javascript - 在Underscore/Ramda/函数式语言/库中组合多个filter()谓词

Question

对于数据列表，我想使用多个谓词对其进行过滤，然后对每个过滤器执行操作。

例如。如果我的数据是:

var people = [
    {name: 'Sachin',    profession: 'doctor',   cases: 12},
    {name: 'Djokovic',  profession: 'lawyer',   cases: 14},
    {name: 'Paes',      profession: 'doctor',   cases: 36},
    {name: 'Jordan',    profession: 'lawyer',   cases: 78},
    {name: 'Williams',  profession: 'doctor',   cases: 30},
    {name: 'Nehwal',    profession: 'lawyer',   cases: 75}
]

我想将其转换为:

var peopleWithoutCases = [
    {name: 'Sachin',    profession: 'doctor',   patients:   12, cases: 12},
    {name: 'Djokovic',  profession: 'lawyer',   courtcases: 14, cases: 14},
    {name: 'Paes',      profession: 'doctor',   patients:   36, cases: 36},
    {name: 'Jordan',    profession: 'lawyer',   courtcases: 78, cases: 78},
    {name: 'Williams',  profession: 'doctor',   patients:   30, cases: 30},
    {name: 'Nehwal',    profession: 'lawyer',   courtcases: 75, cases: 75}
]

有没有像这样优雅的函数方法？

    people
    .filter (person => person.profession == 'doctor')
    .map    (person => {
                person.patients = person.cases
                return person;
            })
    .filter (person => person.profession == 'lawyer')
    .map    (person => {
                person.courtcases = person.cases
                return person;
            })

问题是第一个 map 返回一个只有 doctor 的数组。因此第二个过滤器返回[]。

我知道我能做到:

_.union(
    people
        .filter (person => person.profession == 'doctor')
        .map    (person => {
                    person.patients = person.cases
                    return person;
                }),
    people
        .filter (person => person.profession == 'lawyer')
        .map    (person => {
                    person.courtcases = person.cases
                    return person;
                })
)

如果我错了，请纠正我，但是，这需要采用多遍方法来解决问题，在我看来，这种方法效率低下数组列表增长，谓词数量增长。

使用命令式方法编写此代码非常容易。具有多个 if 语句的单个 for 循环。高效但不优雅:)

请建议使用功能性 javascript(如 Underscore、LoDash 或 Excellent RamdaJS 库)的最佳方法。纯函数式语言是如何完成的？

注意:

1. 在这种情况下，数组顺序并不重要。
2. 请不要按字面意思理解示例并建议替代解决方案，我想要一个通用解决方案来过滤和映射多个谓词的列表。

Answer 1

尝试这样:

var el = document.getElementById('dbg');
var $l = function( val ){
  el.innerHTML = el.innerHTML + '<div class="line"><pre>' + val + '</pre></div>';
}



var people = [
    {name: 'Sachin',    profession: 'doctor',   cases: 12},
    {name: 'Djokovic',  profession: 'lawyer',   cases: 14},
    {name: 'Paes',      profession: 'doctor',   cases: 36},
    {name: 'Jordan',    profession: 'lawyer',   cases: 78},
    {name: 'Williams',  profession: 'doctor',   cases: 30},
    {name: 'Nehwal',    profession: 'lawyer',   cases: 75},
    {name: 'Nawak',     profession: 'developper',   cases: 750}
]

var modifiers = {
  'doctor' : function(  ){ this.patients =this.cases ; return this} , 
  'lawyer' : function(  ){ this.courtcases = this.cases; return this; } , 
  'developper' : function(  ){ this.program = this.cases; return this; } , 
}
var result = people.map(function( p){
      modifiers[p.profession].call( p );
      return p;
    })
console.dir(result);
$l( JSON.stringify( result , null , ' ' ) );

.line {
  border : solid 1px #AAA;
  padding : 3px;
  margin : 3px;
  }

<div id='dbg'></div>