compare list

Shi Mu samrobertsmith at gmail.com
Tue Nov 15 10:49:12 CET 2005


On 11/15/05, Brian van den Broek <broek at cc.umanitoba.ca> wrote:
> Shi Mu said unto the world upon 2005-11-15 01:30:
> >>Hey Ben,
> >>
> >>first, as expected, the other two answers you received are better. :-)
> >>
> >>Sets are much better optimized for things like membership testing than
> >>are lists. I'm not competent to explain why; indeed, I keep
> >>overlooking them myself :-(
>
> <snip discussion of my (Brian) first list-based code>
>
> >>
> >>Best,
> >>
> >>Brian vdB
> >
> > is it possible to modify the codes to compare the two lists with not
> > necessarily consecutive numbers?
> > For example,
> > lisA=[1,2,5,9]
> > lisB=[9,5,0]
> > compare A and b will get true because the two lists have 5 and 9
> > together though they are indifferent order.
> > Best Regards,
> > Robert
> >
>
> Sure, but rather than repair a less than best first approach, better
> to do it with sets:
>
>  >>> lisA=[1,2,5,9]
>  >>> lisB=[9,5,0]
>  >>> lisC=[9,5,0,1]
>  >>> def intersection_has_two(sequence1, sequence2):
>         set1, set2 = set(sequence1), set(sequence2)
>         return len(set1.intersection(set2)) == 2
>
>  >>> intersection_has_two(lisA, lisB)
> True
>  >>> intersection_has_two(lisA, lisC)
> False
>
> Sets make stuff easy. (I have got to remember that :-) The only reason
> to make that a function is the one liner gets a bit long for my taste:
>
>  >>> len(set(lisA).intersection(set(lisB))) == 2
> True
>  >>>
>
> Best,
>
> Brian vdB
it does not work.
>>> len(set(lisA).intersection(set(lisB))) == 2
Traceback (most recent call last):
  File "<interactive input>", line 1, in ?
NameError: name 'set' is not defined



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