list comprehention

[bonono at gmail.com]

Tim Chase wrote:
> > 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]
>
>
won't set remove duplicates which he wants to preserve ? He is not just
looking for the 'values' that is in common, but the occurence as well,
if I understand the requirement correctly.

I would just build a dict with the value as the key and occurence as
the value then loop the list and lookup.