如何仅在 javascript 中执行相当于 LINQ SelectMany() 的操作

Question

不幸的是，我没有 JQuery 或 Underscore，只有纯 javascript（兼容 IE9）。

我想要 LINQ 功能中的 SelectMany() 等效项。

// SelectMany flattens it to just a list of phone numbers.
IEnumerable<PhoneNumber> phoneNumbers = people.SelectMany(p => p.PhoneNumbers);

我可以吗？

编辑：

感谢回答，我得到了这个工作：

var petOwners = 
[
    {
        Name: "Higa, Sidney", Pets: ["Scruffy", "Sam"]
    },
    {
        Name: "Ashkenazi, Ronen", Pets: ["Walker", "Sugar"]
    },
    {
        Name: "Price, Vernette", Pets: ["Scratches", "Diesel"]
    },
];

function property(key){return function(x){return x[key];}}
function flatten(a,b){return a.concat(b);}

var allPets = petOwners.map(property("Pets")).reduce(flatten,[]);

console.log(petOwners[0].Pets[0]);
console.log(allPets.length); // 6

var allPets2 = petOwners.map(function(p){ return p.Pets; }).reduce(function(a, b){ return a.concat(b); },[]); // all in one line

console.log(allPets2.length); // 6

Answer 1

对于简单的选择，您可以使用 Array 的 reduce 函数。
假设你有一个数字数组：

var arr = [[1,2],[3, 4]];
arr.reduce(function(a, b){ return a.concat(b); }, []);
=>  [1,2,3,4]

var arr = [{ name: "name1", phoneNumbers : [5551111, 5552222]},{ name: "name2",phoneNumbers : [5553333] }];
arr.map(function(p){ return p.phoneNumbers; })
   .reduce(function(a, b){ return a.concat(b); }, [])
=>  [5551111, 5552222, 5553333]

编辑：
由于 es6 flatMap 已添加到 Array 原型中。 SelectMany 是 flatMap 的同义词。
该方法首先使用映射函数映射每个元素，然后将结果展平为新数组。 它在 TypeScript 中的简化签名是：

function flatMap<A, B>(f: (value: A) => B[]): B[]

为了完成任务，我们只需要将每个元素 flatMap 到 phoneNumbers

arr.flatMap(a => a.phoneNumbers);

Answer 2

作为一个更简单的选项Array.prototype.flatMap() 或Array.prototype.flat()

const data = [
{id: 1, name: 'Dummy Data1', details: [{id: 1, name: 'Dummy Data1 Details'}, {id: 1, name: 'Dummy Data1 Details2'}]},
{id: 1, name: 'Dummy Data2', details: [{id: 2, name: 'Dummy Data2 Details'}, {id: 1, name: 'Dummy Data2 Details2'}]},
{id: 1, name: 'Dummy Data3', details: [{id: 3, name: 'Dummy Data3 Details'}, {id: 1, name: 'Dummy Data3 Details2'}]},
]

const result = data.flatMap(a => a.details); // or data.map(a => a.details).flat(1);
console.log(result)

Answer 3

稍后，了解 javascript 但仍希望在 Typescript 中使用简单的 Typed SelectMany 方法：

function selectMany<TIn, TOut>(input: TIn[], selectListFn: (t: TIn) => TOut[]): TOut[] {
  return input.reduce((out, inx) => {
    out.push(...selectListFn(inx));
    return out;
  }, new Array<TOut>());
}

Answer 4

Sagi 使用 concat 方法来展平数组是正确的。但是要获得与此示例类似的内容，您还需要选择部分的地图 https://msdn.microsoft.com/library/bb534336(v=vs.100).aspx

/* arr is something like this from the example PetOwner[] petOwners = 
                    { new PetOwner { Name="Higa, Sidney", 
                          Pets = new List<string>{ "Scruffy", "Sam" } },
                      new PetOwner { Name="Ashkenazi, Ronen", 
                          Pets = new List<string>{ "Walker", "Sugar" } },
                      new PetOwner { Name="Price, Vernette", 
                          Pets = new List<string>{ "Scratches", "Diesel" } } }; */

function property(key){return function(x){return x[key];}}
function flatten(a,b){return a.concat(b);}

arr.map(property("pets")).reduce(flatten,[])

