Based on what Jozef said, I could write (defn inplace-xor-hh [^bytes a ^bytes b ^bytes out] (hiphip.array/afill! Byte/TYPE [_ out x a y b] (bit-xor (long x) (long y))))
It took 2 ms on my machine, vs. 80 ms for the 'dotimes' solution.' I think that matches java's speed, but if not, I'm guessing java is your only option. Again, you saw how fiddly this all is: even with a library taking care of most of it, there was still tweaking to do (for the simplest possible operation). That's why I think you should consider building on what Prismatic (or someone else with a byte-munging specific library, if that exists) has done. --Leif On Thursday, March 13, 2014 7:42:35 AM UTC-4, Leif wrote: > > Hi, Ignacio. > > Performance tuning in clojure being somewhat complicated, I would look for > prior art here. For instance, the suggestions above give me a 6x speedup, > but it's still *way* slower than the equivalent java code. So I used > Prismatic's hiphip (array)! library on your problem (but with longs) and > got close to java for-loop speed: > > https://github.com/Prismatic/hiphip > > They are focused on numerics, so they don't have support for bytes/shorts, > but it looks like you could add that fairly easily. Or maybe someone has > already written a equivalent library for bit-twiddling. I would look/ask > around. > > If you're interested, the main slowdown in the above suggestions is > probably the extra work from safe typecasts clojure does. The hiphip > library uses the unsafe equivalents from clojure.lang.RT. > > Cheers, > Leif > > On Thursday, March 13, 2014 1:26:33 AM UTC-4, Ignacio Corderi wrote: >> >> Hey guys, here is a huge performance problem I'm trying to figure out: >> >> ;; Say you have 2 data arrays sitting out there of 1 MB >> >> (def one-mb (byte-array (* 1024 1024))) >> (def another-mb (byte-array (* 1024 1024))) >> >> ;; and another one that should have the byte-by-byte XOR of the previous >> two >> >> (def out-mb (byte-array (* 1024 1024))) >> >> ;; question is... how do you code this guy, so that it doesn't take >> forever >> >> (defn inplace-xor [a b out] >> (def ln (count a)) >> (loop [x 0] >> (if (< x ln) >> (do >> (aset-byte out x (bit-xor (nth a x) (nth b x))) >> (recur (+ x 1)) >> )))) >> >> ;; checking the time >> >> (time (inplace-xor one-mb another-mb out-mb)) >> >> ;; takes about ~400ms which is.... well... A LOT >> >> ;; I'm happy to receive a solution that involves calling some java >> library... >> >> Thanks in advance! >> -Ignacio >> >> >> > -- You received this message because you are subscribed to the Google Groups "Clojure" group. To post to this group, send email to clojure@googlegroups.com Note that posts from new members are moderated - please be patient with your first post. To unsubscribe from this group, send email to clojure+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/clojure?hl=en --- You received this message because you are subscribed to the Google Groups "Clojure" group. To unsubscribe from this group and stop receiving emails from it, send an email to clojure+unsubscr...@googlegroups.com. For more options, visit https://groups.google.com/d/optout.
