Most efficient solution?

[Christopher A. Craig]

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alf at leo.logilab.fr (Alexandre Fayolle) writes:

> 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)

I doubt this would save you anything because the cost of that lambda
function will most likely be greater than the savings from the for
loop.

- -- 
Christopher A. Craig <com-nospam at ccraig.org>
"How does a project get to be a year late? 
- -- One day at a time" - Frederick Brooks, Jr.
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