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

data = [{id:1, name:'BMW'}, {id:2, name:'AN'}, {id:3, name:'BMW'}, {id:1, name:'NNN'}]
key = 'id'//try by id or name
data.reduce((previous, current)=>{
    previous[current[key]] && previous[current[key]].length != 0 ? previous[current[key]].push(current) : previous[current[key]] = new Array(current)
    return previous;
}, {})

其他回答

我从underscore.js fiddler那里借用了这个方法

window.helpers=(function (){
    var lookupIterator = function(value) {
        if (value == null){
            return function(value) {
                return value;
            };
        }
        if (typeof value === 'function'){
                return value;
        }
        return function(obj) {
            return obj[value];
        };
    },
    each = function(obj, iterator, context) {
        var breaker = {};
        if (obj == null) return obj;
        if (Array.prototype.forEach && obj.forEach === Array.prototype.forEach) {
            obj.forEach(iterator, context);
        } else if (obj.length === +obj.length) {
            for (var i = 0, length = obj.length; i < length; i++) {
                if (iterator.call(context, obj[i], i, obj) === breaker) return;
            }
        } else {
            var keys = []
            for (var key in obj) if (Object.prototype.hasOwnProperty.call(obj, key)) keys.push(key)
            for (var i = 0, length = keys.length; i < length; i++) {
                if (iterator.call(context, obj[keys[i]], keys[i], obj) === breaker) return;
            }
        }
        return obj;
    },
    // An internal function used for aggregate "group by" operations.
    group = function(behavior) {
        return function(obj, iterator, context) {
            var result = {};
            iterator = lookupIterator(iterator);
            each(obj, function(value, index) {
                var key = iterator.call(context, value, index, obj);
                behavior(result, key, value);
            });
            return result;
        };
    };

    return {
      groupBy : group(function(result, key, value) {
        Object.prototype.hasOwnProperty.call(result, key) ? result[key].push(value) :              result[key] = [value];
        })
    };
})();

var arr=[{a:1,b:2},{a:1,b:3},{a:1,b:1},{a:1,b:2},{a:1,b:3}];
 console.dir(helpers.groupBy(arr,"b"));
 console.dir(helpers.groupBy(arr,function (el){
   return el.b>2;
 }));

在Joseph Nields的回答之后,有一个polyfill用于将对象分组https://github.com/padcom/array-prototype-functions#arrayprototypegroupbyfieldormapper.因此,您可能希望使用现有的内容,而不是一次又一次地编写这些内容。

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>

我对公认的答案进行了扩展,包括按多个财产分组,然后再加上,使其完全起作用,没有变异。观看演示https://stackblitz.com/edit/typescript-ezydzv

export interface Group {
  key: any;
  items: any[];
}

export interface GroupBy {
  keys: string[];
  thenby?: GroupBy;
}

export const groupBy = (array: any[], grouping: GroupBy): Group[] => {
  const keys = grouping.keys;
  const groups = array.reduce((groups, item) => {
    const group = groups.find(g => keys.every(key => item[key] === g.key[key]));
    const data = Object.getOwnPropertyNames(item)
      .filter(prop => !keys.find(key => key === prop))
      .reduce((o, key) => ({ ...o, [key]: item[key] }), {});
    return group
      ? groups.map(g => (g === group ? { ...g, items: [...g.items, data] } : g))
      : [
          ...groups,
          {
            key: keys.reduce((o, key) => ({ ...o, [key]: item[key] }), {}),
            items: [data]
          }
        ];
  }, []);
  return grouping.thenby ? groups.map(g => ({ ...g, items: groupBy(g.items, grouping.thenby) })) : groups;
};