懒惰的 juxt 函数有什么优点吗？

在回答中一个问题 https://stackoverflow.com/questions/10044254/is-there-a-reverse-map-function关于一个使用相同参数映射多个函数的函数（A：juxt），我想出了一个基本上采用与 juxt 相同形式的函数，但使用了映射：

(defn could-be-lazy-juxt
  [& funs]
  (fn [& args]
    (map apply funs (repeat args))))

=> ((juxt inc dec str) 1)
[2 0 "1"]
=> ((could-be-lazy-juxt inc dec str) 1)
(2 0 "1")

=> ((juxt * / -) 6 2)
[12 3 4]
=> ((could-be-lazy-juxt * / -) 6 2)
(12 3 4)

我对它的惰性或性能知之甚少，但 REPL 中的时间确实表明正在发生一些惰性的事情。

=> (time (apply (juxt + -) (range 1 100)))
"Elapsed time: 0.097198 msecs"
[4950 -4948]
=> (time (apply (could-be-lazy-juxt + -) (range 1 100)))
"Elapsed time: 0.074558 msecs"
(4950 -4948)

=> (time (apply (juxt + -) (range 10000000)))
"Elapsed time: 1019.317913 msecs"
[49999995000000 -49999995000000]
=> (time (apply (could-be-lazy-juxt + -) (range 10000000)))
"Elapsed time: 0.070332 msecs"
(49999995000000 -49999995000000)

我确信这个函数并不是真的那么快（结果的打印“感觉”在两者中都一样长）。在函数上执行“take x”只会限制评估的函数数量，这可能会限制其适用性，并且通过“take”限制其他参数应该与正常的 juxt 一样懒惰。

这真的是懒惰吗？惰性 juxt 是否会给表带来任何有用的东西，例如作为其他惰性函数之间的合成步骤？对性能（内存/CPU/对象计数/编译）有何影响？ 为什么 Clojure juxt 使用reduce 实现并返回一个向量，这打破了惰性？

是的，你的 juxt 实现是懒惰的，因为只调用了懒惰的 map 。

如果不实现调用者中的参数（通过使用 apply. 所以我对其进行了一些更改以采用一系列函数：

user> (defn could-be-lazy-juxt
  [funs]
  (fn [& args]
    (map #(apply %1 %2) funs (repeat args))))
#'user/could-be-lazy-juxt

然后定义一个惰性函数序列，在实现时会产生大量输出

user> (defn loud-seq [len] (take len (map #(do (println "produced a function") %) (cycle [inc dec]))))
#'user/loud-seq

然后使用 juxt 从这个惰性函数序列中创建一个函数

user> (def f (could-be-lazy-juxt (loud-seq 50)))
#'user/f

正如您所看到的，该列表仍然是惰性的，当调用结果函数时，它会意识到它是函数列表。

所以我们称它为：

user> (f 1)
(produced a function
produced a function
2 produced a function
produced a function
0 2 produced a function
produced a function
0 2 produced a function
produced a function
0 2 produced a function
produced a function
0 2 produced a function
produced a function
0 2 produced a function
produced a function
0 2 produced a function
produced a function
0 2 produced a function
produced a function
0 2 produced a function
produced a function
0 2 produced a function
produced a function
0 2 produced a function
produced a function
0 2 produced a function
produced a function
0 2 produced a function
produced a function
0 2 produced a function
produced a function
0 2 produced a function
produced a function
0 2 produced a function
produced a function
0 2 produced a function
produced a function
0 2 produced a function
produced a function
0 2 produced a function
produced a function
0 2 produced a function
produced a function
0 2 produced a function
produced a function
0 2 produced a function
produced a function
0 2 produced a function
produced a function
0 2 produced a function
produced a function
0 2 0)
user> 

我将这样做的原因留给你；）