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

通常,我使用Lodash JavaScript实用程序库和预先构建的groupBy()方法。它非常容易使用,请在此处查看更多详细信息。

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"]}

我不认为给出的答案是对问题的回应,我认为以下内容应回答第一部分:

常量arr=[{阶段:“阶段1”,步骤:“步骤1”,任务:“任务1”,值:“5”},{阶段:“阶段1”,步骤:“步骤1”,任务:“任务2”,值:“10”},{阶段:“阶段1”,步骤:“步骤2”,任务:“任务1”,值:“15”},{阶段:“阶段1”,步骤:“步骤2”,任务:“任务2”,值:“20”},{阶段:“阶段2”,步骤:“步骤1”,任务:“任务1”,值:“25”},{阶段:“阶段2”,步骤:“步骤1”,任务:“任务2”,值:“30”},{阶段:“阶段2”,步骤:“步骤2”,任务:“任务1”,值:“35”},{阶段:“阶段2”,步骤:“步骤2”,任务:“任务2”,值:“40”}]const groupBy=(key)=>arr.sort((a,b)=>a[key].localeCompare(b[key])).reduce((total,currentValue)=>{const newTotal=总计;如果(总长度&&总计[total.length-1][key]==当前值[key])新总计[total.length-1]={…总计[总长度-1],…当前值,值:parseInt(total[total.length-1].Value)+parseInt(currentValue.Value,};else newTotal[total.length]=当前值;return newTotal;}, []);console.log(groupBy(“阶段”));//=>[{阶段:“阶段1”,值:50},{阶段“阶段2”,值130}]console.log(groupBy(“步骤”));//=>[{步骤:“步骤1”,值:70},{步骤“步骤2”,值为110}]

解释相同的代码。喜欢它我喜欢这里

const groupBy = (array, key) => {
  return array.reduce((result, currentValue) => {
    (result[currentValue[key]] = result[currentValue[key]] || []).push(
      currentValue
    );
    console.log(result);
    return result;
  }, {});
};

USE

 let group =   groupBy(persons, 'color');

现在有点晚了,但也许有人喜欢这个。

ES6:

常量用户=[{name:“Jim”,颜色:“蓝色”},{name:“Sam”,颜色:“蓝色”},{name:“Eddie”,颜色:“绿色”},{姓名:“Robert”,颜色:“绿色”},];常量groupBy=(arr,key)=>{const initialValue={};返回arr.reduce((acc,cval)=>{const myAttribute=cval[key];acc[myAttribute]=[…(acc[myAattribute]| |[]),cval]返回acc;},initialValue);};const res=groupBy(用户,“颜色”);console.log(“分组依据:”,res);