On 24/01/2006 09:46, Peter Otten uttered:
len(set([2, 2, 4, 1, 1]).intersection([2, 2, 5, 2, 4]))
> 2
> Close, but no cigar.
I need more coffee! (note to self: Always read entire problem set first)
-- Morten
--
http://mail.python.org/mailman/listinfo/python-list
On 23/01/2006 18:41, Mathijs uttered:
>> len([ref.pop(ref.index(x)) for x in lis if x in ref])
> This is the type of solution I was hoping to see: one-liners, with no
> use of local variables. As Tim Chase already wrote, it has only one
> less elegant side: it alters the original ref list.
>
> T
