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根据对象数组中的重复值将对象分组

object loops node.js arrays javascript

Emad Dehnavi · 技术社区 · 6 年前

这是对象的输入数组:

var array = [{
    id: 1,
    name: 'A',
    family: 'B',
    number: 100,
    category: 'test',
    time: '2018-03-02T10:33:00.000Z'
  },
  {
    id: 2,
    name: 'A',
    family: 'B',
    number: 100,
    category: 'test',
    time: '2018-03-02T10:33:50.000Z'
  },
  {
    id: 3,
    name: 'A',
    family: 'B',
    number: 100,
    category: 'test',
    time: '2018-03-02T10:34:30.000Z'
  },
  {
    id: 4,
    name: 'A',
    family: 'B',
    number: 100,
    category: 'test',
    time: '2018-03-02T10:36:00.000Z'
  },
  {
    id: 5,
    name: 'A',
    family: 'C',
    number: 250,
    category: 'other',
    time: '2018-03-02T10:33:00.000Z'
  },
  {
    id: 6,
    name: 'A',
    family: 'C',
    number: 250,
    category: 'other',
    time: '2018-03-02T10:33:05.000Z'
  }
]

我想将其分组为输出:

var output = [
  [{
      id: 1,
      name: 'A',
      family: 'B',
      number: 100,
      category: 'test',
      time: '2018-03-02T10:33:00.000Z'
    },
    {
      id: 2,
      name: 'A',
      family: 'B',
      number: 100,
      category: 'test',
      time: '2018-03-02T10:33:50.000Z'
    },
    {
      id: 3,
      name: 'A',
      family: 'B',
      number: 100,
      category: 'test',
      time: '2018-03-02T10:34:30.000Z'
    },
    {
      id: 4,
      name: 'A',
      family: 'B',
      number: 100,
      category: 'test',
      time: '2018-03-02T10:36:00.000Z'
    }
  ],
  [{
      id: 5,
      name: 'A',
      family: 'C',
      number: 250,
      category: 'other',
      time: '2018-03-02T10:33:00.000Z'
    },
    {
      id: 6,
      name: 'A',
      family: 'C',
      number: 250,
      category: 'other',
      time: '2018-03-02T10:33:05.000Z'
    }
  ]
]

如你所见,只有 id and time ,是唯一的,我希望根据重复项对其进行分组 name, family, number and category ,因此所有具有相同值的对象 姓名、家庭、号码和类别 应该有一个单独的数组。

我尝试了很多东西,我用了两个嵌套的for循环,我用了map,我用了filter,…我尽了我所能,但我做不到!

如果有人能帮上忙,我真的很感激。

3 回复 | 直到 6 年前

Nina Scholz 6 年前

你可以拿一个 Map 并在数组中收集同一组的所有项。

稍后只获取映射的值作为结果。

这只适用于一组 category .

var array = [{ id: 1, name: 'A', family: 'B', number: 100, category: 'test', time: '2018-03-02T10:33:00.000Z' }, { id: 2, name: 'A', family: 'B', number: 100, category: 'test', time: '2018-03-02T10:33:50.000Z' }, { id: 3, name: 'A', family: 'B', number: 100, category: 'test', time: '2018-03-02T10:34:30.000Z' }, { id: 4, name: 'A', family: 'B', number: 100, category: 'test', time: '2018-03-02T10:36:00.000Z' }, { id: 5, name: 'A', family: 'C', number: 250, category: 'other', time: '2018-03-02T10:33:00.000Z' }, { id: 6, name: 'A', family: 'C', number: 250, category: 'other', time: '2018-03-02T10:33:05.000Z' }],
    grouped = Array.from(array.reduce((m, o) =>
        m.set(o.category, (m.get(o.category) || []).concat(o)), new Map).values());

console.log(grouped);

.as-console-wrapper { max-height: 100% !important; top: 0; }

对于多个组,可以使用一个联接键,该键的值为 name , family , number 和 类别 在一个封闭的 key .

然后用关闭时间映射新数组,并将所有数组组合到单个嵌套数组。在末端按长度过滤。

var array = [{ id: 1, sourceAccount: "A", targetAccount: "B", amount: 100, category: "eating_out", time: "2018-03-02T10:33:00.000Z" }, { id: 2, sourceAccount: "A", targetAccount: "B", amount: 100, category: "eating_out", time: "2018-03-02T10:33:50.000Z" }, { id: 3, sourceAccount: "A", targetAccount: "B", amount: 100, category: "eating_out", time: "2018-03-02T10:34:30.000Z" }, { id: 4, sourceAccount: "A", targetAccount: "B", amount: 100, category: "eating_out", time: "2018-03-02T10:36:00.000Z" }, { id: 5, sourceAccount: "A", targetAccount: "C", amount: 250, category: "other", time: "2018-03-02T10:33:00.000Z" }, { id: 6, sourceAccount: "A", targetAccount: "C", amount: 250, category: "other", time: "2018-03-02T10:33:05.000Z" }],
    groupBy = ['name', 'family', 'number', 'category'],
    result = Array
        .from(
            array
                .reduce((m, o) => (k => m.set(k, (m.get(k) || []).concat(o)))(groupBy.map(k => o[k]).join('|')), new Map)
                .values(),
            o => o.sort((a, b) => a.time.localeCompare(b.time))
        )
        .map(g => g.reduce((r, o, i, a) => {
            if (!i || new Date(o.time).getTime() - new Date(a[i - 1].time).getTime() > 60000) {
                r.push([o]);
            } else {
                r[r.length - 1].push(o);
            }
            return r;
        }, []))
        .reduce((r, a) => r.concat(a), [])
        .filter(a => a.length > 1);

console.log(result);

.作为控制台包装最大高度:100%!重要;顶部:0;

amrender singh 6 年前

只需使用 Array.reduce() 带有组合键的 name, family,number and category . 比你能用的还多 Object.values() 在地图上得到想要的结果。

var array =[ { id: 1, name: 'A', family: 'B', number: 100, category: 'test', time: '2018-03-02T10:33:00.000Z' }, { id: 2, name: 'A', family: 'B', number: 100, category: 'test', time: '2018-03-02T10:33:50.000Z' }, { id: 3, name: 'A', family: 'B', number: 100, category: 'test', time: '2018-03-02T10:34:30.000Z' }, { id: 4, name: 'A', family: 'B', number: 100, category: 'test', time: '2018-03-02T10:36:00.000Z' }, { id: 5, name: 'A', family: 'C', number: 250, category: 'other', time: '2018-03-02T10:33:00.000Z' }, { id: 6, name: 'A', family: 'C', number: 250, category: 'other', time: '2018-03-02T10:33:05.000Z' } ];

var result = Object.values(array.reduce((a,curr)=>{
  var key = curr.name + "_" + curr.family + "_" + curr.number + "_" + curr.category;
  (a[key] = a[key] || []).push(curr);
  return a;
},{}));

console.log(result);

Ankit Agarwal 6 年前

您也可以使用 Array.reduce() 要获得该输出:

var array = [ { id: 1,
  name: 'A',
  family: 'B',
  number: 100,
  category: 'test',
  time: '2018-03-02T10:33:00.000Z' },
  { id: 2,
  name: 'A',
  family: 'B',
  number: 100,
  category: 'test',
  time: '2018-03-02T10:33:50.000Z' },
  { id: 3,
  name: 'A',
  family: 'B',
  number: 100,
  category: 'test',
  time: '2018-03-02T10:34:30.000Z' },
  { id: 4,
  name: 'A',
  family: 'B',
  number: 100,
  category: 'test',
  time: '2018-03-02T10:36:00.000Z' },
  { id: 5,
  name: 'A',
  family: 'C',
  number: 250,
  category: 'other',
  time: '2018-03-02T10:33:00.000Z' },
  { id: 6,
  name: 'A',
  family: 'C',
  number: 250,
  category: 'other',
  time: '2018-03-02T10:33:05.000Z' } 

];

var res = array.reduce((acc, obj)=>{
  var existObj;
  for(var i = 0 ; i < acc.length; i++) {
    let accArray = acc[i];
    if(accArray){
     existObj = accArray.find(({name, family, number, category}) =>  name == obj.name && family == obj.family && number == obj.number && category == obj.category);
     if(existObj) {
        accArray.push(obj);
        return acc;
     }
    }
 }
  acc.push([obj]);
  return acc;
}, []);
console.log(res);