根据属性将对象数组分解为单独的数组

问题描述 投票:0回答:12

假设我有一个像这样的数组:

var arr = [
    {type:"orange", title:"First"},
    {type:"orange", title:"Second"},
    {type:"banana", title:"Third"},
    {type:"banana", title:"Fourth"}
];

我希望将其分成具有相同类型的对象的数组:

[{type:"orange", title:"First"},
{type:"orange", title:"Second"}]

[{type:"banana", title:"Third"},
{type:"banana", title:"Fourth"}]

但我想一般性地执行此操作,因此没有指定橙色或香蕉的 if 语句

// not like this
for (prop in arr){
    if (arr[prop] === "banana"){
       //add to new array
    }
}

想法? JQuery 和 Underscore 都是可以使用的选项。

javascript jquery arrays object underscore.js
12个回答
84
投票

这对于

Array.reduce(...)
来说是一项简单的工作:

function groupBy(arr, property) {
  return arr.reduce(function(memo, x) {
    if (!memo[x[property]]) { memo[x[property]] = []; }
    memo[x[property]].push(x);
    return memo;
  }, {});
}

var o = groupBy(arr, 'type'); // => {orange:[...], banana:[...]}
o.orange; // => [{"type":"orange","title":"First"},{"type":"orange","title":"Second"}]
o.banana; // => [{"type":"banana","title":"Third"},{"type":"banana","title":"Fourth"}]

当然,如果你的目标浏览器不支持 ECMAScript 262 第 5 版,那么你必须自己实现“reduce”,或者使用 polyfill 库,或者选择其他答案。

[更新] 这是一个适用于任何版本的 JavaScript 的解决方案:

function groupBy2(xs, prop) {
  var grouped = {};
  for (var i=0; i<xs.length; i++) {
    var p = xs[i][prop];
    if (!grouped[p]) { grouped[p] = []; }
    grouped[p].push(xs[i]);
  }
  return grouped;
}

56
投票

JQuery 和 Underscore 都是可以使用的选项。

Underscore 的

groupBy
正是您所需要的。

_.groupBy(arr, "type")

16
投票

ES6解决方案:

function groupBy(arr, property) {
  return arr.reduce((acc, cur) => {
    acc[cur[property]] = [...acc[cur[property]] || [], cur];
    return acc;
  }, {});
}

或者完全ES6fy:

const groupBy = (arr, property) => {
    return arr.reduce((acc, cur) => {
      acc[cur[property]] = [...acc[cur[property]] || [], cur];
      return acc;
    }, {});
}

希望对你有帮助!


14
投票

这假设有一个对象数组:

function groupBy(array, property) {
    var hash = {};
    for (var i = 0; i < array.length; i++) {
        if (!hash[array[i][property]]) hash[array[i][property]] = [];
        hash[array[i][property]].push(array[i]);
    }
    return hash;
}

groupBy(arr,'type')  // Object {orange: Array[2], banana: Array[2]}
groupBy(arr,'title') // Object {First: Array[1], Second: Array[1], Third: Array[1], Fourth: Array[1]}

10
投票

只需构建一个字典,根据标题保存对象即可。你可以这样做:

js

var arr = [
{type:"orange", title:"First"},
 {type:"orange", title:"Second"},
 {type:"banana", title:"Third"},
 {type:"banana", title:"Fourth"}
];
var sorted = {};
for( var i = 0, max = arr.length; i < max ; i++ ){
 if( sorted[arr[i].type] == undefined ){
  sorted[arr[i].type] = [];
 }
 sorted[arr[i].type].push(arr[i]);
}
console.log(sorted["orange"]);
console.log(sorted["banana"]);

jsfiddle 演示:http://jsfiddle.net/YJnM6/


7
投票

打字稿版本。

/**
* Group object array by property
 * Example, groupBy(array, ( x: Props ) => x.id );
 * @param array
 * @param property
 */
export const groupBy = <T>(array: Array<T>, property: (x: T) => string): { [key: string]: Array<T> } =>
  array.reduce((memo: { [key: string]: Array<T> }, x: T) => {
    if (!memo[property(x)]) {
      memo[property(x)] = [];
    }
    memo[property(x)].push(x);
    return memo;
  }, {});

export default groupBy;

1
投票

例如,如果我们不想提及类型,那么我们可以这样实现

var arr = [
    {type:"orange", title:"First"},
    {type:"orange", title:"Second"},
    {type:"banana", title:"Third"},
    {type:"banana", title:"Fourth"}
];

const fun = (ar)=>{
  let temp ={}
for(e of ar){
  !temp[e.type] ? temp[e.type]=[e] : temp[e.type].push(e)
}
  return temp
}
console.log(fun(arr))


0
投票

您还可以使用 https://lodash.com/docs/4.17.15#groupBy

它将满足您的目的


0
投票

对@Watchmaker的答案进行一些调整,返回数组而不是对象。:

function groupBy(arr, key) {
  return arr.reduce((acc, cur) => {
    acc[cur[key]] = [...acc[cur[key]] || [], cur];
    return acc;
  }, []).filter(Boolean);
}

0
投票

另一个typescript版本

最初受到@denolsson 答案的启发,但我删除了用作键的属性,因为它现在在内部数组中是多余的。

还更改并重命名了其他一些内容,以提高可读性并使 linter 满意)

/**
 * Inspired by https://stackoverflow.com/a/53632546/7869582
 */
private groupBy<T>(array: Array<T>, property: keyof T): { [key: string]: Array<T> } {
    return array.reduce(
        (objectToBeBuilt: { [key: string]: Array<T> }, arrayElem: T) => {
            
            const newOuterIdx = arrayElem[property] as unknown as string;
            
            if (!objectToBeBuilt[newOuterIdx]) {
                objectToBeBuilt[newOuterIdx] = [];
            }
            
            if (arrayElem[property]) {
                delete arrayElem[property];    
            }

            objectToBeBuilt[newOuterIdx]?.push(arrayElem);
            return objectToBeBuilt;
        },
        {}  // initial value of objectToBeBuild
    );
}

使用示例:

const resultGroupedByCustomer = this.groupBy(result.rows, "customer_id");

0
投票

@denolssonTypeScript 实现 的功能的改进:

/**
 * Group an array of objects by a specified property.
 * @param {Array<T>} array - The array of objects to group.
 * @param {string} property - The property to group the objects by.
 * @returns {Object} An object where the keys are the unique values of the specified property and the values are arrays of objects with that property value.
 * @template T
 *
 * @example
 * const arr = [{type:"A"}, {type:"A"}, {type:"B"}];
 * const result = groupBy(arr, "type");
 * console.log(result); // Output: { A: [{type: "A"}, {type: "A"}], B: [{type: "B"}] }
 */
export function groupBy<T>(array: Array<T>, property: string): { [key: string]: Array<T> } {
    return array.reduce((memo: { [key: string]: Array<T> }, x: T) => {
        memo[x[property]] ||= [];
        memo[x[property]].push(x);
        return memo;
    }, {});
}

和 JavaScript 版本:

export function groupBy(array, property) {
    return array.reduce((memo, x) => {
        memo[x[property]] ||= [];
        memo[x[property]].push(x);
        return memo;
    }, {});
}

0
投票

现在由内置

Object.groupBy
方法处理。

const arr = [
    {type:"orange", title:"First"},
    {type:"orange", title:"Second"},
    {type:"banana", title:"Third"},
    {type:"banana", title:"Fourth"}
];
const res = Object.groupBy(arr, o => o.type);
console.log(res);

© www.soinside.com 2019 - 2024. All rights reserved.