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POList comprehension in Haskell, Python and Ruby
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copied!<p>I have started looking at the <a href="http://projecteuler.net/problems" rel="nofollow">project Euler site</a> as a way to learn Haskell, and improve my Python and Ruby. I think the Haskell and Python versions are ok, but I'm sure there must be a cleaner way for Ruby. </p> <p>This is not about how can I make one language look like another one.</p> <p>This is <a href="https://projecteuler.net/problem=1" rel="nofollow">Problem 1</a>:</p> <blockquote> <p>Q: Add all the natural numbers below one thousand that are multiples of 3 or 5.</p> </blockquote> <p>Haskell:</p> <pre><code>sum [ x | x <- [1..999], mod x 3 == 0 || mod x 5 == 0 ] </code></pre> <p>Python:</p> <pre><code>sum ( [ x for x in range(1,1000) if x % 3 == 0 or x % 5 == 0 ] ) </code></pre> <p>Ruby:</p> <pre><code>(1..999) . map {|x| x if x % 3 == 0 || x % 5 == 0 } . compact . inject(:+) </code></pre> <p>They all give the same answer.</p> <hr> <p>OK, so Python can become:</p> <pre><code>sum ( x for x in range(1,1000) if x % 3 == 0 or x % 5 == 0 ) </code></pre> <p>it is now a generator (a good thing as we are not storing the list)</p> <p>but even more fun is:</p> <pre><code>sum( set(range(0,1000,3)) | set(range(0,1000,5)) ) </code></pre> <p>For some reason I was looking at this again and tried a summation approach which should be constant time. In Python 3:</p> <pre><code>def step_sum(mn,mx,step): amax = mx - (mx - mn) % step return (mn + amax) * ((1 + ((amax - mn) / step)) / 2) step_sum(3,999,3) + step_sum(5,999,5) - step_sum(15,999,15) </code></pre> <p>Ruby can become:</p> <pre><code>(1..999) . select {|x| x % 3 == 0 || x % 5 == 0} . inject(:+) </code></pre> <p>or</p> <pre><code>(1..999) . select {|x| x % 3 == 0 or x % 5 == 0} . reduce(:+) </code></pre> <p>I am presuming as unlike <code>map</code>, <code>select</code> doesn't produce 'nul' and therefore there is no need to call <code>compact</code>. nice.</p> <p>Haskell can also be:</p> <pre><code>let ƒ n = sum [0,n..999] in ƒ 3 + ƒ 5 - ƒ 15 </code></pre> <p>or to be clearer:</p> <pre><code>let ƒ n = sum [ 0 , n .. 999 ] in ƒ 3 + ƒ 5 - ƒ (lcm 3 5) </code></pre> <p>as a function that lets us provide the two numbers ourselves:</p> <pre><code>ƒ :: (Integral a) => a -> a -> a ƒ x y = let ƒ n = sum [0,n..999] in ƒ x + ƒ y - ƒ (lcm x y) </code></pre>

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