javascript - Ramda 的类型检查助手

Question

我想编写一个函数，其规范在下面的代码段中进行了描述，这是我当前的实现。它确实有效。然而，我一直在尝试将它写成 pointfree 并且完全作为 ramda 函数的组合，但找不到解决方案。该问题链接到 obj => map(key => recordSpec[key](obj[key])我无法以可以无意义地编写整个内容的方式减少它。

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，
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

实现此目的的一种方法是使用 R.where ，它采用像您的 recordSpec 这样的规范对象，并将每个谓词与来自第二个对象的相应键的值一起应用。

您的函数将如下所示:

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