将嵌套对象转换为对象数组的JS方法

Question

我有这个：

var dataSet = { 
data: {
    department: [
        { 
          "Marketing": [
            { "employee_1": "1210" },
            { "employee_2": "3241" },
            { "employee_3": "6510" },
            { "employee_4": "9510" }

          ]
        },
        { 
          "Sales": [
          { "employee_1": "5210" },
          { "employee_2": "1241" },
          { "employee_3": "8510" },
          { "employee_4": "2510" }
          ]
        },
        { 
          "RD": [
          { "employee_1": "1010" },
          { "employee_2": "9241" },
          { "employee_3": "9430" },
          { "employee_4": "3510" }
          ]
        },
        ...
    ]
}

我需要这个：

const data = [
          {name: 'Marketing', Low: 2, Medium: 4, High: 3},
          {name: 'Sales', Low: 3, Medium: 1, High: 5},
          {name: 'RD', Low: 6, Medium: 1, High: 2},
          ...
    ];

地点：

LOW     employee_x < 2000 
MEDIUM  employee_x >= 2000 && employee_x < 5000
HIGH    employee_x >= 5000

我的上一次尝试是这样的：

    var low_arr_dept = [];
    var medium_arr_dept = [];
    var high_arr_dept = [];

    for (var i in dataSet.data) {
                var department = departments_data[i];

                        for (var d in department) {
                            var score = department[d];
                            for (var j in score) {

                                if (Object.values(department[d][j]) >= 5000) {
                                    console.log("Object.values(department[d][j]) ", Object.values(department[d][j]))
                                    low_arr_dept.push({

                                        score: Object.values(department[d][j])
                                    });
                                } else if (Object.values(department[d][j]) >= 2000 && Object.values(department[d][j]) < 5000) {
                                    medium_arr_dept.push({

                                        score: Object.values(department[d][j])
                                    });
                                } else if (Object.values(department[d][j])  <2000) {
                                    high_arr_dept.push({

                                        score: Object.values(department[d][j])
                                    });
                                }


                            } 
                        }

            }



            // Creating arrays and then getting their length
            low_arr_dept_marketing.length.toString()
            medium_arr_dept_marketing.length.toString()
            high_arr_dept_marketing.length.toString()

// For creating this:
const data = [
          {name: 'Marketing', Low: low_arr_dept_marketing.length, Medium: medium_arr_dept_marketing.length, High: high_arr_dept_marketing.length},
          {name: 'Sales', Low: low_arr_dept_sales.length, Medium: medium_arr_dept_sales.length, High: high_arr_dept_sales.length},
          {name: 'RD'Low: low_arr_dept_rd.length, Medium: medium_arr_dept_rd.length, High: high_arr_dept_rd.length},
          ...
    ];

- Dubliner

---

你当前的实现有什么问题？ - Prerak Sola

我建议使用 reduce。 - evolutionxbox

我不知道如何遍历数据集并为每个部门获取分离的数组（高、中、低）。在我的一次尝试中，我只得到了聚合的数组（所有部门在一起）。 - Dubliner

Answer 1

也许这不是最佳方式，虽然代码非常简短，但我根据你所说的需求得到了你需要的内容。请查看。

var data = { 
data: {
    department: [
        { 
          "Marketing": [
            { "employee_1": "1210" },
            { "employee_2": "241" },
            { "employee_3": "6510" },
            { "employee_4": "9510" }

          ]
        },
        { 
          "Sales": [
          { "employee_1": "5210" },
          { "employee_2": "1241" },
          { "employee_3": "8510" },
          { "employee_4": "2510" }
          ]
        },
        { 
          "RD": [
          { "employee_1": "1010" },
          { "employee_2": "9241" },
          { "employee_3": "9430" },
          { "employee_4": "3510" }
          ]
        }
    ]
}
};

var new_data = data.data.department.map(function(el, ind) {
    var item = {low: 0, medium: 0, high: 0};
    // set the name of department for new array
    for(var prop in el) item.name = prop; 
    // go and check ids increasing counters
    for(var i = 0; i < el[prop].length; i++) for(var prop2 in el[prop][i]) if(el[prop][i][prop2] < 2000) item.low++; else if(el[prop][i][prop2] >= 5000) item.high++; else item.medium++;  return item;
});

console.log(new_data);

