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    <p>In addition to Guido's solution, there is more than one way to do it in haskell.</p> <p>Please take a look at <code>fst</code> and <code>snd</code>, which takes the first / second element out of a pair, respectively.</p> <pre><code>GHCi&gt; :t fst fst :: (a, b) -&gt; a GHCi&gt; :t snd snd :: (a, b) -&gt; b </code></pre> <p>You should be familiar with <code>map</code> if you are playing with functional programming languages, which takes a function and a list, applies the function on every element of that list, and gives you all the results in another list: </p> <pre><code>GHCi&gt; :t map map :: (a -&gt; b) -&gt; [a] -&gt; [b] </code></pre> <p>Given a list of pairs, you want two lists, one contains all first elements in order, and the other contains all second elements: </p> <pre><code>GHCi&gt; let split xs = (map fst xs, map snd xs) GHCi&gt; split [(1,2),(3,4),(5,6)] ([1,3,5],[2,4,6]) GHCi&gt; </code></pre> <p>One step further, as @jozefg has pointed out in the comment, that this method is not efficient as @Guido 's one, but we can make some changes to improve it (which is exactly what @Guido 's solution):</p> <p>Now it's time to take a look at how <code>map</code> is implemented <a href="http://hackage.haskell.org/package/base-4.6.0.1/docs/src/GHC-Base.html#map" rel="nofollow">here</a></p> <pre><code>map :: (a -&gt; b) -&gt; [a] -&gt; [b] map _ [] = [] map f (x:xs) = f x : map f xs </code></pre> <p>so we can try to change our <code>split</code> a little:</p> <p>we still need the base case, (i.e. what if <code>xs</code> is empty):</p> <pre><code>split [] = ([], []) split ls = (map fst ls, map snd ls) -- attention! </code></pre> <p>and we break the list into head and tail, just like <code>map</code>:</p> <pre><code>split (x:xs) = (fst x: map fst xs, snd x: map snd xs) </code></pre> <p>Now we can do a pattern matching, <code>(a,b) = x</code>, so we don't have to call two individual functions to break a pair into two:</p> <pre><code>split (x:xs) = (a: map fst xs, b: map snd xs) where (a,b) = x </code></pre> <p>Compare the code here with the line I commented "attention!", have you realized that if we know the result of <code>(map fst xs, map snd xs)</code>, we can simply reuse that result to speed up. Luckily, we already have <code>split ls = (map fst ls, map snd ls)</code>!</p> <p>Using this fact, we finally come up with this version:</p> <pre><code>split [] = ([], []) split (x:xs) = (a:as , b:bs) where (a,b) = x (as,bs) = split xs </code></pre> <p>So there are essentially the same! (but as you can see, the last version we have is more efficient.)</p>
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    1. COThis traverses the list twice though, something to keep in mind.
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    2. CO@jozefg I admit that, just thought it's easier to come up with and let `map` handle the recursion part.
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    3. COWay more elegant :).
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