按数组中的对象分组最有效的方法是什么?

例如,给定此对象数组:

[ 
    { Phase: "Phase 1", Step: "Step 1", Task: "Task 1", Value: "5" },
    { Phase: "Phase 1", Step: "Step 1", Task: "Task 2", Value: "10" },
    { Phase: "Phase 1", Step: "Step 2", Task: "Task 1", Value: "15" },
    { Phase: "Phase 1", Step: "Step 2", Task: "Task 2", Value: "20" },
    { Phase: "Phase 2", Step: "Step 1", Task: "Task 1", Value: "25" },
    { Phase: "Phase 2", Step: "Step 1", Task: "Task 2", Value: "30" },
    { Phase: "Phase 2", Step: "Step 2", Task: "Task 1", Value: "35" },
    { Phase: "Phase 2", Step: "Step 2", Task: "Task 2", Value: "40" }
]

我正在表格中显示这些信息。我想通过不同的方法进行分组,但我想对值求和。

我将Undercore.js用于其groupby函数,这很有用,但并不能完成全部任务,因为我不希望它们“拆分”,而是“合并”,更像SQL groupby方法。

我要找的是能够合计特定值(如果需要)。

因此,如果我按阶段分组,我希望收到:

[
    { Phase: "Phase 1", Value: 50 },
    { Phase: "Phase 2", Value: 130 }
]

如果我组了阶段/步骤,我会收到:

[
    { Phase: "Phase 1", Step: "Step 1", Value: 15 },
    { Phase: "Phase 1", Step: "Step 2", Value: 35 },
    { Phase: "Phase 2", Step: "Step 1", Value: 55 },
    { Phase: "Phase 2", Step: "Step 2", Value: 75 }
]

是否有一个有用的脚本,或者我应该坚持使用Undercore.js,然后遍历生成的对象,自己计算总数?


当前回答

具有排序功能

export const groupBy = function groupByArray(xs, key, sortKey) {
      return xs.reduce(function(rv, x) {
        let v = key instanceof Function ? key(x) : x[key];
        let el = rv.find(r => r && r.key === v);

        if (el) {
          el.values.push(x);
          el.values.sort(function(a, b) {
            return a[sortKey].toLowerCase().localeCompare(b[sortKey].toLowerCase());
          });
        } else {
          rv.push({ key: v, values: [x] });
        }

        return rv;
      }, []);
    };

示例:

var state = [
    {
      name: "Arkansas",
      population: "2.978M",
      flag:
  "https://upload.wikimedia.org/wikipedia/commons/9/9d/Flag_of_Arkansas.svg",
      category: "city"
    },{
      name: "Crkansas",
      population: "2.978M",
      flag:
        "https://upload.wikimedia.org/wikipedia/commons/9/9d/Flag_of_Arkansas.svg",
      category: "city"
    },
    {
      name: "Balifornia",
      population: "39.14M",
      flag:
        "https://upload.wikimedia.org/wikipedia/commons/0/01/Flag_of_California.svg",
      category: "city"
    },
    {
      name: "Florida",
      population: "20.27M",
      flag:
        "https://upload.wikimedia.org/wikipedia/commons/f/f7/Flag_of_Florida.svg",
      category: "airport"
    },
    {
      name: "Texas",
      population: "27.47M",
      flag:
        "https://upload.wikimedia.org/wikipedia/commons/f/f7/Flag_of_Texas.svg",
      category: "landmark"
    }
  ];
console.log(JSON.stringify(groupBy(state,'category','name')));

其他回答

我会检查一下lodash组,它似乎正是你想要的。它也很轻,非常简单。

Fiddle示例:https://jsfiddle.net/r7szvt5k/

如果数组名称为arr,则带有lodash的groupBy仅为:

import groupBy from 'lodash/groupBy';
// if you still use require:
// const groupBy = require('lodash/groupBy');

const a = groupBy(arr, function(n) {
  return n.Phase;
});
// a is your array grouped by Phase attribute

