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

使用ES6 Map对象:

/***@描述*采用Array<V>和分组函数,*并返回由分组函数分组的数组的Map。**@param list V类型的数组。*@param keyGetter一个函数,将数组类型V作为输入,并返回类型K的值。*K通常是V的属性键。**@返回由分组函数分组的数组的映射。*///导出函数组By<K,V>(列表:Array<V>,keyGetter:(输入:V)=>K):Map<K,Array<V>>{//const-map=new map<K,Array<V>>();函数groupBy(list,keyGetter){const-map=new map();list.forEach((项)=>{const key=keyGetter(项);constcollection=map.get(key);if(!集合){map.set(键,[项]);}其他{collection.push(项);}});回归图;}//示例用法常量宠物=[{type:“Dog”,name:“Spot”},{类型:“猫”,名称:“老虎”},{类型:“狗”,名称:“路虎”},{类型:“猫”,名称:“利奥”}];const grouped=groupBy(宠物,宠物=>宠物类型);console.log(分组.get(“Dog”));//->〔{类型:“狗”,名称:“斑点”},{类型“狗”、名称:“漫游者”}〕console.log(分组.get(“Cat”));//->〔{类型:“猫”,名称:“老虎”},{类型“猫”、名称:“狮子座”}〕const奇数=符号();const even=符号();常量=[1,2,3,4,5,6,7];const oddEven=groupBy(数字,x=>(x%2===1?奇数:偶数));console.log(oddEven.get(奇数));//->[1,3,5,7]console.log(oddEven.get(偶数));//->[2,4,6]

关于地图:https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Map

其他回答

我已经改进了答案。此函数获取组字段数组并返回分组对象,该对象的键也是组字段的对象。

function(xs, groupFields) {
        groupFields = [].concat(groupFields);
        return xs.reduce(function(rv, x) {
            let groupKey = groupFields.reduce((keyObject, field) => {
                keyObject[field] = x[field];
                return keyObject;
            }, {});
            (rv[JSON.stringify(groupKey)] = rv[JSON.stringify(groupKey)] || []).push(x);
            return rv;
        }, {});
    }



let x = [
{
    "id":1,
    "multimedia":false,
    "language":["tr"]
},
{
    "id":2,
    "multimedia":false,
    "language":["fr"]
},
{
    "id":3,
    "multimedia":true,
    "language":["tr"]
},
{
    "id":4,
    "multimedia":false,
    "language":[]
},
{
    "id":5,
    "multimedia":false,
    "language":["tr"]
},
{
    "id":6,
    "multimedia":false,
    "language":["tr"]
},
{
    "id":7,
    "multimedia":false,
    "language":["tr","fr"]
}
]

groupBy(x, ['multimedia','language'])

//{
//{"multimedia":false,"language":["tr"]}: Array(3), 
//{"multimedia":false,"language":["fr"]}: Array(1), 
//{"multimedia":true,"language":["tr"]}: Array(1), 
//{"multimedia":false,"language":[]}: Array(1), 
//{"multimedia":false,"language":["tr","fr"]}: Array(1)
//}

Ceasar的答案很好,但只适用于数组中元素的内部财产(字符串的长度)。

这个实现的工作方式更像:这个链接

const groupBy = function (arr, f) {
    return arr.reduce((out, val) => {
        let by = typeof f === 'function' ? '' + f(val) : val[f];
        (out[by] = out[by] || []).push(val);
        return out;
    }, {});
};

希望这有帮助。。。

基于@Ceasar Bautista的原始想法,我修改了代码并使用typescript创建了一个groupBy函数。

static groupBy(data: any[], comparator: (v1: any, v2: any) => boolean, onDublicate: (uniqueRow: any, dublicateRow: any) => void) {
    return data.reduce(function (reducedRows, currentlyReducedRow) {
      let processedRow = reducedRows.find(searchedRow => comparator(searchedRow, currentlyReducedRow));

      if (processedRow) {
        // currentlyReducedRow is a dublicateRow when processedRow is not null.
        onDublicate(processedRow, currentlyReducedRow)
      } else {
        // currentlyReducedRow is unique and must be pushed in the reducedRows collection.
        reducedRows.push(currentlyReducedRow);
      }

      return reducedRows;
    }, []);
  };

此函数接受一个回调(比较器)和一个第二个回调(onDuplicate),该回调比较行并查找副本。

用法示例:

data = [
    { name: 'a', value: 10 },
    { name: 'a', value: 11 },
    { name: 'a', value: 12 },
    { name: 'b', value: 20 },
    { name: 'b', value: 1 }
  ]

  private static demoComparator = (v1: any, v2: any) => {
    return v1['name'] === v2['name'];
  }

  private static demoOnDublicate = (uniqueRow, dublicateRow) => {
    uniqueRow['value'] += dublicateRow['value'];    
  };

使命感

groupBy(data, demoComparator, demoOnDublicate) 

将执行计算值和的分组。

{name: "a", value: 33}
{name: "b", value: 21}

我们可以根据项目的需要创建任意多个回调函数,并根据需要聚合这些值。在一个例子中,我需要合并两个数组,而不是求和数据。

此解决方案采用任意函数(而不是键),因此比上述解决方案更灵活,并允许箭头函数,这与LINQ中使用的lambda表达式类似:

Array.prototype.groupBy = function (funcProp) {
    return this.reduce(function (acc, val) {
        (acc[funcProp(val)] = acc[funcProp(val)] || []).push(val);
        return acc;
    }, {});
};

注意:是否要扩展Array的原型取决于您。

大多数浏览器支持的示例:

[{a:1,b:"b"},{a:1,c:"c"},{a:2,d:"d"}].groupBy(function(c){return c.a;})

使用箭头函数(ES6)的示例:

[{a:1,b:"b"},{a:1,c:"c"},{a:2,d:"d"}].groupBy(c=>c.a)

以上两个示例都返回:

{
  "1": [{"a": 1, "b": "b"}, {"a": 1, "c": "c"}],
  "2": [{"a": 2, "d": "d"}]
}

MDN的Array.reduce()文档中有此示例。

// Grouping objects by a property
// https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/Reduce#Grouping_objects_by_a_property#Grouping_objects_by_a_property

var people = [
  { name: 'Alice', age: 21 },
  { name: 'Max', age: 20 },
  { name: 'Jane', age: 20 }
];

function groupBy(objectArray, property) {
  return objectArray.reduce(function (acc, obj) {
    var key = obj[property];
    if (!acc[key]) {
      acc[key] = [];
    }
    acc[key].push(obj);
    return acc;
  }, {});
}

var groupedPeople = groupBy(people, 'age');
// groupedPeople is:
// { 
//   20: [
//     { name: 'Max', age: 20 }, 
//     { name: 'Jane', age: 20 }
//   ], 
//   21: [{ name: 'Alice', age: 21 }] 
// }