Javascript 重新格式化或将对象数组映射到新对象的新数组？就像重新格式化它一样

Question

我对数组和对象很陌生，所以我一整天都在做这个任务，我有一个 JSON 文件或数组 lang (JavaScript)，我也在 StackOverflow 上搜索，但不能很好地理解它们：[  { "date": "1959:01", this"1": 138.89, "2": 139.39, "3": 139.74, "4": 139.69, "5": 140.68, "6": 141.17 },  { "date": "1959:07", "1": 141.7, "2": 141.9, "3": 141.01, "4": 140.47, "5": 140.38, "6": 139.95 },  { "date": "1960:01", "1": 139.98, "2": 139.87, "3": 139.75, "4": 139.56, "5": 139.61, "6": 139.58 },  { "date": "1960:07", "1": 140.18, "2": 141.31, "3": 141.18, "4": 140.92, "5": 140.86, "6": 140.69 },  { "date": "1961:01", "1": 141.06, "2": 141.6, "3": 141.87, "4": 142.13, "5": 142.66, "6": 142.88 },  { "date": "1961:07", "1": 142.92, "2": 143.49, "3": 143.78, "4": 144.14, "5": 144.76, "6": 145.2 },  { "date": "1962:01", "1": 145.24, "2": 145.66, "3": 145.96, "4": 146.4, "5": 146.84, "6": 146.58 },  { "date": "1962:07", "1": 146.46, "2": 146.57, "3": 146.3, "4": 146.71, "5": 147.29, "6": 147.82 },  { "date": "1963:01", "1": 148.26, "2": 148.9, "3": 149.17, "4": 149.7, "5": 150.39, "6": 150.43 },  { "date": "1963:07", "1": 151.34, "2": 151.78, "3": 151.98, "4": 152.55, "5": 153.65, "6": 153.29 },  { "date": "1964:01", "1": 153.74, "2": 154.31, "3": 154.48, "4": 154.77, "5": 155.33, "6": 155.62 },  { "date": "1964:07", "1": 156.8, "2": 157.82, "3": 158.75, "4": 159.24, "5": 159.96, "6": 160.3 },  { "date": "1965:01", "1": 160.71, "2": 160.94, "3": 161.47, "4": 162.03, "5": 161.7, "6": 162.19 },  { "date": "1965:07", "1": 163.05, "2": 163.68, "3": 164.85, "4": 165.97, "5": 166.71, "6": 167.85 },  { "date": "1966:01", "1": 169.08, "2": 169.62, "3": 170.51, "4": 171.81, "5": 171.33, "6": 171.57 },  { "date": "1966:07", "1": 170.31, "2": 170.81, "3": 171.97, "4": 171.16, "5": 171.38, "6": 172.03 }]这些就像公司在指定年数内的收入，例如 1959:01 年初，从 1 到 6 是该年的前 6 个月，同年 1956:07 但“:07”中的差异是值同年最后 6 个月（1 到 6）的数据，所以我想每个月输出如下，其中包含它的值和年份，请使用最容易理解的代码。  [  {"year": 1996,"month": 1,"value": 88512.12  },  {"year": 1996,"month": 2,"value": 71212.12  }, 

Answer 1

如果我正确理解你的问题，那么你可以这样做：

const financialResults = [

  { "date": "1959:01", "1": 138.89, "2": 139.39, "3": 139.74, "4": 139.69, "5": 140.68, "6": 141.17 },

  { "date": "1959:07", "1": 141.7, "2": 141.9, "3": 141.01, "4": 140.47, "5": 140.38, "6": 139.95 },

  { "date": "1960:01", "1": 139.98, "2": 139.87, "3": 139.75, "4": 139.56, "5": 139.61, "6": 139.58 },

  { "date": "1960:07", "1": 140.18, "2": 141.31, "3": 141.18, "4": 140.92, "5": 140.86, "6": 140.69 },

  { "date": "1961:01", "1": 141.06, "2": 141.6, "3": 141.87, "4": 142.13, "5": 142.66, "6": 142.88 },

  { "date": "1961:07", "1": 142.92, "2": 143.49, "3": 143.78, "4": 144.14, "5": 144.76, "6": 145.2 },

  { "date": "1962:01", "1": 145.24, "2": 145.66, "3": 145.96, "4": 146.4, "5": 146.84, "6": 146.58 },

  { "date": "1962:07", "1": 146.46, "2": 146.57, "3": 146.3, "4": 146.71, "5": 147.29, "6": 147.82 },

  { "date": "1963:01", "1": 148.26, "2": 148.9, "3": 149.17, "4": 149.7, "5": 150.39, "6": 150.43 },

  { "date": "1963:07", "1": 151.34, "2": 151.78, "3": 151.98, "4": 152.55, "5": 153.65, "6": 153.29 },

  { "date": "1964:01", "1": 153.74, "2": 154.31, "3": 154.48, "4": 154.77, "5": 155.33, "6": 155.62 },

  { "date": "1964:07", "1": 156.8, "2": 157.82, "3": 158.75, "4": 159.24, "5": 159.96, "6": 160.3 },

  { "date": "1965:01", "1": 160.71, "2": 160.94, "3": 161.47, "4": 162.03, "5": 161.7, "6": 162.19 },

  { "date": "1965:07", "1": 163.05, "2": 163.68, "3": 164.85, "4": 165.97, "5": 166.71, "6": 167.85 },

  { "date": "1966:01", "1": 169.08, "2": 169.62, "3": 170.51, "4": 171.81, "5": 171.33, "6": 171.57 },

  { "date": "1966:07", "1": 170.31, "2": 170.81, "3": 171.97, "4": 171.16, "5": 171.38, "6": 172.03 }

]

// Extract semester and year: assign 2 if entry corresponds to second semester or 1 if it corresponds to the first

// semester in the year

const extractYearAndSemester = date => ({ year: date.split(":")[0], semester: date.split(":")[1] > 6 ? 2 : 1 })

// For each entry, return an array of objects with the desired properties

// Finally, flatten the array, so you don't end up with an array of arrays

const formatResults = arr => arr.

  flatMap(entry => {

  const { year, semester } = extractYearAndSemester(entry.date)

  let semesterResults = []

  for (let key in entry) {

    if (parseInt(key)) {

      semesterResults.push({

        year,

        month: semester === 1 ? parseInt(key) : parseInt(key) + 6,

        value: entry[key]

      })

    }

  }

  return semesterResults

  })

console.log(formatResults(financialResults))

Answer 2

尝试循环遍历数据并解析每个记录的属性，如下所示。在外循环结束时，您将在 MonthData 变量中获得具有所需格式的数据；

var data = [

      { "date": "1959:01", "1": 138.89, "2": 139.39, "3": 139.74, "4": 139.69, "5": 140.68, "6": 141.17 },

      { "date": "1959:07", "1": 141.7, "2": 141.9, "3": 141.01, "4": 140.47, "5": 140.38, "6": 139.95 },

    ];

     var monthData = [];

    

      for (let index = 0; index < data.length; index++) {

        var item =data[index];

        var dateArray = item.date.split(":");

        var year = dateArray[0];

        var baseMonth = Number(dateArray[1]);

        for (var key in item) {

          if (item.hasOwnProperty(key)) {

              var val = item[key];

              if (String(val).indexOf(":")== -1) {

    

              monthData.push({

                year:year ,

                month:  baseMonth-1 + Number(key),

                value: val,

              });

            }

          }

        }

        

      }

Answer 3

阅读有关此软件包的信息，它将让您的生活更轻松：use of _map - Lodash