Most efficient solution?

[Jay Parlar]

> > for eachItem in A:
> >     if eachItem in B:
> >         A.remove(eachItem)
> >
> > Now, this will work fine,
> 
> Are you sure about that?  Normally, modifying the list
> you're iterating on does NOT work fine.  Have you tested
> a decent variety of cases?  For example:
> 
> >>> a=['ciao']*7
> >>> for x in a: a.remove(x)
> ....
> >>> a
> ['ciao', 'ciao', 'ciao']
> >>>
> 
> as you see, NOT all seven occurrences of 'ciao' have
> been removed.  Before worrying about speed, I would
> suggest you first make sure your code is CORRECT... I
> think this loop will misbehave if any two successive
> items of A are both in B (only the first one will be
> removed).  Are you sure this can't occur...?

Hmm... That's very interesting. Well, I guess you learn something new every day ;-) I suppose the simple solution (to my 
original code) is to place all non-offending items in a new list, ie. 

>>> for eachItem in a:
... 	if each not in b:
... 		final.append(eachItem)
... 

I've never noticed the list behaviour before that you pointed out, what causes that, if you don't mind my asking?

I also want to take this chance to really thank everyone for the wealth of replies to this topic. That's what I love about this 
list, find an interesting enough question (which I guess this one is) and everyone throws themselves at it.


Jay Parlar

PS. (If this messages shows up on the list a bunch of times, please forgive me, my e-mail client is acting a little strangely)

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Software Engineering III
McMaster University
Hamilton, Ontario, Canada

"Though there are many paths
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