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beginner, idiomatic python

bambam david at asdf.asdf
Sun Aug 26 21:19:49 EDT 2007 
That looks good, and perhaps a difference operator
would be too simple to be useful anyway.

Steve.

"Mikael Olofsson" <mikael at isy.liu.se> wrote in message 
news:mailman.238.1187948359.32294.python-list at python.org...
>
>
> bambam wrote:
>>
>> In this case it doesn't matter - my lists don't contain
>> duplicate elements this time - but I have worked with lists in
>> money market and in inventory, and finding the intersection
>> and difference for matching off and netting out are standard
>> operations.
>
> I would use a list comprehension for that case:
>
> A = ['a','b','c','a','c','d']
> U = ['a','b','e']
> B = [x for x in A if x in U]
>
> The result would be B==['a','b','a']
>
> /MiO
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