list comprehention

[Tim Chase]

 > Python beginner here and very much enjoying it. I'm looking
 > for a pythonic way to find how many listmembers are also
 > present in a reference list. Don't count duplicates (eg. if
 > you already found a matching member in the ref list, you can't
 > use the ref member anymore).
 >
 > Example1:
 > ref=[2, 2, 4, 1, 1]
 > list=[2, 3, 4, 5, 3]
 > solution: 2
 >
 > Example2:
 > ref=[2, 2, 4, 1, 1]
 > list=[2, 2, 5, 2, 4]
 > solution: 3 (note that only the first two 2's count, the third
 > 2 in the list should not be counted)

It sounds like you're looking for "set" operations: (using "ell" 
for clarity)

 >>> from sets import Set
 >>> a = [2,2,4,1,1]
 >>> b = [2,3,4,5,3]
 >>> setA = Set(a)
 >>> setB = Set(b)
 >>> results = setA.intersection(setB)
 >>> results
Set([2,4])
 >>> intersection = [x for x in results]
 >>> intersection
[2,4]


I'm a tad confused by the help, as it sounds like sets are 
supposed to be first-class citizens, but in ver2.3.5 that I'm 
running here (or rather "there", on a friend's box), I have to 
"import sets" which I didn't see mentioned in the reference manual.

-one of many tims on the list
tim = Set(["bald", "vegetarian", "loving husband"])

:)