Ben Wolfson <[email protected]> writes:
Hi Ben,
>> However, the what's bad with the juxt approach above is that the
>> collection is iterated twice. If somebody comes up with a version
>> that returns a vector of two lazy seqs and which iterates the input
>> collection only once, she'd get my warmest thank you, and my vote for
>> putting it into core.
>
> here's a first shot, kind of ugly but seems to work:
>
> user> (defn seqmr
> [f g a bs]
> (lazy-seq
> (if (seq bs)
> (g (f (first bs)) (seqmr f g a (rest bs)))
> a)))
> #'user/seqmr
> user> (defn lazy-partition [p xs]
> (seqmr (fn [x]
> [(p x) x])
> (fn [[t? x] r]
> (if t?
> [(lazy-seq (cons x (first r))) (lazy-seq (second r))]
> [(lazy-seq (first r)) (lazy-seq (cons x (second r)))]))
> [() ()]
> xs))
> #'user/lazy-partition
Hey, cool, my warmest thank you! :-)
But although it iterates the seq just ones and thus applies the
predicate just once to every element, it's way slower than the juxt
approach.
user> (time (let [[e o] (lazy-partition even? (range 1000000))]
(dorun e)
(dorun o)))
"Elapsed time: 4806.654139 msecs"
nil
user> (time (let [[e o] ((juxt filter remove) even? (range 1000000))]
(dorun e)
(dorun o)))
"Elapsed time: 424.327826 msecs"
nil
Well, if the predicate becomes expensive, it'll start becoming faster,
though.
user> (time (let [[e o] (lazy-partition
(fn [x] (Thread/sleep 5) (even? x))
(range 100))]
(dorun e)
(dorun o)))
"Elapsed time: 513.543943 msecs"
nil
user> (time (let [[e o] ((juxt filter remove)
(fn [x] (Thread/sleep 5) (even? x))
(range 100))]
(dorun e)
(dorun o)))
"Elapsed time: 1025.856917 msecs"
nil
Bye,
Tassilo
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