Answer 5

// you can save this function in a common js file of your project
function selectMany(f){ 
    return function (acc,b) {
        return acc.concat(f(b))
    }
}

var ex1 = [{items:[1,2]},{items:[4,"asda"]}];
var ex2 = [[1,2,3],[4,5]]
var ex3 = []
var ex4 = [{nodes:["1","v"]}]

开始吧

ex1.reduce(selectMany(x=>x.items),[])

=> [1, 2, 4, "asda"]

ex2.reduce(selectMany(x=>x),[])

=> [1, 2, 3, 4, 5]

ex3.reduce(selectMany(x=> "this will not be called" ),[])

=> []

ex4.reduce(selectMany(x=> x.nodes ),[])

=> ["1", "v"]

注意：在 reduce 函数中使用有效数组（非 null）作为初始值

Answer 6

试试这个（使用 es6）：

 Array.prototype.SelectMany = function (keyGetter) {
 return this.map(x=>keyGetter(x)).reduce((a, b) => a.concat(b)); 
 }

示例数组：

 var juices=[
 {key:"apple",data:[1,2,3]},
 {key:"banana",data:[4,5,6]},
 {key:"orange",data:[7,8,9]}
 ]

使用：

juices.SelectMany(x=>x.data)

Answer 7

我会这样做（避免使用 .concat()）：

function SelectMany(array) {
    var flatten = function(arr, e) {
        if (e && e.length)
            return e.reduce(flatten, arr);
        else 
            arr.push(e);
        return arr;
    };

    return array.reduce(flatten, []);
}

var nestedArray = [1,2,[3,4,[5,6,7],8],9,10];
console.log(SelectMany(nestedArray)) //[1,2,3,4,5,6,7,8,9,10]

如果你不想使用 .reduce()：

function SelectMany(array, arr = []) {
    for (let item of array) {
        if (item && item.length)
            arr = SelectMany(item, arr);
        else
            arr.push(item);
    }
    return arr;
}

如果你想使用 .forEach():

function SelectMany(array, arr = []) {
    array.forEach(e => {
        if (e && e.length)
            arr = SelectMany(e, arr);
        else
            arr.push(e);
    });

    return arr;
}

Answer 8

这里是 joel-harkes 在 TypeScript 中的答案的重写版本，作为扩展，可用于任何数组。所以你可以像somearray.selectMany(c=>c.someprop)一样使用它。转堆，这是javascript。

declare global {
    interface Array<T> {
        selectMany<TIn, TOut>(selectListFn: (t: TIn) => TOut[]): TOut[];
    }
}

Array.prototype.selectMany = function <TIn, TOut>( selectListFn: (t: TIn) => TOut[]): TOut[] {
    return this.reduce((out, inx) => {
        out.push(...selectListFn(inx));
        return out;
    }, new Array<TOut>());
}


export { };

Answer 9

您可以尝试实现所有 C# LINQ 方法并保留其语法的 manipula 包：

Manipula.from(petOwners).selectMany(x=>x.Pets).toArray()

https://github.com/litichevskiydv/manipula

https://www.npmjs.com/package/manipula

Answer 10

对于更高版本的 JavaScript，您可以这样做：

  var petOwners = [
    {
      Name: 'Higa, Sidney',
      Pets: ['Scruffy', 'Sam']
    },
    {
      Name: 'Ashkenazi, Ronen',
      Pets: ['Walker', 'Sugar']
    },
    {
      Name: 'Price, Vernette',
      Pets: ['Scratches', 'Diesel']
    }
  ];

  var arrayOfArrays = petOwners.map(po => po.Pets);
  var allPets = [].concat(...arrayOfArrays);

  console.log(allPets); // ["Scruffy","Sam","Walker","Sugar","Scratches","Diesel"]

见example StackBlitz。