javascript - 将现有阵列转换为新阵列

我有一个要变成适合我工作的新数组。这是我原来拥有的Array:

注意:在我的代码中，“jayanagar”，“mallashwaram”，“kolar”被称为城市。

var array1=[
  ["Year", "2018-8", "2017-8"],
  ["JAYANAGAR", "2018-8", 10910665],
  ["MALLESHWARAM", "2018-8", 2018451],
  ["KOLAR", "2018-8", 2277562],
  ["JAYANAGAR", "2017-8", 1134]
]

然后我将其转换为:

var array2=[
  ["Year", "Jul 2018", "Jul 2017"],
  ["JAYANAGAR", 10910665, 1134],
  ["MALLESHWARAM", 2018451],
  ["KOLAR", 2277562]
]

我想做的是

正如您在arra1中看到的那样，我有两年，即2018-8和2017-8，对于日期2018-8我拥有所有城市的数据，但对于日期2017-08来说只有一个城市数据，即jayanagar

所以我想做的是每当前一个日期的城市数较少时，我想显示为该城市的0

例如，在array2中，我没有malleshwaram和kolar的数据，因此我想在

中显示为0

所选的年份用户会给我所有城市的名称，因此据此我想在此处循环显示其他日期数据，即array2

最高日期总是将具有所有城市的名称

我做了什么

var input = [
  ["Year", "2018-8", "2017-8"],
  ["JAYANAGAR", "2018-8", 10910665],
  ["MALLESHWARAM", "2018-8", 2018451],
  ["KOLAR", "2018-8", 2277562],
  ["JAYANAGAR", "2017-8", 1134]
]
var e = input[0]
var a = new Date(e[1]);

a.setMonth(a.getMonth());
a = a.toUTCString();

var c = a.split(' ');
e[1] = c[2] + " " + c[3];
a = new Date(e[2]);

a.setMonth(a.getMonth());
a = a.toUTCString();
c = a.split(' ');
e[2] = c[2] + " " + c[3];

const merged = input.reduce((acc, arr) => {
  const [city, year, value] = arr;

  if (city === "Year")
    acc[city] = arr
  else {
    acc[city] = acc[city] || [city]
    acc[city].push(value)
  }

  return acc;
}, {})

const output = Object.values(merged)
console.log(output)

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

我的预期输出

[
  [
    "Year",
    "Aug 2018",
    "Aug 2017"
  ],
  [
    "JAYANAGAR",
    10910665,
    1134
  ],
  [
    "MALLESHWARAM",
    2018451,
    0
  ],
  [
    "KOLAR",
    2277562,
    0
  ]
]

最佳答案

此解决方案创建对象以按相同的键(其中键为城市)存储所有数据，然后创建另一个在所需输出中格式化数组的数组。

    var input = [
      ["Year", "2018-8", "2017-8"],
      ["JAYANAGAR", "2018-8", 10910665],
      ["MALLESHWARAM", "2018-8", 2018451],
      ["KOLAR", "2018-8", 2277562],
      ["JAYANAGAR", "2017-8", 1134]
    ];



  var cities = {};
  var years = input[0];


function formatDate(date, index){

if(index === 0){ //skip formatting, because it is not a date
  return date;
}
  dates = {
  '1' : 'Jan',
  '2' : 'Feb',
  '3' : 'Mar',
  '4' : 'Apr',
  '5' : 'May',
  '6' : 'Jun',
  '7' : 'Jul',
  '8' : 'Aug',
  '9' : 'Sep',
  '10': 'Oct',
  '11': 'Nov',
  '12': 'Dec'
  };

  let dateArr = date.split('-');

  return dates[dateArr[1]] + ' ' + dateArr[0];

}


years = years.map(formatDate); //run through dates array.



  //store data in object items, where each key is city , and items are arrays of data
  input.forEach(function(element,index) {
    if(index !== 0){
    if( cities[element[0]] === undefined){
       cities[element[0]] = [];
    }

      cities[element[0]].push(element[2]); // add to city item new data

    }
  });


 var citiesData = [];
 for (var city in cities) { //reformat the data in format [city, ...data]
    if (cities.hasOwnProperty(city)) {
      //  citiesData.push
     cities[city].unshift(city); //add city as first item in array
     if(years.length > cities[city].length){ //assuming that data should be available for each year, check the length and fill with zeroes
     let fillsLength =  years.length -  cities[city].length;
     let fills = Array(fillsLength).fill(0, 0, fillsLength); //fill with zeros x times of missing data per each year
     cities[city] = cities[city].concat(fills); //concat array of zeros with city data
     }
     citiesData.push(cities[city]);
    }
}

citiesData.unshift(years);

console.log(citiesData); 

关于日期的注释:与其使用Date和Date方法，不如使用更容易的方法，即使用每个月的缩写表示值的对象，将数字表示月份的键值的对象。

关于javascript - 将现有阵列转换为新阵列，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/55116331/