将多个谓词功能组合为一个 | 将多个谓词功能组合为一个

本文介绍了将多个谓词功能组合为一个的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

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))

变为：

multiple-of-three-or-five?

所以我可以将其用于过滤：

so I can use it for filtering:

(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可以实现为：

For example with Javascript module Ramda it would be achieved as: http://ramdajs.com/docs/#either

const multipleOfThreeOrFive = R.either(multipleOfThree, multipleOfFive)

可以，在Clojure中为。

Sure, in Clojure this is 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

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