Time 惰性序列评估中的运行时间
鉴于此代码:Time 惰性序列评估中的运行时间,time,clojure,lazy-evaluation,lazy-sequences,Time,Clojure,Lazy Evaluation,Lazy Sequences,鉴于此代码: (reduce my-fun my-lazy-seq) 要测量整个操作的运行时间,请执行以下操作: (time (reduce my-fun my-lazy-seq)) ;;Elapsed time: 1000.1234 msecs 如何在完成之前的各个阶段测量此循环的运行时间?例如: Elapsed time to process next 1000 samples in my-lazy-seq: 100.1234 msecs Elapsed time to proce
(reduce my-fun my-lazy-seq)
要测量整个操作的运行时间,请执行以下操作:
(time (reduce my-fun my-lazy-seq)) ;;Elapsed time: 1000.1234 msecs
如何在完成之前的各个阶段测量此循环的运行时间?例如:
Elapsed time to process next 1000 samples in my-lazy-seq: 100.1234 msecs
Elapsed time to process next 1000 samples in my-lazy-seq: 99.1234 msecs
Elapsed time to process next 1000 samples in my-lazy-seq: 101.1234 msecs
...
这个怎么样:
(defn seq-counter [n coll]
(let [t0 (System/currentTimeMillis)
f (fn [i x]
(let [i (inc i)]
(if (= 0 (rem i n))
(println i "items processed in" (- (System/currentTimeMillis) t0) "ms."))
x))]
(map-indexed f coll)))
地图索引
用于检查进度。上述功能将打印每个n
元素的计数和处理时间
user=> (reduce + (seq-counter 10 (range 100)))
10 items processed in 0 ms.
20 items processed in 0 ms.
...
100 items processed in 1 ms.
4950
请参阅一个非常小的调整,因为
如果只有一个分支,您可以在时放入一个,我认为解决方案是将延迟序列转换为大小为1000的批次的延迟序列,并将每个批次的处理时间。。。但我不知道如何以干净的方式完成第一步,因为源代码已经如此简单和优雅。
(doseq [thousand (partition 1000 my-lazy-seq)]
(time (reduce my-fun thousand)))