如何通過對象屬性的迭代一個數組，在對象

Question

我有對象的數組的數組，看起來像這樣：

data = [ 
    { 
    title: 'John Doe', 
    departments: [ 
     { name: 'Marketing', slug: 'marketing'}, 
     { name: 'Sales', slug: 'sales'}, 
     { name: 'Administration', slug: 'administration'}, 
    ] 
    }, 
    { 
    title: 'John Doe Junior', 
    departments: [ 
     { name: 'Operations', slug: 'operations'}, 
     { name: 'Sales', slug: 'sales'}, 
    ] 
    }, 
    { 
    title: 'Rick Stone', 
    departments: [ 
     { name: 'Operations', slug: 'operations'}, 
     { name: 'Marketing', slug: 'marketin'}, 
    ] 
    }, 
] 

我如何可以遍歷每個對象的部門數組，並創建新的陣列在那裏我會按部門分類的員工，從而使最終的結果會是這樣的：

operations = [ 
    { 
    title: 'John Doe Junior', 
    departments: [ 
     { name: 'Operations', slug: 'operations'}, 
     { name: 'Sales', slug: 'sales'}, 
    ] 
    }, 
    { 
    title: 'Rick Stone', 
    departments: [ 
     { name: 'Operations', slug: 'operations'}, 
     { name: 'Marketing', slug: 'marketin'}, 
    ] 
    }, 
] 

marketing = [ 
    { 
    title: 'John Doe', 
    departments: [ 
     { name: 'Marketing', slug: 'marketing'}, 
     { name: 'Sales', slug: 'sales'}, 
     { name: 'Administration', slug: 'administration'}, 
    ] 
    }, 
    { 
    title: 'Rick Stone', 
    departments: [ 
     { name: 'Operations', slug: 'operations'}, 
     { name: 'Marketing', slug: 'marketin'}, 
    ] 
    }, 
] 

什麼是動態地創建這種陣列的方法是什麼？

更新

我試圖想出了使用從答案，在那裏我會動態地創建部門對象的數組，將有員工組成的數組建議的解決方案：

const isInDepartment = departmentToCheck => employer => employer.departments.find(department => department.slug == departmentToCheck); 

var departments = []; 
function check(departments, name) { 
    return departments.some(object => name === object.department); 
} 

employees.forEach((employee) => { 
    employee.departments.forEach((department) => { 
     let found = check(departments, department.slug); 
     if (!found) { 
      departments.push({ department: department.slug }); 
     } 
     }); 
}); 

departments.forEach((department) => { 
    // push an array of employees to each department 
    //employees.filter(isInDepartment(department)); 
}); 

但是，我不知道如何將員工數組推到數組中的最後一個循環中的對象？ 這是fiddle。

Answer 1

這個怎麼樣？我使用Array.protoype.filter操作，並且使用高階函數（在本例中爲返回函數的函數）創建謂詞（返回布爾值的函數），以檢查員工是否在特定部門。我在代碼中添加了一些（希望）澄清評論。

編輯：使用您提供的新代碼和上下文this JSFiddle demo顯示它將如何協同工作。

const employees = [ 
 
    { 
 
    title: 'John Doe', 
 
    departments: [ 
 
     { name: 'Marketing', slug: 'marketing'}, 
 
     { name: 'Sales', slug: 'sales'}, 
 
     { name: 'Administration', slug: 'administration'} 
 
    ] 
 
    }, 
 
    { 
 
    title: 'John Doe Junior', 
 
    departments: [ 
 
     { name: 'Operations', slug: 'operations'}, 
 
     { name: 'Sales', slug: 'sales'} 
 
    ] 
 
    }, 
 
    \t { 
 
    title: 'Rick Stone', 
 
    departments: [ 
 
     { name: 'Operations', slug: 'operations'}, 
 
     { name: 'Marketing', slug: 'marketin'} 
 
    ] 
 
    } 
 
]; 
 

 
// given a department, this returns a function that checks 
 
// whether an employee is in the specified department 
 
// NOTE: the "find" returns the found object (truthy) 
 
// or undefined (falsy) if no match was found. 
 
const isInDepartment = 
 
\t departmentToCheck => 
 
    \t employee => employee.departments.find(dep => dep.name == departmentToCheck); 
 

 
const employeesInMarketing = employees.filter(isInDepartment('Marketing')); 
 
const employeesInOperations = employees.filter(isInDepartment('Operations')); 
 

 
console.log('Employees in marketing', employeesInMarketing); 
 
console.log('Employees in operations', employeesInOperations);