Sorting lists

Discussion in 'Python' started by asc, Nov 17, 2008.

  1. asc

    asc Guest

    Hi all,
    I have a problem and I'm not sure whether sort() can help me.
    I understand that if I have a list; say L = ['b', 'c', 'a']
    I can use L.sort() and I will then have; L = ['a', 'b', 'c']

    But my problem is this. I have a list, that contains a number of
    embeded lists;
    e.g. L2 = [['something', 'bb'], ['somethingElse', 'cc'],
    ['anotherThing', 'aa']]
    Now I want to sort this list by the second item of each sublist. So
    the outcome I would like is;
    L2 = [['anotherThing', 'aa'], ['something', 'bb'], ['somethingElse',

    Is there a way I can use sort for this? Or am I going to have write a
    custom function to sort this out?

    Many thanks.
    asc, Nov 17, 2008
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  2. asc

    John Machin Guest

    Yes. Read the manual. Look for the key argument. You need to make up a
    function (using def or lambda) that will pluck the desired sort-key
    out of a sub-list.
    John Machin, Nov 17, 2008
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  3. asc

    Chris Rebert Guest

    You use the `key` argument to .sort():

    L2.sort(key=lambda item: item[1])

    This sorts L2 by the result of applying the `key` function to each of
    the items. Behind the scenes, I believe it does a Schwartzian

    Chris Rebert, Nov 17, 2008
  4. Chris Rebert:
    I like the lambda because it's a very readable solution that doesn't
    require the std lib and it doesn't force the programmer (and the
    person that reads the code) to learn yet another thing/function.

    But I can suggest to also look for operator.itemgetter.

    In C# a syntax for lambdas may become a little shorter, but the
    parsing may become less easy:
    L2.sort(key = sub => sub[1])

    In Mathematica they use the & to define lambdas, and #1, #2, etc, to
    denote the arguments.

    bearophileHUGS, Nov 17, 2008
  5. asc

    Terry Reedy Guest

    Since itemgetter is builtin, it will probably run faster, though the
    O(nlogn) time for sorting will override the O(n) time for key calls.
    Terry Reedy, Nov 17, 2008
  6. Well, eventually, for large enough lists, sure. But if the constants are
    sufficiently different, the key calls may be the bottleneck. O(foo)
    analysis is valuable, but it isn't the full story. A fast enough O(n**2)
    algorithm might be preferable to a slow O(log n) algorithm for any data
    you're interested in.

    To give an extreme example, suppose your itemgetter function (not the
    Python built-in!) had to query some remote database over the internet. It
    might be O(n) but the multiplicative constant would be so large that the
    time taken to get items far dominates the time to sort the list, unless
    the list is *seriously* huge:

    (Say) sorting takes O(n*log n) with multiplicative constant of 1e-5.
    Item getting takes O(n) with multiplicative constant of 1.

    Then the time taken to sort doesn't become larger than the time taken to
    get the items until approximately:

    1e-5*n*log(n) > n
    1e-5*log(n) > 1
    log(n) > 1e5
    n > 2**100000

    which is pretty big... for any reasonable-sized list, the O(n) part
    dominates despite the asymptotic behaviour being O(n*log n).
    Steven D'Aprano, Nov 18, 2008
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