将多个谓词函数组合成一个

Question

是否可以撰写例如：

(defn- multiple-of-three? [n] (zero? (mod n 3))
(defn- multiple-of-five? [n] (zero? (mod n 5))

成：

multiple-of-three-or-five?

所以我可以用它来过滤：

(defn sum-of-multiples [n]
  (->> (range 1 n)
       (filter multiple-of-three-or-five?)
       (reduce +)))

另外我不想像这样定义它：

(defn- multiple-of-three-or-five? [n]
  (or (multiple-of-three? n)
      (multiple-of-five? n)))

例如，使用Javascript模块Ramda，它将实现为： http ： //ramdajs.com/docs/#either

const multipleOfThreeOrFive = R.either(multipleOfThree, multipleOfFive)

Answer 1

当然，在Clojure中，这是some-fn 。

(def multiple-of-three-or-five?
  (some-fn multiple-of-three? multiple-of-five?))

(multiple-of-three-or-five? 3)  ; => true
(multiple-of-three-or-five? 4)  ; => false
(multiple-of-three-or-five? 5)  ; => true