Question

好的，我一直试图创建正确的算法来解决这个问题一周，我似乎没有采用正确的方法。我有一个对象数组，我需要通过多个属性进行分组，但也有能力在x prop处打破。例如：

var arr = [
 {car: 'ford', model: 'mustang', color: 'black', price: 400},
 {car: 'ford', model: 'focus', color: 'yellow', price: 300},
 {car: 'toyota', model: 'corolla', color: 'blue', price: 100},
 {car: 'toyota', model: 'camry', color: 'green', price: 200},
 {car: 'toyota', model: 'camry', color: 'black', price: 250},
 {car: 'toyota', model: 'camry', color: 'black', price: 350},
]

然后groupBy(arr, ['car', 'model', 'color'], {break: 'model'})

这会打印类似

的内容

[
 {
  "ford - mustang": [
    "black": [
      {car: 'ford', model: 'mustang', color: 'black', price: 400}
     ]
   ]
 },
 {
  "ford - focus": [
    "yellow": [
     {car: 'ford', model: 'focus', color: 'yellow', price: 300}
    ]
  ]
 },
 {
  "toyota - corolla": [
    "blue": [
      {car: 'toyota', model: 'corolla', color: 'blue', price: 100},
    ]
  ]
 },
 {
  "toyota - camry": [
    "green": [
     {car: 'toyota', model: 'camry', color: 'green', price: 200}
    ],
    "black": [
      {car: 'toyota', model: 'camry', color: 'black', price: 250},
      {car: 'toyota', model: 'camry', color: 'black', price: 350}
    ]
  ]
 }
]

编辑1：这是我创建的分组功能。这个功能考虑了三个道具＆＃34;标准1，标准2和标准3＆＃34;，这个功能完成了分组工作但是我还没有弄清楚如何做破坏部分

const criteriaMapping = {
 Lenders: "lenderStr",
 Branches: "branchStr",
 Officers: "officerStr",
 Programs: "programStr",
 Stages: "stageStr"
}
private createNewArray(data){
let keys1 = []
let keys2 = []
let keys3 = []
let result = []
let orderBy = 'reservationNo'
let orderByDirection1 = ''
let orderByDirection2 = ''
let orderByDirection3 = ''

if(this.query.criteria1.description){
  keys1 = Object.keys(_.groupBy(data, criteriaMapping[this.query.criteria1.description]))
  orderByDirection1 = this.query.criteria1.desc ? 'desc' : 'asc'
}
if(this.query.criteria1.description && this.query.criteria2.description){
  keys2 = Object.keys(_.groupBy(data, criteriaMapping[this.query.criteria2.description]))
  orderByDirection2 = this.query.criteria2.desc ? 'desc' : 'asc'
}
if(this.query.criteria2.description && this.query.criteria3.description){
  keys3 = Object.keys(_.groupBy(data, criteriaMapping[this.query.criteria3.description]))
  orderByDirection3 = this.query.criteria3.desc ? 'desc' : 'asc'
}

let i = 0
do{
  let j = 0
  do{
    let k = 0
    do{
      let orderProp = criteriaMapping[this.query.criteria1.description]
      let order = 'asc'
      if(keys3[k])
        order = orderByDirection3
      else if(keys2[j])
        order = orderByDirection2
      else if(keys1[i])
        order = orderByDirection1
      let concat = _.orderBy(data.filter(x => {
        return (keys1[i] ? x[criteriaMapping[this.query.criteria1.description]] === keys1[i] : true)
          && (keys2[j] ? x[criteriaMapping[this.query.criteria2.description]] === keys2[j] : true)
          && (keys3[k] ? x[criteriaMapping[this.query.criteria3.description]] === keys3[k] : true)
      }), ['reservationNo'], [order])
      if(concat && concat.length){
        if(keys1[i]) result.push({"header": true, 0: keys1[i], 1: keys2[j], 2: keys3[k]})
        result = result.concat(concat)
      }
      k++
    }while(keys3.length > k)
    j++
  }while (keys2.length > j)
  i++
}while (keys1.length > i)

return result

}

所以我不认为我让自己清楚了（我的错）。休息部分很难解释我会尝试更好地解释它。所以如果我要打电话，我先前写过的数组 groupBy(arr, ['car', 'model', 'color'], {break: 'model'}) 它会打印

{
 "ford - mustang": {
  "black": Array(1)
 },
 "ford - focus": {
   "yellow": Array(1)
 },
 "toyota - corolla": {
  "blue": Array(1)
 },
 "toyota - camry": {
  "green": Array(1),
  "black": Array(2)
 }
}

现在，如果我打电话

groupBy(arr, ['car', 'model', 'color'], {break: 'car'})

它会打印

{
 "ford": {
  "mustang - black": Array(1),
  "focus - yellow": Array(1)
 },
 "toyota": {
  "corolla - blue": Array(1),
  "camry - green": Array(1),
  "camry - black": Array(2)
 }
}

Answer 1

您可以使用reduce()方法。

var arr = [
 {car: 'ford', model: 'mustang', color: 'black', price: 400},
 {car: 'ford', model: 'focus', color: 'yellow', price: 300},
 {car: 'toyota', model: 'corolla', color: 'blue', price: 100},
 {car: 'toyota', model: 'camry', color: 'green', price: 200},
 {car: 'toyota', model: 'camry', color: 'black', price: 250},
 {car: 'toyota', model: 'camry', color: 'black', price: 350},
]

var result = arr.reduce(function(r, e) {
  let cm = `${e.car} - ${e.model}`;
  if (!r[cm]) r[cm] = {}
  if (!r[cm][e.color]) r[cm][e.color] = [];
  r[cm][e.color].push(e);
  return r;
}, {})

console.log(result)

您还可以使用reduce()创建自定义功能以获得更加动态的解决方案。

var arr = [
 {car: 'ford', model: 'mustang', color: 'black', price: 400},
 {car: 'ford', model: 'focus', color: 'yellow', price: 300},
 {car: 'toyota', model: 'corolla', color: 'blue', price: 100},
 {car: 'toyota', model: 'camry', color: 'green', price: 200},
 {car: 'toyota', model: 'camry', color: 'black', price: 250},
 {car: 'toyota', model: 'camry', color: 'black', price: 350},
]

function groupBy(data, group, {divide} = {}) {
  return arr.reduce(function(r, e) {
    let d = e[divide], cm = group
    .filter(a => a != divide)
    .map(a => e[a]).join(' - ')
    
    if(!r[cm]) r[cm] = divide ? {} : []
    if(!r[cm][d] && divide) r[cm][d] = [];
    divide ? r[cm][d].push(e) : r[cm].push(e)
    return r;
  }, {})
}

let result = groupBy(arr, ['car', 'model', 'color'], {divide: 'model'})
let result2 = groupBy(arr, ['car', 'model', 'color'], {divide: 'car'})
let result3 = groupBy(arr, ['car', 'color'])


console.log(result)
console.log(result2)
console.log(result3)

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 2

这是你在找什么？

function groupBy(arr, keys, breakVal){
    return arr.reduce( (grouped, next) => {
        var idx = keys.reduce( (idx, nextkey) => {
            idx.push(next[nextkey])
            return idx;
        }, [] ).join(' - ')
        if(breakVal){
            if(!grouped[idx]) {
                grouped[idx] = {}
            }
            if(!grouped[idx][next[breakVal]]) {
                grouped[idx][next[breakVal]] = [];
            }
            grouped[idx][next[breakVal]].push(next);
        }else{
            if(!grouped[idx]) {
                grouped[idx] = [];
            }
            grouped[idx].push(next);
        }
        return grouped;
    }, {})
}

console.log(groupBy(arr,  ['car', 'color'], 'model'))

Answer 3

感谢@Nenad和@asosnovsky，我能够提出一个满足我需求的功能

function groupBy(arr, props, dividers){
return arr.reduce((r, e) => {
    var prop = []
    var divide = []
    for(var i = 0; i < props.length; i++){
        prop.push(e[props[i]])
    }
    for(var i = 0; i < dividers.length; i++){
        divide.push(e[dividers[i]])
    }
    prop = prop.join(' - ')
    divide = divide.join(' - ')
    if(!r[prop]) r[prop] = divide ? {} : []
    if(divide){
        if(!r[prop][divide]) r[prop][divide] = []
        r[prop][divide].push(e)
    }else
        r[prop].push(e)
    return r
}, {})

}

非常感谢你。

数组按属性进入数组数组

3 个答案: