On Thursday 17 March 2005 20:08, Leeds, Mark wrote:
But, I also want it to get rid of the AAA KP because there are two AAA's even though the last two letters are different. It doesn't matter to me which one is gotten rid of but I don't know how to change the function to handle this ? I have a feeling it's not that hard though ? Thanks.
Doing the same thing Brian van den Brook did with sets (also for 2.4 only):
def uniqueItems(oldlist,comppos=3): rv = {} for i in reversed(oldlist): rv[i[:comppos]] = i return rv.values()
uniqueItems(["AAA BC","BBB KK","CCC TD","AAA KP","CCC TD"])
['AAA BC', 'BBB KK', 'CCC TD']
[EMAIL PROTECTED] ~ $ python2.4 /usr/local/lib/python2.4/timeit.py -s "import test; uniqueItems = test.uniqueItems; uniqueItemsBrian = test.uniqueItemsBrian" "uniqueItemsBrian(uniqueItems)"
100000 loops, best of 3: 13.8 usec per loop
[EMAIL PROTECTED] ~ $ python2.4 /usr/local/lib/python2.4/timeit.py -s "import test; uniqueItems = test.uniqueItems; uniqueItemsHeiko = test.uniqueItemsHeiko" "uniqueItemsHeiko(uniqueItems)"
100000 loops, best of 3: 9.28 usec per loop
Seems like the dictionary solution is faster, at least for n=3. Do your own tests... ;)
I'm not surprised the dict approach is faster than mine. Mine was maintaining two data structures. But that was done so as to guarantee the output's ordering was the same as the input's, while also taking advantage of the lookup speed that sets and dicts provide. Am I not right in thinking that with the dict approach there is no guarantee that the order from the original list will be preserved?
Also, Heiko, I wonder what is the reason for reversed(oldlist)? Since the list isn't being mutated, there isn't any danger in forward iteration over it. (Plus, unless I'm mistaken, its the only thing making yours a 2.4-only solution.)
Best to all,
Brian vdB
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