如何将具有相同属性的对象合并到一个数组中？

Question

我正在收到下一个数据

    [
      { id: "1", name: "test1", rName: "the1" },
      { id: "1", name: "test1", rName: "the2" },
      { id: "1", name: "test1", rName: "the3" },
      { id: "2", name: "test2", rName: "the1" },
      { id: "2", name: "test2", rName: "the2" },
      { id: "3", name: "test3", rName: "the1" }
    ]

我想通过id合并它并将rName推入一个数组以获得这个结构

    [
      { id: "1", name: "test1", rName: ["the1", "the2","the3"] },
      { id: "2", name: "test2", rName: ["the1", "the2"] },
      { id: "3", name: "test3", rName: ["the1"] }
    ]

我认为这样做会减少，但没有成功，如果有人能指出我正确的方向，将不胜感激。

Answer 1

这是一个非常简单的数据重新格式化案例。 代码如下。

 var data = [ { id: "1", name: "test1", rName: "the1" }, { id: "1", name: "test1", rName: "the2" }, { id: "1", name: "test1", rName: "the3" }, { id: "2", name: "test2", rName: "the1" }, { id: "2", name: "test2", rName: "the2" }, { id: "3", name: "test3", rName: "the1" } ]; var reduced = Object.values(data.reduce(function(accumulator, element) { if (!accumulator[element.id]) { accumulator[element.id] = { id: element.id, name: element.name, rName: [] }; } accumulator[element.id].rName.push(element.rName); return accumulator; }, {})); console.log(reduced); 

累加器检查按element.id中的键是否存在于累加器中。 如果没有，则创建它。 然后它在现有堆栈上推送新的rName 。 然后使用Object.values()将转换返回到数组。

Answer 2

您可以使用reduce并find如下：

在reduce accumulator ，检查是否已存在与正在迭代的当前项具有相同id的项。 如果是，请将当前项的rName推送到rName数组。 否则，将新项目推送到accumulator

 var data = [ { id: "1", name: "test1", rName: "the1" }, { id: "1", name: "test1", rName: "the2" }, { id: "1", name: "test1", rName: "the3" }, { id: "2", name: "test2", rName: "the1" }, { id: "2", name: "test2", rName: "the2" }, { id: "3", name: "test3", rName: "the1" } ]; const newArray = data.reduce((acc, {id,name,rName}) => { const existing = acc.find(a => a.id == id); if (existing) existing["rName"].push(rName); else acc.push({id,name,rName: [rName]}) return acc }, []); console.log(newArray) 

这是一个代码高尔夫的答案。 （从@Sébastien的回答中得到了想法）：

 var data = [ { id: "1", name: "test1", rName: "the1" }, { id: "1", name: "test2", rName: "the2" }, { id: "1", name: "test2", rName: "the3" }, { id: "2", name: "test2", rName: "the1" }, { id: "2", name: "test1", rName: "the2" }, { id: "3", name: "test3", rName: "the1" } ] const anotherArray = Object.values(data.reduce((acc, {id,name,rName}) => ((acc[id] = acc[id] || {id,name,rName:[]})["rName"].push(rName), acc), {})); console.log(anotherArray) 

Answer 3

纯ES6

let result = obj.reduce((acc, item) => {
  let found = acc.find(i => i.id === item.id);
  found ? (found.rName = [...found.rName, item.rName]) : (acc = [...acc, { ...item, rName: [item.rName] }]);
  return acc;
}, []);

Answer 4

如果每个id都可以确定唯一的name ，那么这将起作用：

let data = [
  { id: "1", name: "test1", rName: "the1" },
  { id: "1", name: "test1", rName: "the2" },
  { id: "1", name: "test1", rName: "the3" },
  { id: "2", name: "test2", rName: "the1" },
  { id: "2", name: "test2", rName: "the2" },
  { id: "3", name: "test3", rName: "the1" }
];

let result = [];

for (let item of data) 
{
    const index = result.findIndex(i => i.id === item.id);
    if (index < 0) {
        result.push({ id: item.id, name: item.name, rName: [item.rName] });
    }
    else {
        result[index].rName.push(item.rName);
    }
}

请注意，当以下数据接受时，这将不起作用：

[
  { id: "1", name: "test1", rName: "the1" },
  { id: "1", name: "test2", rName: "the2" },
  { id: "1", name: "test2", rName: "the3" },
  { id: "2", name: "test2", rName: "the1" },
  { id: "2", name: "test1", rName: "the2" },
  { id: "3", name: "test3", rName: "the1" }
]

Answer 5

如果你坚持使用'reduce'方法，你可以这样做：

var arr = [
        { id: "1", name: "test1", rName: "the1" },
        { id: "1", name: "test1", rName: "the2" },
        { id: "1", name: "test1", rName: "the3" },
        { id: "2", name: "test2", rName: "the1" },
        { id: "2", name: "test2", rName: "the2" },
        { id: "3", name: "test3", rName: "the1" }
    ]
    var arr1 = arr.reduce(function (a1, a2) {
        if (a1 instanceof Array) {
            let lastItem = a1[a1.length - 1]
            if (lastItem.id == a2.id) {
                lastItem.rName.push(a2.rName)
            } else {
                a1.push({ ...a2, rName: [a2.rName] })
            }
            return a1
        } else {
            let result = []
            if (a1.id == a2.id) {
                result.push({ ...a1, rName: [a1.rName, a2.rName] })
            } else {
                result.push({ ...a1, rName: [a1, rName] })
                result.push({ ...a2, rName: [a2, rName] })
            }
            return result

        }

    })
    console.log(JSON.stringify(arr1))

但我认为你应该这样做，因为代码更清晰：

var arr = [
    { id: "1", name: "test1", rName: "the1" },
    { id: "1", name: "test1", rName: "the2" },
    { id: "1", name: "test1", rName: "the3" },
    { id: "2", name: "test2", rName: "the1" },
    { id: "2", name: "test2", rName: "the2" },
    { id: "3", name: "test3", rName: "the1" }
]
var map = new Map()
arr.forEach(function(item){
    if(map.has(item.id)){
        map.get(item.id).rName.push(item.rName)
    }else{
        map.set(item.id, {...item, rName: [item.rName]})
    }
})
var arr1 = Array.from(map.values())
console.log(JSON.stringify(arr1))