Chris Rebert wrote: > On Wed, Sep 24, 2008 at 2:02 PM, David Di Biase <[EMAIL PROTECTED]> wrote: >> Hi, >> >> I have a rather large list structure with tuples contained in them (it's >> part of a specification I received) looks like so: >> [(x1,y1,r1,d1),(x2,y2,r2,d2)...] >> >> The list can range from about 800-1500 tuples in size and I'm currently >> sorting it with this: >> >> a_list.sort(lambda a, b: cmp(b[3], a[3])) > > You'd probably be better off using the 'key' keyword argument to > .sort(), which is made for the Schwartzian Transform: > > a_list.sort(key=lambda x: x[3]) > > This sorts the list items by the value produced by the key function > for each item. It's also (IIRC) faster than using a comparison > function like you're currently doing. > > Regards, > Chris
Yes, using 'key' is faster than 'cmp'. If you have Python 2.4 or newer, it would also be slightly faster to use operator.itemgetter() instead of a lambda: >>> import operator >>> a_list.sort(key=operator.itemgetter(3)) >> I'm actually sorting it by the last value in the tuple (d2). I have been >> researching more efficient sorting algorithms and came across Schwartzian >> transform via these links: >> >> http://www.biais.org/blog/index.php/2007/01/28/23-python-sorting-efficiency >> http://dev.fyicenter.com/Interview-Questions/Python/Can_you_do_a_Schwartzian_Transform_in_Python_.html >> >> I get what's happening (sorta...errr...lol) but I'm not sure if it is more >> efficient in my scenario, if it is then I have no idea how to implement it >> properly :-/ >> >> Would be great if a true expert would offer a suggestion for me... >> >> Thanks! >> >> David -- -- http://mail.python.org/mailman/listinfo/python-list
