Most efficient solution?

Alex new_name at
Mon Jul 16 17:02:12 CEST 2001

> C = {}
> for item in B:
>     C[item]=None
> A = filter(lambda e, dic = C: dic.has_key(e), A)

I guess you can even do filter(C.has_key, A), which is probably a little
faster still.

> You may get some speedup by making B a dictionary, and using has_key()
> to see if the word is there. This should get you a O(log(n)) instead
> of O(n) inside the loop. To gain further performance, use filter to
> skim A.

What's n, here?  This looks linear in the length of A, and constant time
in the number of unique elements in B, to me.


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