将多个谓词函数组合为一个

是否可以编写例如：

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

into:

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 +)))

我也不想这样定义它：

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

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

const multipleOfThreeOrFive = R.either(multipleOfThree, multipleOfFive)

当然，在 Clojure 中这是some-fn https://clojure.github.io/clojure/clojure.core-api.html#clojure.core/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