将多个谓词函数组合成一个

Question

是否可以组成例如：

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

进入：

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:

(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

Compose multiple predicate functions into one