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POIterating Lists in Python, Ruby, Haskell (or whatever)
 

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    <p><strong>Update: I realize that I put the question very badly. Here's a second run.</strong></p> <p>Consider the following function:</p> <pre><code>myList = [] optimumList = [] def findOptimumListItems(): n = 5 for i in range (n + 1): for j in range (n + 1 - i): myList.append((i, j, n-i-j)) for i in myList: win = 0.0 draw = 0.0 for j in myList: score = 0 if (i[0] &gt; j[0]): score += 1 if (i[0] == j[0]): score += 0.5 if (i[1] &gt; j[1]): score += 1 if (i[1] == j[1]): score += 0.5 if (i[2] &gt; j[2]): score += 1 if (i[2] == j[2]): score += 0.5 if (score == 2): win += 1 if (score == 1.5): draw += 1 if (win/(len(myList)-win-draw) &gt; 1.0): optimumList.append(i) return optimumList </code></pre> <p>First I make a list. For n = 5 the generated list is:</p> <pre><code>[(0, 0, 5), (0, 1, 4), (0, 2, 3), (0, 3, 2), (0, 4, 1), (0, 5, 0), (1, 0, 4), (1, 1, 3), (1, 2, 2), (1, 3, 1), (1, 4, 0), (2, 0, 3), (2, 1, 2), (2, 2, 1), (2, 3, 0), (3, 0, 2), (3, 1, 1), (3, 2, 0), (4, 0, 1), (4, 1, 0), (5, 0, 0)] </code></pre> <p>Then, the function takes each element of the list and compares it with the list itself. This is how you do it: Say I'm comparing [0, 0, 5] against [3, 1, 1]. 0 loses to 3 (so no points), 0 loses to 1, so no points, 5 wins against 1 (1 point for that). A draw gets 0.5 points, a win gets 1 point. For any item, if wins are more than loses then that item is considered optimum and is added to the optimum list.</p> <p>For n = 5, the optimum list is:</p> <pre><code>[(0, 2, 3), (0, 3, 2), (1, 1, 3), (1, 2, 2), (1, 3, 1), (2, 0, 3), (2, 1, 2), (2, 2, 1), (2, 3, 0), (3, 0, 2), (3, 1, 1), (3, 2, 0)] </code></pre> <p>My question is: How can I write the above function in a <strong>concise</strong> way? I'm especially interested in functional algorithms. Python, Ruby, Java, Haskell answers will be appreciated. (Having said that, if you have a neat solution in any language; that's okay.)</p> <p>Sorry for repeating the same question. I agree that the original question was messy and hard to understand. I hope it's clear now.</p> <p><strong>Update (upon rampion's comment):</strong> Is there an efficient algorithm for this (or this type) problem?</p>
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    1. COOk, I'm still confused. What are `item_`, `item_0`, and `item_n`?
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    2. COThis would be clearer if you said things like `myList[0][0]`, `myList[0][1]`, etc -- using normal (for Python) nested list syntax.
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    3. CONo score for when the different items are equal?
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