Simpler and faster: (count (clojure.string/replace s " " ""))
On 2010/12/22 18:52, Rayne wrote: > I have a piece of code, and I'd like to see how fast it can be. > > (defn count-num-chars [^String s] > (loop [s s acc 0] > (if (seq s) > (recur (rest s) (if (= (first s) \space) acc (inc acc))) > acc))) > > This is the fastest I've been able to get it. The function is very > simple. It takes a string and counts the number of non-space > characters inside of that string. > > I've been testing this code against a 460 non-space character string. > Here is the entire source, benchmarking and all: > > (def s (apply str (repeat 20 "This is a really long string"))) > > (defn count-num-chars [^String s] > (loop [s s acc 0] > (if (seq s) > (recur (rest s) (if (= (first s) \space) acc (inc acc))) > acc))) > > (println "Chars outputted:" (count-num-chars s)) > > (let [before (System/nanoTime)] > (dotimes [_ 1000] > (count-num-chars s)) > (let [after (System/nanoTime)] > (println "Time (in nanoseconds):" (/ (- after before) 1000.0)))) > > Running it gives me around 137343.295 nanoseconds. I've seen some Java > algorithms that could run at just under 3000 nanoseconds. > > Hide your children and hide your women; I want to see the direst, > nastiest, rawest, fastest possible implementation of this function. -- 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
