使用双数据集进行数据操作

Question

返回一个对象，其中每个键是教师姓名，每个值是 他们可以根据自己的技能教授一系列模块。

以下是我应该操作的数据集：

const instructors = [
  { name: 'Pam', module: 2, teaches: ['scope', 'recursion', 'node'] },
  { name: 'Brittany', module: 2, teaches: ['oop', 'pwas'] },
  { name: 'Nathaniel', module: 2, teaches: ['oop', 'scope', 'mobile'] },
  { name: 'Robbie', module: 4, teaches: ['node', 'pwas'] },
  { name: 'Leta', module: 4, teaches: ['pwas', 'node', 'recursion'] },
  { name: 'Travis', module: 1, teaches: ['javascript', 'html', 'css'] },
  { name: 'Louisa', module: 1, teaches: ['javascript', 'html', 'css', 'node', 'pwas'] },
  { name: 'Christie', module: 3, teaches: ['javascript', 'react', 'node'] },
  { name: 'Will', module: 3, teaches: ['javascript', 'redux', 'react', 'oop', 'scope'] }
];

const cohorts = [
  { cohort: 1806, module: 1, studentCount: 30, curriculum: ['html', 'css', 'javascript'] },
  { cohort: 1804, module: 2, studentCount: 21, curriculum: ['javascript', 'css', 'recursion', 'scope', 'oop'] },
  { cohort: 1803, module: 3, studentCount: 20, curriculum: ['react', 'redux', 'html', 'javascript'] },
  { cohort: 1801, module: 4, studentCount: 18, curriculum: ['pwas', 'mobile', 'node', 'javascript', 'css'] }
];

以下是我尝试过的代码，我一直在遍历两个数据集时遇到问题，将它们链接在一起很困难！

这是我尝试过的，但无法让正确的数组在对象中显示为值：

let result1 = instructors.reduce((teacherObj, teacherName) => {
  if(!teacherObj[teacherName.name]) {
    teacherObj[teacherName.name] = []
    // console.log(instructors.map(a => a.module))
  }

  return teacherObj
}, {})
console.log(result1)

预期结果：

{
   Pam: [2, 4],
   Brittany: [2, 4],
   Nathaniel: [2, 4],
   Robbie: [4],
   Leta: [2, 4],
   Travis: [1, 2, 3, 4],
   Louisa: [1, 2, 3, 4],
   Christie: [1, 2, 3, 4],
   Will: [1, 2, 3, 4]
}

Answer 1

首先将cohorts 数组转换为由课程表索引的对象，其值是课程表的关联模块，以便快速查找。也就是说，像这样的对象：

// modulesByCurriculum
{
  "html": [    // html is present in module 1 and 3
    1,
    3
  ],
  "css": [     // css is present in modules 1, 2, and 4
    1,
    2,
    4
  ], ...

这允许您取一个程序名称，例如scope，并快速获取其关联的模块。

然后，遍历讲师并查找每个相关课程以找到模块，通过 Set 进行重复数据删除：

const instructors = [
  { name: 'Pam', module: 2, teaches: ['scope', 'recursion', 'node'] },
  { name: 'Brittany', module: 2, teaches: ['oop', 'pwas'] },
  { name: 'Nathaniel', module: 2, teaches: ['oop', 'scope', 'mobile'] },
  { name: 'Robbie', module: 4, teaches: ['node', 'pwas'] },
  { name: 'Leta', module: 4, teaches: ['pwas', 'node', 'recursion'] },
  { name: 'Travis', module: 1, teaches: ['javascript', 'html', 'css'] },
  { name: 'Louisa', module: 1, teaches: ['javascript', 'html', 'css', 'node', 'pwas'] },
  { name: 'Christie', module: 3, teaches: ['javascript', 'react', 'node'] },
  { name: 'Will', module: 3, teaches: ['javascript', 'redux', 'react', 'oop', 'scope'] }
];

const cohorts = [
  { cohort: 1806, module: 1, studentCount: 30, curriculum: ['html', 'css', 'javascript'] },
  { cohort: 1804, module: 2, studentCount: 21, curriculum: ['javascript', 'css', 'recursion', 'scope', 'oop'] },
  { cohort: 1803, module: 3, studentCount: 20, curriculum: ['react', 'redux', 'html', 'javascript'] },
  { cohort: 1801, module: 4, studentCount: 18, curriculum: ['pwas', 'mobile', 'node', 'javascript', 'css'] }
];

const modulesByCurriculum = cohorts.reduce((a, { module, curriculum }) => {
  curriculum.forEach((currName) => {
    if (!a[currName]) {
      a[currName] = [];
    }
    a[currName].push(module);
  });
  return a;
}, {});
const output = instructors.reduce((a, { name, teaches }) => {
  a[name] = [...new Set(
    teaches.flatMap(currName => modulesByCurriculum[currName])
  )];
  return a;
}, {});
console.log(output);

