Most efficient solution?

Jay Parlar jparlar at home.com
Mon Jul 16 09:19:09 EDT 2001


I have a simple problem, but one that, without some optimization, might hurt the performance of my project. To sum it up, I 
have two lists, we'll call them list A and list B, both lists contain only one-word strings (ie. both were generated by string.split() 
performed on a large amount of text)

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)

Now, this will work fine, however, the size of list A varies, and while list B is constant, it is quite large (more than 750 items). 
List A can also, should the situation warrant, become quite huge as well. The main problem is that I have no idea of the 
efficiency of "in" for large lists like this. Can anyone think of a possibly quicker way, or is this the best?
One other note: List A may (and usually does) contain duplicate words. If those duplicate words appear in list B, then I want 
them both removed, but if they don't appear in list B, then I want them to remain separate (ie. if the word "Python" shows up 
five times in A, then I want the final version of A to still contain five occurrences of "Python")

I really don't know if this can be made any quicker, but any insight would be appreciated. For the cases I've been running, 
it's been quick enough so far, but there's a good chance that the amount of data in list A will be getting much larger, and I'll 
have to perform this entire operation (for different versions of list A) multiple times in one program execution.



Jay Parlar
----------------------------------------------------------------
Software Engineering III
McMaster University
Hamilton, Ontario, Canada

"Though there are many paths
At the foot of the mountain
All those who reach the top
See the same moon."






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