interleave, take-nth and map or a combination of mapcat and partition when I thought of this:(map #(%1 %2) (cycle [f identity]) coll)I really love clojure's map parallel processing. (I should ask if
every? and some could be allowed to take several colls.)
Hey CHRISTOPHE,
ReplyDeleteI really enjoy reading the examples you post.
Question: Suppose 'coll' is a collection of 10,000 items, this line of code will create another collection using 'cycle' of 10,000 items. Isn't it a bit expensive from performance point of view?
Hi Michael,
ReplyDeletecycle would not create a 10,000-items collection but would lazily realize the first 10,000 conses of an infinite seq (and would not keep a reference to them, making them GCable as soon as they are processed so they have a very short life-span).
I didn't benchmark them but one can count how many transient objects are created (since it's what worries you) by each other approach:
(interleave (map f (take-nth 2 coll)) (take-nth 2 (rest coll))) would create 3*5,000 transient conses (each take-nth + map) -- and I'm not counting seq being called twice on coll.
(mapcat (fn [[a b]] [(f a) b]) (partition 2 coll)) would create 3*5,000 conses (partition yields a seq of 5000 seqs of 2 conses), the map part of mapcat would create 5000 conses and 5000 vectors. The concat part would create 10000 transient conses (seq on each vector)
NB: I used "cons" rather liberally to mean "objects which implement clojure.lang.ISeq".
NB2: A quick pointless benchmark on (range 100000) seems to show that (map #(%1 %2) (cycle [...]) coll) is indeed faster but what I was after when I wrote this post is conciseness.
Michael,
ReplyDeleteYou can redefine cycle using seq-utils/rec-seq:
(defn my-cycle [coll] (rec-seq c (concat coll c))) which create a cyclic seq of only two conses.
(map #(%1 %2) (my-cycle [...]) coll) is obviously faster: no garbage trumps easily collectable garbage.