Question

我一直试图通过将相同的数据组合在一起来最小化我的代码多维数组。

这是我的数组：

data:[
    0:{
        id:"1"
        date:"2017-11-27"
        type:"Meeting"
        log_in:"06:30:00"
        log_out:"11:50:00"
    }
    1:{
        id:"1"
        date:"2017-11-27"
        type:"Lunch"
        log_in:"12:00:00"
        log_out:"01:00:00"
    }
    2:{
        id:"1"
        date:"2017-11-27"
        type:"Seminar"
        log_in:"1:15:00"
        log_out:"03:05:00"
    }
    3:{
        id:"1"
        date:"2017-11-28"
        type:"Lunch"
        log_in:"12:00:00"
        log_out:"01:00:00"
    }
    4:{
        id:"1"
        date:"2017-12-28"
        type:"Out of the Office"
        log_in:"03:00:00"
        log_out:"05:00:00"
    } ]

我一直试图将我的数组转换为：

new_data:[
    0:{
     id:"1"
     date:"2017-11-27"
     logs:[
        {type:"Meeting",log_in:"06:30:00",log_out:"11:50:00"},
        {type:"Lunch",log_in:"12:00:00",log_out:"01:00:00"},
        {type:"Seminar",log_in:"1:15:00",log_out:"03:05:00"}
     ]
    }
    1:{
     id:"1"
     date:"2017-11-28"
     logs:[
        {type:"Lunch",log_in:"12:00:00",log_out:"01:00:00"},
        {type:"Out of the Office",log_in:"03:00:00",log_out:"05:00:00"}
     ]
    }
]

我试过这个代码是link，但是我无法操纵它好。我只是在该链接中使用了正确的答案。

Answer 1

根据您的数据判断，您希望按日期分组日志。考虑到这一点，您首先要通过map创建一个新数组。然后在每次迭代中，通过filter()获取日期等于当前对象的所有对象。最后，为当前对象log属性创建另一个数组，其中包含您问题中描述的对象。

＆＃13;

const data = [
  {
    id: "1",
    date: "2017-11-27",
    type: "Meeting",
    log_in: "06:30:00",
    log_out: "11:50:00"
  },
  {
    id: "1",
    date: "2017-11-27",
    type: "Lunch",
    log_in: "12:00:00",
    log_out: "01:00:00"
  },
  {
    id: "1",
    date: "2017-11-27",
    type: "Seminar",
    log_in: "1:15:00",
    log_out: "03:05:00",
  },
  {
    id: "1",
    date: "2017-11-28",
    type: "Lunch",
    log_in: "12:00:00",
    log_out: "01:00:00"
  },
  {
    id: "1",
    date: "2017-12-28",
    type: "Out of the Office",
    log_in: "03:00:00",
    log_out: "05:00:00"
  }
];

const newData = data.map(obj => {
  const { id, date } = obj;
  
  const currObj = { id, date };

  currObj['logs'] = data.filter(dataObj => {
    return date === dataObj['date'];
  }).map(obj => {
    const {type, log_in, log_out } = obj;
    
    return { type, log_in, log_out };
  });
  
  return currObj;
});

console.log(newData);

＆＃13;

Answer 2

这是我的解决方案。如果具有id和date的元素已经存在，则使用reduce迭代初始数组的元素并将事件添加到日志中，否则创建新元素。

＆＃13;

const data = [
    {
        id:"1",
        date:"2017-11-27",
        type:"Meeting",
        log_in:"06:30:00",
        log_out:"11:50:00"
    },
    {
        id:"1",
        date:"2017-11-27",
        type:"Lunch",
        log_in:"12:00:00",
        log_out:"01:00:00"
    },
    {
        id:"1",
        date:"2017-11-27",
        type:"Seminar",
        log_in:"1:15:00",
        log_out:"03:05:00"
    },
    {
        id:"1",
        date:"2017-11-28",
        type:"Lunch",
        log_in:"12:00:00",
        log_out:"01:00:00"
    },
    {
        id:"1",
        date:"2017-12-28",
        type:"Out of the Office",
        log_in:"03:00:00",
        log_out:"05:00:00"
    } ];



const result = data.reduce((acc, elt) => {
  const {id, date, type: typeValue, log_in, log_out} = elt;
  // Check if the element exists
  const existingEltIndex = acc.findIndex(e =>
    e.id === elt.id && e.date === elt.date);
  const newEvent = {type: typeValue, log_in, log_out};
  
  if (existingEltIndex < 0) { // new element
    const newElement = {
      id,
      date,
      logs: [newEvent]
    };
    return [...acc, newElement];
  } else { // Existing element
    acc[existingEltIndex].logs.push(newEvent);
    return acc;
  }
}, []);

console.log(result);

＆＃13;

Answer 3

使用for loop的简单解决方案，

user@host ~/Desktop $ ./a.out myFile.txt
[DEBUG] myFile
[DEBUG] 'm' == 'm'
[DEBUG] 'y' == 'y'
[DEBUG] 'F' == 'F'
[DEBUG] 'i' == 'i'
[DEBUG] 'l' == 'l'
[DEBUG] 'e' == 'e'
[DEBUG] equal
[DEBUG] ret_val: 0
[DEBUG] 'myFile'
[ERR] Could not read file myFile.
user@host ~/Desktop $

通过组合相同的元素转换数组

3 个答案: