Question

假设我有这个javascript数组。

var events = [
  {id: 1, name: 'Football Game', invitees: ['Gary', 'Jane']},
  {id: 2, name: 'Football Game', invitees: ['Gary', 'Jane']},
  {id: 2, name: 'Football Game', invitees: ['Gary', 'Jane','Bella']},
  {id: 3, name: 'Football Game', invitees: ['Gary', 'Jane']},
  {id: 4, name: 'Football Game', invitees: ['Gary', 'Jane']},
  {id: 5, name: 'Football Game', invitees: ['Gary', 'Jane']},
  {id: 5, name: 'Football Game', invitees: ['Ted', 'Laura']},
  {id: 5, name: 'Football Game', invitees: ['Jim', 'Samantha']},
];

我需要的是将此数组减少为：

var events = [
  {id: 1, name: 'Football Game', invitees: ['Gary', 'Jane']},
  {id: 2, name: 'Football Game', invitees: ['Gary', 'Jane','Bella']},
  {id: 3, name: 'Football Game', invitees: ['Gary', 'Jane']},
  {id: 4, name: 'Football Game', invitees: ['Gary', 'Jane']},
  {id: 5, name: 'Football Game', invitees: ['Gary', 'Jane', 'Ted', 'Laura', 'Jim', 'Samantha']}
];

在减少数组的情况下，每个id只有一个对象，每个对象中的被邀请者数组显示所有被邀请者。

我在这里开始使用基本逻辑时遇到了麻烦。

Answer 1

您可以使用reduce()和forEach()这样做。

var events = [
  {id: 1, name: 'Football Game', invitees: ['Gary', 'Jane']},
  {id: 2, name: 'Football Game', invitees: ['Gary', 'Jane']},
  {id: 2, name: 'Football Game', invitees: ['Gary', 'Jane','Bella']},
  {id: 3, name: 'Football Game', invitees: ['Gary', 'Jane']},
  {id: 4, name: 'Football Game', invitees: ['Gary', 'Jane']},
  {id: 5, name: 'Football Game', invitees: ['Gary', 'Jane']},
  {id: 5, name: 'Football Game', invitees: ['Ted', 'Gary', 'Laura']},
  {id: 5, name: 'Football Game', invitees: ['Jim', 'Samantha']}
];

var o = {}
var result = events.reduce(function(r, e) {
  if (!o[e.id]) {
    o[e.id] = e;
    r.push(o[e.id]);
  } else {
    e.invitees.forEach(function(a) {
      if (o[e.id].invitees.indexOf(a) == -1) o[e.id].invitees.push(a);
    })
  }
  return r;
}, [])

console.log(result)

您也可以使用Set代替forEach()循环删除重复项。

var events = [
  {id: 1, name: 'Football Game', invitees: ['Gary', 'Jane']},
  {id: 2, name: 'Football Game', invitees: ['Gary', 'Jane']},
  {id: 2, name: 'Football Game', invitees: ['Gary', 'Jane','Bella']},
  {id: 3, name: 'Football Game', invitees: ['Gary', 'Jane']},
  {id: 4, name: 'Football Game', invitees: ['Gary', 'Jane']},
  {id: 5, name: 'Football Game', invitees: ['Gary', 'Jane']},
  {id: 5, name: 'Football Game', invitees: ['Ted', 'Gary', 'Laura']},
  {id: 5, name: 'Football Game', invitees: ['Jim', 'Samantha']}
];

var o = {}
var result = events.reduce(function(r, e) {
  if (!o[e.id]) {
    o[e.id] = e;
    r.push(o[e.id]);
  } else {
    o[e.id].invitees = [...new Set(o[e.id].invitees.concat(e.invitees))];
  }
  return r;
}, [])

console.log(result)

Answer 2

您可以使用此ES6 函数式编程样式代码：

var result = Array.from (
    events.reduce ( (acc, obj) => acc.set(obj.id,
        Object.assign({}, obj, {invitees: [...new Set(
            [...(acc.get(obj.id) || obj).invitees, ...obj.invitees] )]})
    ), new Map() ), ([id, obj]) => obj );

var events = [
  {id: 1, name: 'Football Game', invitees: ['Gary', 'Jane']},
  {id: 2, name: 'Football Game', invitees: ['Gary', 'Jane']},
  {id: 2, name: 'Football Game', invitees: ['Gary', 'Jane','Bella']},
  {id: 3, name: 'Football Game', invitees: ['Gary', 'Jane']},
  {id: 4, name: 'Football Game', invitees: ['Gary', 'Jane']},
  {id: 5, name: 'Football Game', invitees: ['Gary', 'Jane']},
  {id: 5, name: 'Football Game', invitees: ['Ted', 'Laura']},
  {id: 5, name: 'Football Game', invitees: ['Jim', 'Samantha']},
];

var result = Array.from (
    events.reduce ( (acc, obj) => acc.set(obj.id,
        Object.assign({}, obj, {invitees: [...new Set(
            [...(acc.get(obj.id) || obj).invitees, ...obj.invitees] )]})
    ), new Map() ), ([id, obj]) => obj );
  
console.log(result);

循环遍历集合并返回一个新集合，其中组合了具有相同id的对象

2 个答案: