Most efficient solution?

Simon Brunning SBrunning at trisystems.co.uk
Mon Jul 16 16:19:53 CEST 2001


> From:	Jeffery D. Collins [SMTP:jcollins at boulder.net]
> > 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)
 
Clever! Unfortunately, it dies (giving a ValueError) if any of the tokens in
B are not present in A. Nasty, but how about:

map((A + B).remove, B)

Cheers,
Simon Brunning
TriSystems Ltd.
sbrunning at trisystems.co.uk




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