/***数组分组依据*@类别数组*@function arrayGroupBy*@return{object}{“fieldName”:〔{…}〕,…}*@静态*@作者hht*@param{string}}密钥组密钥*@param{array}数据数组**@示例01* --------------------------------------------------------------------------*从“@xx/utils”导入{arrayGroupBy};*常量数组=[* {*type:'资产',*name:'zhangsan',*年龄:33岁,* },* {*类型:'config',*name:“a”,*年龄:13岁,* },* {*类型:'run',*名称:'lisi',*年龄:“3”,* },* {*类型:'xx',*name:'timo',*年龄:'4',* },*];*arrayGroupBy(array,'type',);**结果:{*资产:[{年龄:'33',名称:'zhangsan',类型:'assets'}],*config:[{age:“13”,名称:“a”,类型:“config”}],*运行:[{age:“3”,名称:“lisi”,类型:“run”}],*xx:[{age:“4”,名称:“timo”,类型:“xx”}],* };**@example示例02 null* --------------------------------------------------------------------------*常量数组=空;*arrayGroupBy(数组,“类型”);**结果:{}**@example示例03键取消绑定* --------------------------------------------------------------------------*常量数组=[* {*type:'资产',*name:'zhangsan',*年龄:33岁,* },* {*类型:'config',*name:“a”,*年龄:13岁,* },* {*类型:'run',*名称:'lisi',*年龄:“3”,* },* {*类型:'xx',*name:'timo',*年龄:'4',* },*];*arrayGroupBy(数组,“xx”);** {}**/const arrayGroupBy=(data,key)=>{if(!data||!Array.isArray(data))返回{};常量groupObj={};data.forEach((项)=>{if(!item[key])返回;const fieldName=项[key];if(!groupObj[fieldName]){groupObj[fieldName]=[item];回来}groupObj[fieldName].push(项);});返回groupObj;};常量数组=[{type:'资产',name:'zhangsan',年龄:33岁,},{类型:'config',name:“a”,年龄:13岁,},{类型:'run',名称:'lisi',年龄:“3”,},{类型:'run',名称:“wangmazi”,年龄:“3”,},{类型:'xx',name:'timo',年龄:'4',},];console.dir(arrayGroupBy(array,'type'))<p>description('arrayGroupBy match',()=>{常量数组=[{type:'资产',name:'zhangsan',年龄:33岁,},{类型:'config',name:“a”,年龄:13岁,},{类型:'run',名称:'lisi',年龄:“3”,},{类型:'xx',name:'timo',年龄:'4',},];测试('arrayGroupBy…',()=>{常量结果={资产:[{年龄:'33',名称:'zhangsan',类型:'assets'}],config:[{age:“13”,名称:“a”,类型:“config”}],运行:[{age:“3”,名称:“lisi”,类型:“run”}],xx:[{age:“4”,名称:“timo”,类型:“xx”}],};expect(arrayGroupBy(array,'type')).toEqual(result);});test('arrayGroupBy不匹配..',()=>{//结果expect(arrayGroupBy(array,'xx')).toEqual({});});test('arrayGroupBy null',()=>{let数组=空;expect(arrayGroupBy(array,'type')).toEqual({});});test('arrayGroupBy undefined',()=>{let array=未定义;expect(arrayGroupBy(array,'type')).toEqual({});});test('arrayGroupBy空',()=>{let数组=[];expect(arrayGroupBy(array,'type')).toEqual({});});});</p>

基于以前的答案

const groupBy = (prop) => (xs) =>
  xs.reduce((rv, x) =>
    Object.assign(rv, {[x[prop]]: [...(rv[x[prop]] || []), x]}), {});

如果您的环境支持,使用对象扩展语法会更好一些。

const groupBy = (prop) => (xs) =>
  xs.reduce((acc, x) => ({
    ...acc,
    [ x[ prop ] ]: [...( acc[ x[ prop ] ] || []), x],
  }), {});

在这里,我们的reducer接受部分形成的返回值(从一个空对象开始),并返回一个由上一个返回值的展开成员组成的对象,以及一个新成员,该成员的键是从prop处的当前iteree值计算的,其值是该prop的所有值以及当前值的列表。

常量动物=[{type:'狗',品种:“水坑”},{type:'狗',品种:“拉布拉多”},{类型:'cat',品种:“暹罗”},{type:'狗',品种:“法国斗牛犬”},{类型:'cat',品种:'泥'}];var groupBy=(arr,prop)=>{return arr.reduce((objs,obj)=>{const key=obj[prop];if(键){让fi=objs.findIndex(x=>x.key==key);如果(fi>=0){objs[fi].values=[…objs[fi].values,obj];}其他{对象.推送({key:键,值:[obj]})}}返回对象;}, []);}console.log(groupBy(动物,“类型”))

从@mortb、@jmarceli的回答和这篇文章中,

我利用JSON.stringify()作为分组依据的PRIMITIVE VALUE多列的标识。

无第三方

function groupBy(list, keyGetter) {
    const map = new Map();
    list.forEach((item) => {
        const key = keyGetter(item);
        if (!map.has(key)) {
            map.set(key, [item]);
        } else {
            map.get(key).push(item);
        }
    });
    return map;
}

const pets = [
    {type:"Dog", age: 3, name:"Spot"},
    {type:"Cat", age: 3, name:"Tiger"},
    {type:"Dog", age: 4, name:"Rover"}, 
    {type:"Cat", age: 3, name:"Leo"}
];

const grouped = groupBy(pets,
pet => JSON.stringify({ type: pet.type, age: pet.age }));

console.log(grouped);

使用Lodash第三方

const pets = [
    {type:"Dog", age: 3, name:"Spot"},
    {type:"Cat", age: 3, name:"Tiger"},
    {type:"Dog", age: 4, name:"Rover"}, 
    {type:"Cat", age: 3, name:"Leo"}
];

let rslt = _.groupBy(pets, pet => JSON.stringify(
 { type: pet.type, age: pet.age }));

console.log(rslt);