根据另一个嵌套对象数组的内容过滤对象数组

Question

我想过滤positions数组并删除数组中表示的所有位置people。我尝试了_.forEach和的几种组合，_.filter但似乎无法弄清楚。console.log(position)var test = _.filter(position, function(pos) {    _.forEach(people, function(peo) {        _.forEach(peo.position, function(peoplePos) {            if(peoplePos.value == pos.value){                return false;            }        });    });});console.log(test)我认为，我的主要问题是职位嵌套在每个人的对象中var positions = [{    val: 'CEO',    label: 'CEO XXX'}, {    val: 'CTO',    label: 'CTO XXX'}, {    val: 'CBO',    label: 'CBO XXX'}, {    val: 'CLO',    label: 'CLO XXX'}]var people = [{    id: 'AAA',    positions: [{        val: 'CEO',        label: 'CEO XXX'    }]},{    id: 'BBB',    positions: [{        val: 'CXO',        label: 'CXO XXX'    },{        val: 'CEO',        label: 'CEO XXX'    }]},{    id: 'CCC',    positions: [{        val: 'CTO',        label: 'CTO XXX'    }]}]在这种情况下，我的目标是以下结果：var positions = [{    val: 'CBO',    label: 'CBO XXX'}, {    val: 'CLO',    label: 'CLO XXX'}]由于CBO和CLO没有由people数组中的任何对象表示。

Answer 1

一种快速的方法是同时对people数组进行字符串化并检查字符串中的位置。

这免除了您遍历嵌套结构的麻烦。

var positions = [{ val: 'CEO', label: 'CEO XXX' }, { val: 'CTO', label: 'CTO XXX' }, { val: 'CBO', label: 'CBO XXX' }, { val: 'CLO', label: 'CLO XXX' }]

var people = [{ id: 'AAA', positions: [{ val: 'CEO', label: 'CEO XXX' }] }, { id: 'BBB', positions: [{ val: 'CXO', label: 'CXO XXX' }, { val: 'CEO',

    label: 'CEO XXX' }] }, { id: 'CCC', positions: [{ val: 'CTO', label: 'CTO XXX' }] }];

var stringifiedPeople = JSON.stringify(people)

var newPositions = positions.filter((position) =>

  !stringifiedPeople.includes(JSON.stringify(position))

);

console.log(newPositions)

或者，您可以创建一个包含所有占用位置的地图，并过滤掉可用的位置。

var positions = [{ val: 'CEO', label: 'CEO XXX' }, { val: 'CTO', label: 'CTO XXX' }, { val: 'CBO', label: 'CBO XXX' }, { val: 'CLO', label: 'CLO XXX' }]

var people = [{ id: 'AAA', positions: [{ val: 'CEO', label: 'CEO XXX' }] }, { id: 'BBB', positions: [{ val: 'CXO', label: 'CXO XXX' }, { val: 'CEO',

    label: 'CEO XXX' }] }, { id: 'CCC', positions: [{ val: 'CTO', label: 'CTO XXX' }] }];

var mappedPositions = {}

people.forEach((p) =>

  p.positions.forEach((position) =>

    mappedPositions[position.val] = true

  )

);

var newPositions = positions.filter((position) => !mappedPositions[position.val]);

console.log(newPositions)

Answer 2

执行我的评论。

为了.reduce()提高效率，可以将整个事情写成一个很大的数组，但是我更喜欢显示确切的步骤，以使每个步骤的工作更加清晰。

var positions = [{val:'CEO',label:'CEOXXX'},{val:'CTO',label:'CTOXXX'},{val:'CBO',label:'CBOXXX'},{val:'CLO',label:'CLOXXX'}];

var people = [{id:'AAA',positions:[{val:'CEO',label:'CEOXXX'}]},{id:'BBB',positions:[{val:'CXO',label:'CXOXXX'},{val:'CEO',label:'CEOXXX'}]},{id:'CCC',positions:[{val:'CTO',label:'CTOXXX'}]}];

const occupied_positions = people

  .map( person => person.positions )

  .flat()

  .map( position => position.val );

  

const all_positions = positions

  .map( position => position.val );

  

const open_positions = all_positions

  .filter( position => !occupied_positions.includes( position ))

  .map( position => positions.find( source => source.val === position ));

  

console.log( open_positions );

Answer 3

您可以使用filter，find和some过滤掉那些不在people数组的positions数组中的对象。

var positions = [{val:'CEO',label:'CEOXXX'},{val:'CTO',label:'CTOXXX'},{val:'CBO',label:'CBOXXX'},{val:'CLO',label:'CLOXXX'}];

var people = [{id:'AAA',positions:[{val:'CEO',label:'CEOXXX'}]},{id:'BBB',positions:[{val:'CXO',label:'CXOXXX'},{val:'CEO',label:'CEOXXX'}]},{id:'CCC',positions:[{val:'CTO',label:'CTOXXX'}]}];

const out = positions.filter(position => {

  return !people.find(person => {

    return person.positions.some(({ val, label }) => {

      return val === position.val && label === position.label;

    });

  });

});

console.log(out);