clojure - 在 Clojure 中减少

Question

有人可以解释一下下面的匿名函数是如何求值的吗？

(defn min-by [f coll] 
  (when (seq coll)
    (reduce (fn [min this]
        (if (> (f min) (f this)) this min))
      coll)))

(min-by :cost [{:cost 100} {:cost 36} {:cost 9}])
;=> {:cost 9}

我不明白参数 min 和 this 从何而来。似乎 coll 可能正在被隐式地破坏。

我怎样才能更好地理解这个函数在做什么？

Answer 1

Reduce 需要一个函数作为它的第一个参数。这个函数有两个参数，第一个参数是“到目前为止的结果”，第二个参数是“用来改变它的数据”。在上面的例子中，reduce 正在接受一个函数，该函数接收迄今为止找到的最小的东西，以及下一个要与之比较的元素。然后它决定它们中哪个更小，并将其作为目前的结果。

(defn min-by [f   ;; this if a function that will be passed the smallest value 
                  ;; yet found and called again with the current value. 
                  ;; the result of these two calls will decide which is the min.
              coll] 
  (when (seq coll)
    (reduce 

      ;; first arg to reduce: a function to add something to the result

      ;; this function will be called once for each of the elements in the
      ;; collection passed as the second argument
      (fn [min     ; the result thus far 
           this]   ; the next element to include in the result

        ;; here we decide which is smaller by passing each to the function f
        ;; that was passed to min-by and compare is results for each of them.
        (if (> (f min) (f this)) this min))

      ;; second arg to reduce: the collection to work on

      ;; each element in this collection will be one of the values of
      ;; the "this" argument to the function above
      coll)))

在这两个中间还有一个可选参数，用于指定结果的初始值。如果省略此可选参数，如上例所示，则前两个参数用于生成结果中的第一个值。所以 reducing 函数实际上被调用的次数比输入集合中的元素数量少一次。

user> (defn print+ [& numbers] (println "called print+") (apply + numbers))
#'builder.upload/print+
user> (reduce print+ [1 2 3])
called print+
called print+
6
user> (reduce print+ 0 [1 2 3])
called print+
called print+
called print+
6