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'
}
]
我想像这样将其分组为输出：
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'
}
]
]
如您所见，只有id and time是唯一的，我想根据重复项name, family, number and category对其进行分组，因此，name, family, number and category具有相同值的所有对象都应具有一个单独的数组。
我尝试了很多事情，我使用了两个嵌套的for循环，我使用了map，我对Filter进行了起诉，...我尝试了所有我能想到的，但是我却做不到！
如果有人可以帮助您，我真的很感激。
答案 0 :(得分：2)
只需使用Array.reduce()和组合键name, family,number and category制作地图。然后您可以在该地图上使用Object.values()来获得所需的结果。
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);
答案 1 :(得分：1)
您可以使用Map并将同一组中的所有项目收集到一个阵列中。
以后仅获取地图的值作为结果。
这仅适用于一组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，您可以在一个闭包中使用一个连接键，其中包含name，family，number和category的实际对象的值key。
然后映射具有关闭时间的新数组，并将所有数组组合为单个嵌套数组。在末尾按长度过滤。
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);.as-console-wrapper { max-height: 100% !important; top: 0; }
答案 2 :(得分：1)
您还可以使用Array.reduce()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);