如果你不能使用flatMap，你可以改为concat：

const instructors = [
  { name: 'Pam', module: 2, teaches: ['scope', 'recursion', 'node'] },
  { name: 'Brittany', module: 2, teaches: ['oop', 'pwas'] },
  { name: 'Nathaniel', module: 2, teaches: ['oop', 'scope', 'mobile'] },
  { name: 'Robbie', module: 4, teaches: ['node', 'pwas'] },
  { name: 'Leta', module: 4, teaches: ['pwas', 'node', 'recursion'] },
  { name: 'Travis', module: 1, teaches: ['javascript', 'html', 'css'] },
  { name: 'Louisa', module: 1, teaches: ['javascript', 'html', 'css', 'node', 'pwas'] },
  { name: 'Christie', module: 3, teaches: ['javascript', 'react', 'node'] },
  { name: 'Will', module: 3, teaches: ['javascript', 'redux', 'react', 'oop', 'scope'] }
];

const cohorts = [
  { cohort: 1806, module: 1, studentCount: 30, curriculum: ['html', 'css', 'javascript'] },
  { cohort: 1804, module: 2, studentCount: 21, curriculum: ['javascript', 'css', 'recursion', 'scope', 'oop'] },
  { cohort: 1803, module: 3, studentCount: 20, curriculum: ['react', 'redux', 'html', 'javascript'] },
  { cohort: 1801, module: 4, studentCount: 18, curriculum: ['pwas', 'mobile', 'node', 'javascript', 'css'] }
];

const modulesByCurriculum = cohorts.reduce((a, { module, curriculum }) => {
  curriculum.forEach((currName) => {
    if (!a[currName]) {
      a[currName] = [];
    }
    a[currName].push(module);
  });
  return a;
}, {});
const output = instructors.reduce((a, { name, teaches }) => {
  a[name] = [...new Set(
    [].concat(...teaches.map(currName => modulesByCurriculum[currName]))
  )];
  return a;
}, {});
console.log(output);

Answer 2

自己想出答案后，结果与answer of CertainPerformance基本相同。

我选择使用Set 作为查找哈希中的值，以便在添加重复项时直接消除它们。我的版本还使用另一种技术来展平阵列。我将concat 数组使用reduce，而不是spreading 数组转换成concat 调用。

const instructors = [{name: 'Pam', module: 2, teaches: ['scope', 'recursion', 'node']}, {name: 'Brittany', module: 2, teaches: ['oop', 'pwas']}, {name: 'Nathaniel', module: 2, teaches: ['oop', 'scope', 'mobile']}, {name: 'Robbie', module: 4, teaches: ['node', 'pwas']}, {name: 'Leta', module: 4, teaches: ['pwas', 'node', 'recursion']}, {name: 'Travis', module: 1, teaches: ['javascript', 'html', 'css']}, {name: 'Louisa', module: 1, teaches: ['javascript', 'html', 'css', 'node', 'pwas']}, {name: 'Christie', module: 3, teaches: ['javascript', 'react', 'node']}, {name: 'Will', module: 3, teaches: ['javascript', 'redux', 'react', 'oop', 'scope']}];
const cohorts = [{cohort: 1806, module: 1, studentCount: 30, curriculum: ['html', 'css', 'javascript']}, {cohort: 1804, module: 2, studentCount: 21, curriculum: ['javascript', 'css', 'recursion', 'scope', 'oop']}, {cohort: 1803, module: 3, studentCount: 20, curriculum: ['react', 'redux', 'html', 'javascript']}, {cohort: 1801, module: 4, studentCount: 18, curriculum: ['pwas', 'mobile', 'node', 'javascript', 'css']}];

// prepare module lookup hash
let moduleLookup = {};
cohorts.forEach(({module, curriculum}) => {
  curriculum.forEach(craft => {
    let modules = moduleLookup[craft] || (moduleLookup[craft] = new Set());
    modules.add(module);
  });
});

// answer
let result = {};
instructors.forEach(({name, teaches}) => {
  let modules = teaches
    .map(craft => Array.from(moduleLookup[craft])) // get modules for each craft
    .reduce((acc, arr) => acc.concat(arr), []); // flatten 1 level

  result[name] = Array.from(new Set(modules)); // remove duplicates
});

console.log(result);