Question

传统方法：

(defn make-people [first-names last-names]
  (loop [first-names first-names last-names last-names people []]
    (if (seq first-names)
     (recur
       (rest first-names)
       (rest last-names)
       (conj people {:first (first first-names) :last (first last-names)}))
 people)))

更短的版本：

(defn shorter-make-people [first-names last-names]
  (for [[first last] (partition 2 (interleave first-names last-names))]
    {:first first :last last}))

但我现在手头没有IDE来测试大量数据的性能。

问题是：

没有＆＃39;对于＆＃39;做同样的事情＆＃39;循环＆＃39;和＆＃39;复发＆＃39;在这个例子中？
是否适用于更一般的案例？

建议使用任何性能测试结果。

core.clj中的参考源代码：for loop

Answer 1

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http://clojure.org/sequences

Answer 2

也许我错过了什么，但为什么会这样？在for理解的情况下，无需任何递归。

关于测试结果：

请-人

(bench (doall (make-people first-names last-names))) Evaluation count : 1581540 in 60 samples of 26359 calls. Execution time mean : 40.210018 µs Execution time std-deviation : 1.838808 µs Execution time lower quantile : 37.110371 µs ( 2.5%) Execution time upper quantile : 44.515176 µs (97.5%) Overhead used : 10.301128 ns Found 2 outliers in 60 samples (3.3333 %) low-severe 2 (3.3333 %) Variance from outliers : 31.9497 % Variance is moderately inflated by outliers

偶数短化妆的人

(bench (doall (shorter-make-people first-names last-names))) Evaluation count : 306180 in 60 samples of 5103 calls. Execution time mean : 204.226064 µs Execution time std-deviation : 5.726497 µs Execution time lower quantile : 196.693866 µs ( 2.5%) Execution time upper quantile : 213.226726 µs (97.5%) Overhead used : 10.301128 ns

偶数短化妆的人

(defn even-shorter-make-people [first-names last-names] (map #(array-map :first %1 :last %2) first-names last-names)) (bench (doall (even-shorter-make-people first-names last-names))) Evaluation count : 1049880 in 60 samples of 17498 calls. Execution time mean : 59.182048 µs Execution time std-deviation : 2.338641 µs Execution time lower quantile : 56.361840 µs ( 2.5%) Execution time upper quantile : 64.056606 µs (97.5%) Overhead used : 10.301128 ns

是＆＃39; for＆＃39;使用像loop-recur模式一样的尾递归？

2 个答案: