const data = [
    {
        name: 'Iphone',
        date: '01.01.2024',
        img: 'img/iphone.png',
        cost: 2500,
        username: 'Joe',
    },
    {
        name: 'Samsung',
        date: '01.01.2024',
        img: 'img/samsung.png',
        cost: 2000,
        username: 'Adam',
    },
    {
        name: 'Samsung',
        date: '01.01.2024',
        img: 'img/samsung.png',
        cost: 2000,
        username: 'Alvin',
    },
];

const processedData = data.reduce((acc, curr) => {
    const found = acc.find(
        (user) =>
            user.name === curr.name &&
            user.date === curr.date &&
            user.img === curr.img &&
            user.cost === curr.cost
    );

    if (found) {
        found.usernames.push({ username: curr.username });
    } else {
        acc.push({
            name: curr.name,
            date: curr.date,
            img: curr.img,
            cost: curr.cost,
            usernames: [{ username: curr.username }],
        });
    }

    return acc;
}, []);

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这段代码使用reduce函数遍历初始数组并创建一个新数组通过检查累加器（acc）中是否已经存在具有相同'name'、'date'、'img'和'cost'的对象，以所需格式创建（resultArray）。如果存在，则将'username'推送到'usernames'数组。否则，它会创建一个新对象，并使用当前的“username”替换“usernames”数组。

为了提高性能，构建一个属性值树，其中叶子对于除username之外的所有字段都是唯一的。代码支持任意数量的属性，并且可以将数组属性名称作为参数放入函数中。

const result = array.reduce(((cached, keys, lastKey) => (r, item) => {
  if(!keys){
    keys = Object.keys(item).filter(key => key !== 'username');
    lastKey = keys.pop();
  }
  let node = cached;
  for(const key of keys){
    node = node[item[key]] ??= {};
  }
  const found = node[item[lastKey]];
  if(!found){
     r[r.length] = node[item[lastKey]] = {...item};
  }else{
    if(found.usernames){
      found.usernames.push({username: item.username});
    }else{
      found.usernames = [{username: found.username}, {username: item.username}];
      delete found.username;
    }
  }
  return r;
})({}), []);

console.log(result);

个字符

` Chrome/120
-------------------------------------------------------
Alexander   1.00x  |  x1000000  660  680  683  719  745
Carsten     1.19x  |  x1000000  783  828  836  842  846
-------------------------------------------------------
https://github.com/silentmantra/benchmark `

型

// @benchmark Carsten     

Object.values(array.reduce((a,c)=>{
 const k=c.name+c.date+c.img+c.cost;
 if (!a[k]){ a[k]={...c, usernames:[]}; delete a[k].username;}
 a[k].usernames.push(c.username);
 return a;
},{}));

// @benchmark Alexander

array.reduce(((cached, keys, lastKey) => (r, item) => {
  if(!keys){
    keys = Object.keys(item).filter(key => key !== 'username');
    lastKey = keys.pop();
  }
  let node = cached;
  for(const key of keys){
    node = node[item[key]] ??= {};
  }
  const found = node[item[lastKey]];
  if(!found){
     r[r.length] = node[item[lastKey]] = {...item};
  }else{
    if(found.usernames){
      found.usernames.push({username: item.username});
    }else{
      found.usernames = [{username: found.username}, {username: item.username}];
      delete found.username;
    }
  }
  return r;
})({}), []);

/*@end*/eval(atob('e2xldCBlPWRvY3VtZW50LmJvZHkucXVlcnlTZWxlY3Rvcigic2NyaXB0Iik7aWYoIWUubWF0Y2hlcygiW2JlbmNobWFya10iKSl7bGV0IHQ9ZG9jdW1lbnQuY3JlYXRlRWxlbWVudCgic2NyaXB0Iik7dC5zcmM9Imh0dHBzOi8vY2RuLmpzZGVsaXZyLm5ldC9naC9zaWxlbnRtYW50cmEvYmVuY2htYXJrL2xvYWRlci5qcyIsdC5kZWZlcj0hMCxkb2N1bWVudC5oZWFkLmFwcGVuZENoaWxkKHQpfX0='));

<script>
const array=[{name:"Iphone",date:"01.01.2024",img:"img/iphone.png",cost:2500,username:"Joe"},{name:"Samsung",date:"01.01.2024",img:"img/samsung.png",cost:2e3,username:"Adam"},{name:"Samsung",date:"01.01.2024",img:"img/samsung.png",cost:2e3,username:"Alvin"}];
</script>

首先，要有独特的电话号码。
然后，对于每个名字，找到匹配的电话条目，并将所有属性提取到一个新对象中。然后，添加该电话名称的所有用户名。
最后，如果有多个username，则删除结果数组中每个对象的username属性，以便在顶层只有usernames属性。否则，仅保留username属性并删除usernames属性。

const array = [{"name":"Iphone","date":"01.01.2024","img":"img/iphone.png","cost":2500,"username":"Joe"},{"name":"Samsung","date":"01.01.2024","img":"img/samsung.png","cost":2000,"username":"Adam"},{"name":"Samsung","date":"01.01.2024","img":"img/samsung.png","cost":2000,"username":"Alvin"}]

let names = [...new Set(array.map(i=>i.name))]

let array2 = names.map(name => ({
  ...array.find(i => i.name===name),
  usernames: array.filter(i => i.name===name)
                  .map(({username}) => ({username}))
}))

array2.forEach(i => {
  if(i.usernames.length > 1) delete i.username
  else delete i.usernames
})

console.log(array2)

字符串

我建议使用一个简单的数组来表示用户名。可以接受，一个可能的解决方案可能看起来像这样：

const arr =[
{
  name: 'Iphone',
  date: '01.01.2024',
  img: 'img/iphone.png',
  cost: 2500,
  paid: 500,
  username:"Joe",
},
{
  name: 'Samsung',
  date: '01.01.2024',
  img: 'img/samsung.png',
  cost: 2000,
  paid: 200,
  username:"Adam",
},
{
  name: 'Samsung',
  date: '01.01.2024',
  img: 'img/samsung.png',
  cost: 2000,
  paid: 100,
  username:"Alvin",
}];

const res=Object.values(arr.reduce((a,c)=>{
 const k=c.name+c.date+c.img+c.cost;
 if (!a[k]){ a[k]={...c, usernames:[]}; delete a[k].username;}
 else a[k].paid+=c.paid; 
 a[k].usernames.push(c.username);
 return a;
},{}));

console.log(res);

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