重新排列 JavaScript 数组

Question

如何以上述格式重新排列数据。我必须重新格式化数组并制作新的：

[
   {
      "cost":"1.3947648891000006",
      "start_date":"2020-06-10",
      "account":"1"
   },
   {
      "cost":"1.4069091170999999",
      "start_date":"2020-06-11",
      "account":"1"
   },
   {
      "cost":"1.401164025099997",
      "start_date":"2020-06-11",
      "account":"2"
   },
   {
      "cost":"2.50928182113",
      "start_date":"2020-06-12",
      "account":"2"
   }
]

到：

[
   {
      "start_date":"2020-06-10",
      "account_1_cost":"1.3947648891000006"
   },
   {
      "start_date":"2020-06-11",
      "account_1_cost":"1.4069091170999999",
      "account_2_cost":"1.401164025099997"
   },
   {
      "start_date":"2020-06-12",
      "account_2_cost":"2.50928182113"
   }
]

reduce函数和map函数我都试过了，没用。

Answer 1

只需使用 array.reduce

const arr = 
  [ { cost: '1.3947648891000006', start_date: '2020-06-10', account: '1' } 
  , { cost: '1.4069091170999999', start_date: '2020-06-11', account: '1' } 
  , { cost: '1.401164025099997',  start_date: '2020-06-11', account: '2' } 
  , { cost: '2.50928182113',      start_date: '2020-06-12', account: '2' } 
  ] 

const res = 
  arr.reduce((acc,{cost, start_date, account})=>
    {
    let el = acc.find(x=>x.start_date===start_date)
    if (!el) acc.push(el = { start_date })
    el[`account_${account}_cost`] = cost
    return acc
    },[])

console.log( res ) 

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

Answer 2

您可以分两步执行此操作，首先使用reduce 按start_date 分组，然后将其映射回数组：

var input = [{"cost":"1.3947648891000006","start_date":"2020-06-10","account":"1"},{"cost":"1.4069091170999999","start_date":"2020-06-11","account":"1"},{"cost":"1.401164025099997","start_date":"2020-06-11","account":"2"},{"cost":"2.50928182113","start_date":"2020-06-12","account":"2"}];

var grouped = input.reduce( (acc,i) => {
   if(acc.hasOwnProperty(i.start_date))
      acc[i.start_date].push(i);
    else
       acc[i.start_date] = [i];
    return acc;
},{});

var result = Object.entries(grouped).map( entry => {
   var [key,value] = entry;
   var obj = {start_date:key};
   for(var i=0;i<value.length;i++)
       obj[`account_${value[i].account}_cost`] = value[i].cost;
    return obj;
});

console.log(result);

Answer 3

这是实现所需输出的简单方法。 只需创建一个start_date 的字典，然后继续将account_i_cost 添加到带有接收日期的字典中。最后，获取字典的所有值。

var arr = [{"cost":"1.3947648891000006","start_date":"2020-06-10","account":"1"},{"cost":"1.4069091170999999","start_date":"2020-06-11","account":"1"},{"cost":"1.401164025099997","start_date":"2020-06-11","account":"2"},{"cost":"2.50928182113","start_date":"2020-06-12","account":"2"}];

let objMap={}
arr.forEach(obj=>{
let field=`account_${obj.account}_cost`
objMap[obj.start_date]=objMap[obj.start_date]?{...objMap[obj.start_date],[field]:obj.cost}:{
      "start_date":obj.start_date,
      [field]:obj.cost
   }
})

let result=Object.values(objMap)
console.log(result)

Answer 4

在这种情况下，正确的工具是Array.prototype.reduce()。

reduce() 方法对数组的每个元素执行（您提供的）reducer 函数，从而产生单个输出值。

---

查看 cmets：

var account_data = [
   {
      "cost":"1.3947648891000006",
      "start_date":"2020-06-10",
      "account":"1"
   },
   {
      "cost":"1.4069091170999999",
      "start_date":"2020-06-11",
      "account":"1"
   },
   {
      "cost":"1.401164025099997",
      "start_date":"2020-06-11",
      "account":"2"
   },
   {
      "cost":"2.50928182113",
      "start_date":"2020-06-12",
      "account":"2"
   }
];

var refactored = account_data.reduce(function(new_object, current_item){

  // Get the essential information from the current item
  let {start_date, cost} = current_item;
  let cost_key = `account_${current_item.account}_cost`;
  
  // If there isn't a key for this date, create it
  if(!new_object[start_date]){
    new_object[start_date] = {start_date};
  }
  
  // Add the relevant info for the current date
  new_object[start_date][cost_key] = cost;
  return new_object;
}, {});

// Flatten it into an array
refactored = Object.values(refactored);

console.log(refactored);

Answer 5

我会强烈建议不要指望一个数组，而是有一个以日期为键的字典。

然后这段代码：

// list = (...)
list.reduce((prev, curr) => {
  if (!prev.hasOwnProperty(curr.start_date)) {
    prev[curr.start_date] = {};
  }
  prev[curr.start_date][`account_${curr.account}_cost`] = curr.cost;
  return prev;
}, {})

会给你这个结果：

{
  "2020-06-10":{
    "account_1_cost":"1.3947648891000006"
  },
  "2020-06-11":{
    "account_1_cost":"1.4069091170999999",
    "account_2_cost":"1.401164025099997"
  },
  "2020-06-12":{
    "account_2_cost":"2.50928182113"
  }
}

但如果你真的想要它，试试这个：

list.reduce((prev, curr) => {
  let elemWithDate = prev.find(el => el['start_date'] === curr['start_date']);
  if (!elemWithDate) {
    prev.push({'start_date': curr['start_date']});
    elemWithDate = prev[prev.length - 1];
  }
  elemWithDate[`account_${curr.account}_cost`] = curr.cost;
  return prev;
}, []);

Answer 6

在数组上使用 foreach 并查看哈希图中是否有该对象日期的条目。如果是这样，则不仅仅是将新属性 account_x_cost 添加到结果数组的相应元素（通过哈希中的索引）。否则，在结果数组中为此对象创建一个新条目，并将日期作为索引添加到哈希中。

let arr = [ 
    { "cost": "1.3947648891000006", "start_date": "2020-06-10", "account": "1" }, 
    { "cost": "1.4069091170999999", "start_date": "2020-06-11", "account": "1" }, 
    { "cost": "1.401164025099997", "start_date": "2020-06-11", "account": "2" }, 
    { "cost": "2.50928182113", "start_date": "2020-06-12", "account": "2" }
];

let result = [];
let hash = [];

arr.forEach(obj => {
let index = hash.indexOf(obj.start_date);
    if (index==-1) {
        result.push({start_date: obj.start_date, ['account' + obj.account + '_cost']: obj.cost});
        hash.push(obj.start_date);
    } else {
        result[index]['account' + obj.account + '_cost'] = obj.cost;
    }
});
console.log(result);