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POGenerating cyclic permutations / reduced Latin Squares in Python
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  1. POGenerating cyclic permutations / reduced Latin Squares in Python
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    copied!<p>Was just wondering what's the most efficient way of generating all the cyclic permutations of a list in Python. In either direction. For example, given a list <code>[1, 2, 3, 4]</code>, I want to generate either:</p> <pre><code>[[1, 2, 3, 4], [4, 1, 2, 3], [3, 4, 1, 2], [2, 3, 4, 1]] </code></pre> <p>where the next permutation is generated by moving the last element to the front, or:</p> <pre><code>[[1, 2, 3, 4], [2, 3, 4, 1], [3, 4, 1, 2], [4, 1, 2, 3]] </code></pre> <p>where the next permutation is generated by moving the first element to the back.</p> <p>The second case is slightly more interesting to me because it results in a reduced Latin square (the first case also gives a Latin square, just not reduced), which is what I'm trying to use to do experimental block design. It actually isn't that different from the first case since they're just re-orderings of each other, but order does still matter.</p> <p>The current implementation I have for the first case is:</p> <pre><code>def gen_latin_square(mylist): tmplist = mylist[:] latin_square = [] for i in range(len(mylist)): latin_square.append(tmplist[:]) tmplist = [tmplist.pop()] + tmplist return latin_square </code></pre> <p>For the second case its:</p> <pre><code>def gen_latin_square(mylist): tmplist = mylist[:] latin_square = [] for i in range(len(mylist)): latin_square.append(tmplist[:]) tmplist = tmplist[1:] + [tmplist[0]] return latin_square </code></pre> <p>The first case seems like it should be reasonably efficient to me, since it uses <code>pop()</code>, but you can't do that in the second case, so I'd like to hear ideas about how to do this more efficiently. Maybe there's something in <code>itertools</code> that will help? Or maybe a double-ended queue for the second case?</p>
 

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