Question

循环假设我将数组存储在对象Map中，并向我展示户外最有利可图的部门。

我想运行循环遍历数组，我只是在地图中再次访问该数据并且获得部门最高利润（销售额）的最高利润值。

E.g将所有户外销售，硬件和木工结合起来给你提供总和和console.log的＆＃34;户外＆＃34;这是最高的。

＆＃13;

    var salesData = [{
      department: 'hardware',
      sales: 4500,
      day: 'Monday'
    }, {
      department: 'outdoor',
      sales: 1500,
      day: 'Monday'
    }, {
      department: 'carpentry',
      sales: 5500,
      day: 'Monday'
    }, {
      department: 'hardware',
      sales: 7500,
      day: 'Tuesday'
    }, {
      department: 'outdoor',
      sales: 2505,
      day: 'Tuesday'
    }, {
      department: 'carpentry',
      sales: 1540,
      day: 'Tuesday'
    }, {
      department: 'hardware',
      sales: 1500,
      day: 'Wednesday'
    }, {
      department: 'outdoor',
      sales: 8507,
      day: 'Wednesday'
    }, {
      department: 'carpentry',
      sales: 8009,
      day: 'Wednesday'
    }, {
      department: 'hardware',
      sales: 12000,
      day: 'Thursday'
    }, {
      department: 'outdoor',
      sales: 18007,
      day: 'Thursday'
    }, {
      department: 'carpentry',
      sales: 6109,
      day: 'Thursday'
    }, {
      department: 'hardware',
      sales: 7005,
      day: 'Friday'
    }, {
      department: 'outdoor',
      sales: 12006,
      day: 'Friday'
    }, {
      department: 'carpentry',
      sales: 16109,
      day: 'Friday'
    }];
    var departmentMap = {};
    var high = "";
    
    function mostProfitableDepartment(salesList) {
      for (var i = 0; i < salesList.length; i++) {
    
        var currentDep = salesList[i];
    
        if (departmentMap[currentDep] === undefined) {
          departmentMap[currentDep] = 0;
        }
        departmentMap[currentDep] = departmentMap[currentDep] + 1
      }
      console.log(departmentMap);
    }
    
    
    mostProfitableDepartment(salesData);
    
    for (var depHigh in departmentMap) {
      if (departmentMap[depHigh] > high) {
        high = departmentMap[depHigh]
      }
    }
    
    console.log(high);

＆＃13;

Answer 1

＆＃13;

var salesData = [
  {department : 'hardware', sales : 4500, day : 'Monday'},
  {department : 'outdoor', sales : 1500, day : 'Monday'},
  {department : 'carpentry', sales : 5500, day : 'Monday'},
  {department : 'hardware', sales : 7500, day : 'Tuesday'},
  {department : 'outdoor', sales : 2505, day : 'Tuesday'},
  {department : 'carpentry', sales : 1540, day : 'Tuesday'},
  {department : 'hardware', sales : 1500, day : 'Wednesday'},
  {department : 'outdoor', sales : 8507, day : 'Wednesday'},
  {department : 'carpentry', sales : 8009, day : 'Wednesday'},
  {department : 'hardware', sales : 12000, day : 'Thursday'},
  {department : 'outdoor', sales : 18007, day : 'Thursday'},
  {department : 'carpentry', sales : 6109, day : 'Thursday'},
  {department : 'hardware', sales : 7005, day : 'Friday'},
  {department : 'outdoor', sales : 12006, day : 'Friday'},
  {department : 'carpentry', sales : 16109, day : 'Friday'},
];

// Build a obj map and sum up the sales of each department
var sales = salesData.reduce((obj, a) => {
  obj[a.department] = (obj[a.department] || 0) + a.sales
  return obj
}, {})

// Figure out which department is the highest
var high = Object.keys(sales).reduce((a, b) => sales[a] > sales[b] ? a : b )

console.log(sales)
console.log(high, sales[high])

＆＃13;

Answer 2

您可以使用以下.reduce来电的组合。

＆＃13;

var salesData = [{
      department: 'hardware',
      sales: 4500,
      day: 'Monday'
    }, {
      department: 'outdoor',
      sales: 1500,
      day: 'Monday'
    }, {
      department: 'carpentry',
      sales: 5500,
      day: 'Monday'
    }, {
      department: 'hardware',
      sales: 7500,
      day: 'Tuesday'
    }, {
      department: 'outdoor',
      sales: 2505,
      day: 'Tuesday'
    }, {
      department: 'carpentry',
      sales: 1540,
      day: 'Tuesday'
    }, {
      department: 'hardware',
      sales: 1500,
      day: 'Wednesday'
    }, {
      department: 'outdoor',
      sales: 8507,
      day: 'Wednesday'
    }, {
      department: 'carpentry',
      sales: 8009,
      day: 'Wednesday'
    }, {
      department: 'hardware',
      sales: 12000,
      day: 'Thursday'
    }, {
      department: 'outdoor',
      sales: 18007,
      day: 'Thursday'
    }, {
      department: 'carpentry',
      sales: 6109,
      day: 'Thursday'
    }, {
      department: 'hardware',
      sales: 7005,
      day: 'Friday'
    }, {
      department: 'outdoor',
      sales: 12006,
      day: 'Friday'
    }, {
      department: 'carpentry',
      sales: 16109,
      day: 'Friday'
    }];

var aggregated = salesData.reduce(function(p,c){
   if(p[c.department] == undefined)
     p[c.department] = 0;
   p[c.department]+= c.sales;
   return p;
 },{});

console.log("total sales",aggregated);

var highest = Object.keys(aggregated).reduce(function(p,c){
  if(aggregated[c]>p.value)
    return {key:c,value:aggregated[c]}
  return p;
},{key:'notset',value:0});

console.log("best department", highest);

＆＃13;

无法显示室外变量

2 个答案: