Ramda类型检查助手

Question

我想写一个函数，其规范在下面的代码片段中描述，这是我目前的实现。它确实有效。不过，我一直试图将它写成无缺陷的，完全是作为ramda函数的组合，并且找不到解决方案。这个问题与obj => map(key => recordSpec[key](obj[key])有关，我无法以一种可以完全无缺点的方式来减少这个问题。

我该怎么办？

/** * check that an object : * - does not have any extra properties than the expected ones (strictness) * - that its properties follow the defined specs * Note that if a property is optional, the spec must include that case * @param {Object.<String, Predicate>} recordSpec * @returns {Predicate} * @throws when recordSpec is not an object */ function isStrictRecordOf(recordSpec) { return allPass([ // 1. no extra properties, i.e. all properties in obj are in recordSpec // return true if recordSpec.keys - obj.keys is empty pipe(keys, flip(difference)(keys(recordSpec)), isEmpty), // 2. the properties in recordSpec all pass their corresponding predicate // For each key, execute the corresponding predicate in recordSpec on the // corresponding value in obj pipe(obj => map(key => recordSpec[key](obj[key]), keys(recordSpec)), all(identity)), ] ) }

例如，为了实现这一目标是使用R.where，这需要一个规范的对象喜欢你recordSpec，并与值从的对应键适用每个谓词 isStrictRecordOf({a : isNumber, b : isString})({a:1, b:'2'}) -> true isStrictRecordOf({a : isNumber, b : isString})({a:1, b:'2', c:3}) -> false isStrictRecordOf({a : isNumber, b : isString})({a:1, b:2}) -> false

Answer 1

的一种方式第二个对象。然后

你的函数看起来像：

const isStrictRecordOf = recordSpec => allPass([ 
    pipe(keys, flip(difference)(keys(recordSpec)), isEmpty), 
    where(recordSpec) 
])