- curveball

---

谢谢@curveball。 - Dubliner

Answer 2

你可以使用数组来表示组，使用哈希表来表示部门。

var dataSet = { data: { department: [{ Marketing: [{ employee_1: "1210" }, { employee_2: "3241" }, { employee_3: "6510" }, { employee_4: "9510" }] }, { Sales: [{ employee_1: "5210" }, { employee_2: "1241" }, { employee_3: "8510" }, { employee_4: "2510" }] }, { RD: [{ employee_1: "1010" }, { employee_2: "9241" }, { employee_3: "9430" }, { employee_4: "3510" }] }] } },
    groups = [{ key: 'LOW', value: 2000 }, { key: 'MEDIUM', value: 5000 }, { key: 'HIGH', value: Infinity }],            
    hash = {},
    result = [];

dataSet.data.department.forEach(function (o) {
    var dept = Object.keys(o)[0];
    o[dept].forEach(function (p) {
        var value = p[Object.keys(p)[0]],
            group = groups.find(g => +value < g.value).key;

        if (!hash[dept]) {
            hash[dept] = { name: dept };
            groups.forEach(g => hash[dept][g.key] = 0);
            result.push(hash[dept]);
        }
        hash[dept][group]++;
    });
});
    
console.log(result);

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

- Nina Scholz

Answer 3

你好，这是我为你准备的可行解决方案。

var dataSet = { 
    data: {
        department: [
            { 
              "Marketing": [
                { "employee_1": "1210" },
                { "employee_2": "3241" },
                { "employee_3": "6510" },
                { "employee_4": "9510" }
    
              ]
            },
            { 
              "Sales": [
              { "employee_1": "5210" },
              { "employee_2": "1241" },
              { "employee_3": "8510" },
              { "employee_4": "2510" }
              ]
            },
            { 
              "RD": [
              { "employee_1": "1010" },
              { "employee_2": "9241" },
              { "employee_3": "9430" },
              { "employee_4": "3510" }
              ]
            }
     ]
    }
    }
    
    var employeenumber = 1;
    var Low = 0;
    var Medium = 0;
    var High = 0;
    var data = [];
    var i = 0;
    for (var departments in dataSet.data) {
     //console.log(JSON.stringify(dataSet.data[departments])); 
     for (var arrays in dataSet.data[departments]) {
      //console.log(dataSet.data[departments][arrays]);
      var department = dataSet.data[departments][arrays];
      for (var employees in department) {
       //console.log(JSON.stringify(department[employees]));
       var employeesArray = department[employees];
       for (employee in employeesArray) {
    
        var employeeScoreObject = employeesArray[employee];
        var employee_score = employeeScoreObject["employee_"+employeenumber];
        //console.log(employee_score);
        if (employee_score < 2000) {
         Low++;
        }
        if (employee_score >= 2000 && employee_score < 5000) {
         Medium++;
        }
        if (employee_score > 5000) {
         High++;
        }
        employeenumber++;
        if (employeenumber>4) {
        employeenumber = 1;
        }
       }
       var departmentName = Object.keys(department)[0];
    
       data[i] = {'name': departmentName, 'Low': Low, 'Medium': Medium, 'High': High};
       i++;
    
       Low = 0;
       Medium = 0;
       High = 0;
      } // employees
      
     } // department 1= marketing 2 =....
     
    } // departments
    console.log(JSON.stringify(data));

- Henrikki

---

谢谢@Henrikki！ - Dubliner