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

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);

其他回答

Array.prototype.groupBy = function(keyFunction) {
    var groups = {};
    this.forEach(function(el) {
        var key = keyFunction(el);
        if (key in groups == false) {
            groups[key] = [];
        }
        groups[key].push(el);
    });
    return Object.keys(groups).map(function(key) {
        return {
            key: key,
            values: groups[key]
        };
    });
};

想象一下,你有这样的东西:

〔{id:1,cat:'sedan'},{id:2,cat:'sport‘},{id:3,cat:'sport‘},{id:4,cat:'sadan‘}〕

通过这样做:const categories=[…new Set(cars.map((car)=>car.cat))]

你会得到这个:[“sadan”,“port”]

说明:1.首先,我们通过传递一个数组来创建一个新的Set。由于Set仅允许唯一值,因此将删除所有重复项。

现在重复项消失了,我们将使用扩展运算符将其转换回数组。。。

设置文档:https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Set排列运算符文档:https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Spread_syntax

var newArr = data.reduce((acc, cur) => {
    const existType = acc.find(a => a.Phase === cur.Phase);
    if (existType) {
        existType.Value += +cur.Value;
        return acc;
    }

    acc.push({
        Phase: cur.Phase,
        Value: +cur.Value
    });
    return acc;
}, []);

无突变:

const groupBy = (xs, key) => xs.reduce((acc, x) => Object.assign({}, acc, {
  [x[key]]: (acc[x[key]] || []).concat(x)
}), {})

console.log(groupBy(['one', 'two', 'three'], 'length'));
// => {3: ["one", "two"], 5: ["three"]}

/***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>