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

让我们生成一个通用的Array.protocol.groupBy()工具。为了多样化,让我们在递归方法上使用ES6 fancyty扩展运算符进行Haskell式模式匹配。同样,让我们让Array.prototype.groupBy()接受一个回调,该回调将项(e)、索引(i)和应用的数组(a)作为参数。

Array.prototype.groupBy=函数(cb){返回函数迭代([x,…xs],i=0,r=[[],[]]){cb(x,i,[x,…xs])?(r[0].推(x),r):(r[1].推(x),r);是否返回xs.length?迭代(xs,++i,r):r;}(本);};var arr=[0,1,2,3,4,5,6,7,8,9],res=arr.groupBy(e=>e<5);console.log(res);

其他回答

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 }] 
// }

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);

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。。。

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
function groupBy(array, groupBy){
        return array.reduce((acc,curr,index,array) => {
           var  idx = curr[groupBy]; 
              if(!acc[idx]){
                    acc[idx] = array.filter(item => item[groupBy] === idx)
              } 
            return  acc; 

        },{})
    }

// call
groupBy(items,'Step')