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

这是一个基于TS的功能,不是性能最好的,但很容易阅读和理解!

function groupBy<T>(array: T[], key: string): Record<string, T[]> {
const groupedObject = {}
for (const item of array) {
  const value = item[key]
    if (groupedObject[value] === undefined) {
  groupedObject[value] = []
  }
  groupedObject[value].push(item)
}
  return groupedObject
}

我们以->

const data = [
{ 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" },
];
console.log(groupBy(data, 'Step'))
{
'Step 1': [
    {
      Phase: 'Phase 1',
      Step: 'Step 1',
      Task: 'Task 1',
      Value: '5'
    },
    {
      Phase: 'Phase 1',
      Step: 'Step 1',
      Task: 'Task 2',
      Value: '10'
    }
  ],
  'Step 2': [
    {
      Phase: 'Phase 1',
      Step: 'Step 2',
      Task: 'Task 1',
      Value: '15'
    },
    {
      Phase: 'Phase 1',
      Step: 'Step 2',
      Task: 'Task 2',
      Value: '20'
    }
  ]
}

其他回答

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。

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

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);

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)
//}
let x  = [
  {
    "id": "6",
    "name": "SMD L13",
    "equipmentType": {
      "id": "1",
      "name": "SMD"
    }
  },
  {
    "id": "7",
    "name": "SMD L15",
    "equipmentType": {
      "id": "1",
      "name": "SMD"
    }
  },
  {
    "id": "2",
    "name": "SMD L1",
    "equipmentType": {
      "id": "1",
      "name": "SMD"
    }
  }
];

function groupBy(array, property) {
  return array.reduce((accumulator, current) => {
    const object_property = current[property];
    delete current[property]

    let classified_element = accumulator.find(x => x.id === object_property.id);
    let other_elements = accumulator.filter(x => x.id !== object_property.id);

   if (classified_element) {
     classified_element.children.push(current)
   } else {
     classified_element = {
       ...object_property, 
       'children': [current]
     }
   }
   return [classified_element, ...other_elements];
 }, [])
}

console.log( groupBy(x, 'equipmentType') )

/* output 

[
  {
    "id": "1",
    "name": "SMD",
    "children": [
      {
        "id": "6",
        "name": "SMD L13"
      },
      {
        "id": "7",
        "name": "SMD L15"
      },
      {
        "id": "2",
        "name": "SMD L1"
      }
    ]
  }
]

*/