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

发帖是因为即使这个问题已经7年了,我仍然没有看到一个符合原始标准的答案:

我不希望它们“拆分”,而是“合并”,更像SQL组方法

我最初发表这篇文章是因为我想找到一种方法来减少对象数组(例如,当您从csv中读取时创建的数据结构),并通过给定索引聚合以生成相同的数据结构。我正在寻找的返回值是另一个对象数组,而不是我在这里看到的嵌套对象或映射。

下面的函数获取一个数据集(对象数组)、一个索引列表(数组)和一个reducer函数,并将reducer功能应用于索引的结果作为一个对象数组返回。

function agg(data, indices, reducer) {

  // helper to create unique index as an array
  function getUniqueIndexHash(row, indices) {
    return indices.reduce((acc, curr) => acc + row[curr], "");
  }

  // reduce data to single object, whose values will be each of the new rows
  // structure is an object whose values are arrays
  // [{}] -> {{}}
  // no operation performed, simply grouping
  let groupedObj = data.reduce((acc, curr) => {
    let currIndex = getUniqueIndexHash(curr, indices);

    // if key does not exist, create array with current row
    if (!Object.keys(acc).includes(currIndex)) {
      acc = {...acc, [currIndex]: [curr]}
    // otherwise, extend the array at currIndex
    } else {
      acc = {...acc, [currIndex]: acc[currIndex].concat(curr)};
    }

    return acc;
  }, {})

  // reduce the array into a single object by applying the reducer
  let reduced = Object.values(groupedObj).map(arr => {
    // for each sub-array, reduce into single object using the reducer function
    let reduceValues = arr.reduce(reducer, {});

    // reducer returns simply the aggregates - add in the indices here
    // each of the objects in "arr" has the same indices, so we take the first
    let indexObj = indices.reduce((acc, curr) => {
      acc = {...acc, [curr]: arr[0][curr]};
      return acc;
    }, {});

    reduceValues = {...indexObj, ...reduceValues};


    return reduceValues;
  });


  return reduced;
}

我将创建一个返回count(*)和sum(Value)的reducer:

reducer = (acc, curr) => {
  acc.count = 1 + (acc.count || 0);
  acc.value = +curr.Value + (acc.value|| 0);
  return acc;
}

最后,使用我们的reducer将agg函数应用于原始数据集会生成一个应用了适当聚合的对象数组:

agg(tasks, ["Phase"], reducer);
// yields:
Array(2) [
  0: Object {Phase: "Phase 1", count: 4, value: 50}
  1: Object {Phase: "Phase 2", count: 4, value: 130}
]

agg(tasks, ["Phase", "Step"], reducer);
// yields:
Array(4) [
  0: Object {Phase: "Phase 1", Step: "Step 1", count: 2, value: 15}
  1: Object {Phase: "Phase 1", Step: "Step 2", count: 2, value: 35}
  2: Object {Phase: "Phase 2", Step: "Step 1", count: 2, value: 55}
  3: Object {Phase: "Phase 2", Step: "Step 2", count: 2, value: 75}
]

其他回答

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);

您可以使用本机JavaScript组数组方法(目前处于第3阶段)。

我认为,与reduce相比,或者与lodash等第三方库相比,解决方案要优雅得多。

常量产品=[{名称:“牛奶”,类型:“乳制品”},{名称:“cheese”,类型:“乳制品”},{名称:“牛肉”,类型:“肉”},{名称:“chicken”,类型:“肉”}];const productsByType=products.group((product)=>product.type);console.log(“按类型分组的产品:”,productsByType);<script src=“https://cdn.jsdelivr.net/npm/core-js-bundle@3.23.2/minified.min.js“></script>

/***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>

使用ES6的简单解决方案:

该方法有一个返回模型,可以比较n个财产。

const compareKey = (item, key, compareItem) => {
    return item[key] === compareItem[key]
}

const handleCountingRelatedItems = (listItems, modelCallback, compareKeyCallback) => {
    return listItems.reduce((previousValue, currentValue) => {
        if (Array.isArray(previousValue)) {
        const foundIndex = previousValue.findIndex(item => compareKeyCallback(item, currentValue))

        if (foundIndex > -1) {
            const count = previousValue[foundIndex].count + 1

            previousValue[foundIndex] = modelCallback(currentValue, count)

            return previousValue
        }

        return [...previousValue, modelCallback(currentValue, 1)]
        }

        if (compareKeyCallback(previousValue, currentValue)) {
        return [modelCallback(currentValue, 2)]
        }

        return [modelCallback(previousValue, 1), modelCallback(currentValue, 1)]
    })
}

const itemList = [
    { type: 'production', human_readable: 'Production' },
    { type: 'test', human_readable: 'Testing' },
    { type: 'production', human_readable: 'Production' }
]

const model = (currentParam, count) => ({
    label: currentParam.human_readable,
    type: currentParam.type,
    count
})

const compareParameter = (item, compareValue) => {
    const isTypeEqual = compareKey(item, 'type', compareValue)
    return isTypeEqual
}

const result = handleCountingRelatedItems(itemList, model, compareParameter)

 console.log('Result: \n', result)
/** Result: 
    [
        { label: 'Production', type: 'production', count: 2 },
        { label: 'Testing', type: 'testing', count: 1 }
    ]
*/
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;
}, {})