Most efficient solution?

[Alexandre Fayolle]

On 16 Jul 2001 16:39:34 +0200, Bernhard Herzog <bh at intevation.de> wrote:

>A bit more elegant, perhaps, and a little faster still would be to use 1
>as the value in C and directly use C.get in filter:
>
>C = {}
>for item in B:
>    C[item] = 1
>
>A = filter(C.get, A)

Much more elegant, of course. 
This is a good example of iterative refactoring ;o)

I suppose that the first loop could also be rewritten using map, in order 
to squeeze some more juice out of the beast:

C={}
map(lambda item,dic = C: dic[item]=1, B)

Alexandre Fayolle
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