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9764364
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9
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2012-03-19T02:45:11.673
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2016-01-20T12:21:37.760
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2017-05-23T12:10:42.083
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2
Score
141
ViewCount
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LastEditorDisplayName
text
Body
One classic approach to this problem is to use the "decorate, sort, undecorate" idiom, which is especially simple using python's built-in <code>zip</code> function: <pre><code>>>> list1 = [3,2,4,1, 1] >>> list2 = ['three', 'two', 'four', 'one', 'one2'] >>> list1, list2 = zip(*sorted(zip(list1, list2))) >>> list1 (1, 1, 2, 3, 4) >>> list2 ('one', 'one2', 'two', 'three', 'four') </code></pre> These of course are no longer lists, but that's easily remedied, if it matters: <pre><code>>>> list1, list2 = (list(t) for t in zip(*sorted(zip(list1, list2)))) >>> list1 [1, 1, 2, 3, 4] >>> list2 ['one', 'one2', 'two', 'three', 'four'] </code></pre> It's worth noting that the above may sacrifice speed for terseness; the in-place version, which takes up 3 lines, is a tad faster on my machine for small lists: <pre><code>>>> %timeit zip(*sorted(zip(list1, list2))) 100000 loops, best of 3: 3.3 us per loop >>> %timeit tups = zip(list1, list2); tups.sort(); zip(*tups) 100000 loops, best of 3: 2.84 us per loop </code></pre> On the other hand, for larger lists, the one-line version could be faster: <pre><code>>>> %timeit zip(*sorted(zip(list1, list2))) 100 loops, best of 3: 8.09 ms per loop >>> %timeit tups = zip(list1, list2); tups.sort(); zip(*tups) 100 loops, best of 3: 8.51 ms per loop </code></pre> As Quantum7 points out, <a href="https://stackoverflow.com/a/9764390/577088">JSF's suggestion</a> is a bit faster still, but it will probably only ever be a little bit faster, because Python uses the <a href="https://hg.python.org/cpython/file/e4c32152b25b/Objects/listobject.c#l1863" rel="noreferrer">very same DSU idiom internally</a> for all key-based sorts. It's just happening a little closer to the bare metal. (This shows just how well optimized the <code>zip</code> routines are!) I think the <code>zip</code>-based approach is more flexible and is a little more readable, so I prefer it.
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