code>generalized :: (Float -> Float -> Float) -> [(Float, Integer)] -> Float -> [(Float, Integer)] generalized _ [] _ = [] generalized f ((x,y):xys) z = (f x z,y):generalized f (xys) z dotProduct :: [(Float, Integer)] -> Float -> [(Float, Integer)] dotProduct xs z = generalized (*) xs z </code></pre> Next, we can observe two things: <code>generalized</code> doesn't depend on arithmetic directly anymore, so it can work on any type; and the only time <code>z</code> is used is as the second argument to <code>f</code>, so we can combine them into a single function argument: <pre><code>generalized :: (a -> b) -> [(a, c)] -> [(b, c)] generalized _ [] = [] generalized f ((x,y):xys) = (f x, y):generalized f (xys) dotProduct :: [(Float, Integer)] -> Float -> [(Float, Integer)] dotProduct xs z = generalized (* z) xs </code></pre> Now, we note that <code>f</code> is only used on the first element of a tuple. This sounds useful, so we'll extract that as a separate function: <pre><code>generalized :: (a -> b) -> [(a, c)] -> [(b, c)] generalized _ [] = [] generalized f (xy:xys) = onFirst f xy:generalized f (xys) onFirst :: (a -> b) -> (a, c) -> (b, c) onFirst f (x, y) = (f x, y) dotProduct :: [(Float, Integer)] -> Float -> [(Float, Integer)] dotProduct xs z = generalized (* z) xs </code></pre> Now we again observe that, in <code>generalized</code>, <code>f</code> is only used with <code>onFirst</code>, so we again combine them into a single function argument: <pre><code>generalized :: ((a, c) -> (b, c)) -> [(a, c)] -> [(b, c)] generalized _ [] = [] generalized f (xy:xys) = f xy:generalized f (xys) dotProduct :: [(Float, Integer)] -> Float -> [(Float, Integer)] dotProduct xs z = generalized (onFirst (* z)) xs </code></pre> And once again, we observe that <code>generalized</code> no longer depends on the list containing tuples, so we let it work on any type: <pre><code>generalized :: (a -> b) -> [a] -> [b] generalized _ [] = [] generalized f (x:xs) = f x : generalized f xs </code></pre> Now, compare the code for <code>generalized</code> to this: <pre><code>map :: (a -> b) -> [a] -> [b] map _ [] = [] map f (x:xs) = f x : map f xs </code></pre> It also turns out that a slightly more general version of <code>onFirst</code> also exists, so we'll replace both that and <code>generalized</code> with their standard library equivalents: <pre><code>import Control.Arrow (first) dotProduct :: [(Float, Integer)] -> Float -> [(Float, Integer)] dotProduct xs z = map (first (* z)) xs </code></pre>
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1. POHaskell: Map function with tuples
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1. USC. A. McCann
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1. COI'm amazed at your explanation!! I will read it carefully. Thank you very much for taking the trouble to write an answer like that!
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