有人可以解释一下下面的匿名函数如何求值吗？
(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}
我不明白这里的参数min和this来自何处。似乎coll可能正在隐式解构。
如何更好地了解此功能的作用？
Reduce期望函数是它的第一个参数。该函数有两个参数，第一个参数是“到目前为止的结果”，第二个参数是“用于更改结果的数据”。在上面的示例中，reduce采取的功能是接收迄今为止找到的最小的事物，并将其与下一个元素进行比较。然后，它决定其中的哪一个较小，并将其保留至今。
(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)))
这两个中间还有一个可选参数，用于指定结果的初始值。如果忽略此可选参数，如上例所示，则前两个参数用于在结果中生成第一个值。因此，实际上，归约函数的调用时间比输入集合中元素的数量少一倍。
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
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