在 RamdaJS 中使用带参数的管道过滤器函数

Question

假设我有这两个过滤器：

const getNamesStartingWith = (letter) => persons => {
  return persons.filter(person => person.name.startsWith(letter);  
}

const getPersonsStartingWithZipCode = (zipCode) => persons => {
  return persons.filter(person => person.address.zipCode.startsWith(zipCode);
}

我想做一个通用的管道过滤器。

const pipeFilter = (funcArr, args) => arr => {
...
}

funcArr 是要运行的函数的数组。 args 是一个双精度数组，其中索引是函数的参数。

所以对于我的示例函数，它将是：

const pipeFilter = ([getNamesStartingWith, getPersonsStartingWithZipCode], [['J'], [4]]) => persons => {..}

getNamesStartingWith 的参数是J。 getPersonsStartingWithZipCode 的参数是 4

如果我要“手动”，我会这样做：

export const doPipeFiltering = (funcArr: any[], args: any[]) => (arr) => {

  return funcArr.reduce((acc, func, index) => {

    let filterdArr;
    if (index === 0) {
      filterdArr = func(...args[index])(arr);
    } else {
      filterdArr = func(...args[index])(acc);
    }
    acc = filterdArr;
    return acc;
  }, []);
};

有效。但我想在 RamdaJs 中做这件事，这样我就可以在那里使用所有简洁的功能。

我没有找到如何在 Ramda 中为不同的过滤器函数应用参数。有可能吗？

Answer 1

您绝对可以使用 Ramda 函数使这变得更加简洁。这是一种方法：

const doPipeFiltering = curry ( (funcArr, args, arr) => 
  reduce ( (acc, func) => func (acc), arr, zipWith (apply, funcArr, args) ))

const getNamesStartingWith = (letter) => (persons) => {
  return persons.filter(person => person.name.startsWith(letter)) 
}

const getPersonsStartingWithZipCode = (zipCode) => persons => {
  return persons.filter(person => person.address.zipCode.startsWith(zipCode))
}
                      
const people = [
  {name: 'Amanda', address: {zipCode: '86753'}}, 
  {name: 'Aaron', address: {zipCode: '09867'}},
  {name: 'Amelia', address: {zipCode: '85309'}}, 
  {name: 'Bob', address: {zipCode: '67530'}}
]

console .log (
  doPipeFiltering (
    [getNamesStartingWith, getPersonsStartingWithZipCode], 
    [['A'], ['8']],
    people
  )
)

<script src="//cdnjs.cloudflare.com/ajax/libs/ramda/0.26.1/ramda.js"></script>
<script>const {curry, reduce, zipWith, apply} = R                    </script>

但我建议这不是一个非常符合人体工程学的 API。首先，如果过滤函数有一个完全通用的接口，它可能不会那么糟糕，但现在你必须提供两个单独的数组参数，它们的值必须同步。这对我来说是灾难的根源。

其次，我们必须为此创建过滤列表的函数。如果代码处理过滤，我会发现它更简洁，而且我们只提供了一组简单的谓词。

所以这是一个替代建议，对于我认为更清洁的 API：

const runFilters = useWith (filter, [allPass, identity] )

// or one of these
// const runFilter = curry((filters, xs) => filter(allPass(filters), xs))
// const runFilters = curry ((filters, xs) => reduce ((a, f) => filter (f, a), xs, filters))

const nameStartsWith = (letter) => (person) => person.name.startsWith (letter)
const zipStartsWith = (digits) => (person) => person.address.zipCode.startsWith (digits)

const myFilter = runFilters ([nameStartsWith ('A'), zipStartsWith ('8')])

const people = [
  {name: 'Amanda', address: {zipCode: '86753'}}, 
  {name: 'Aaron', address: {zipCode: '09867'}},
  {name: 'Amelia', address: {zipCode: '85309'}}, 
  {name: 'Bob', address: {zipCode: '67530'}}
]

console .log (
  myFilter (people)
)

<script src="//cdnjs.cloudflare.com/ajax/libs/ramda/0.26.1/ramda.js"></script>
<script>const {useWith, filter, allPass, identity} = R               </script>

请注意，这里的 main 函数更简单，过滤谓词更简单，调用更明确。特别提高了nameStartsWith ('A')和zipStartsWith ('8')的可读性