重新排列JavaScript数组

如何解决重新排列JavaScript数组

如何重新排列上述格式的数据。我必须重新格式化数组并制作新的数组：

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

收件人：

[
   {
      "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"
   }
]

我尝试过reduce函数和map函数都没有用。

解决方法

这是获得所需输出的简单方法。 只需创建start_date的字典，然后将account_i_cost添加到带有接收日期的字典即可。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",{"cost":"1.401164025099997","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)

,

您可以分两步进行操作，首先是使用N按start_date进行分组，然后将其映射回数组：

reduce

,

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

reduce() 方法在数组的每个元素上执行 reducer 函数（由您提供），从而产生单个输出值。

---

查看评论：

var account_data = [
   {
      "cost":"1.3947648891000006","account":"1"
   },{
      "cost":"1.4069091170999999",{
      "cost":"1.401164025099997","account":"2"
   },{
      "cost":"2.50928182113","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);

,

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',{ cost: '1.401164025099997',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 ) 

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,

我强烈建议不要使用数组，而应该使用以日期为键的字典。

然后输入此代码：

// 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;
},{})

将为您提供此结果：

{
  "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"
  }
}

但是，如果您真的想要它，请尝试以下方法：

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;
},[]);

,

在数组上遍历foreach，并查看哈希图中是否有该对象日期的条目。如果是这样，则只需将新属性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resultArray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",{ "cost": "1.401164025099997","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);