Hi,
Am 15.11.2010 um 23:07 schrieb Ken Wesson:
> (loop [s some-coll o nil]
> (if (empty? s)
> o
> (let [f (first s)]
> (blah blah blah s blah blah blah f blah blah blah)
> (recur (rest s) (conj o foobar)))))
>
> or some similar control flow structure, where s gets first and rest
> used on it, but not (directly) seq. If some-coll is not already a seq,
> using these implicitly generates a seq view of it with seq; on every
> iteration but the first, s is bound to a seq implementation already.
Changing the above code to the following (which is similarly readable) should
give an immediate speed bump. Rich once stated something around 20%, although I
have not verified the numbers and this was quite a while ago...
(loop [s (seq some-coll)
o nil]
(if s
(let [f (first s)]
(bla bla bla f bla bla bla)
(recur (next s) (conj o foobar)))
o))
Sincerely
Meikel
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