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

[snipped]

> won't set remove duplicates which he wants to preserve ? 

My reading was that the solution shouldn't count duplicates 
("Don't count duplicates").  However, once you mentioned it, and 
I saw other folks' responses that looked very diff. from my own, 
I re-read the OP's comments and found that I missed this key bit:

    "note that only the first *two* 2's count, the third 2
     in the list should not be counted"
                                         (*emphasis* mine)

and that the desired result was a count (though a len(Set()) 
would return the right count if the OP wanted the true 
intersection, but that's beside the point).

My error.  Sorry, ladies and gentlemen :)

I'm partial to the elegance of markscala's suggestion of:

    len([ref.pop(ref.index(x)) for x in lis if x in ref])

though it might need to come with a caveat that it doesn't leave 
"ref" in the same state is it was originally, so it should be 
copied and then manipulated thusly:

    r = ref[:]
    len([r.pop(r.index(x)) for x in (lis if x in r])

which would then leave ref undisturbed.  As tested:

   import random
   ref = [random.randint(1,5) for n in range(5)]
   for x in range(1,10):
       lis = [random.randint(1,5) for n in range(5)]
       r = ref[:]
       print repr((r,lis))
       print len([r.pop(r.index(x)) for x in lis if x in r])

seems to give the results the OP describes.

-tim