Most efficient solution?
Jeffery D. Collins
jcollins at boulder.net
Mon Jul 16 10:04:11 EDT 2001
On Mon, Jul 16, 2001 at 09:19:09AM -0400, Jay Parlar wrote:
<snip>
> List B consists of my "stopwords", meaning, the words I don't want included in my final version of list A. So what I need to
> do is check every item in list A, and if it occurs in list B, then I want to remove it from the final version of A. My first thought
> would be:
>
> for eachItem in A:
> if eachItem in B:
> A.remove(eachItem)
>
How about:
map(A.remove, B)
--
Jeffery Collins (http://www.boulder.net/~jcollins)
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