我想使用“id”属性连接2个对象数组（文件x计数），然后，我想按“folder”属性分组，将每个“folder”的所有计数相加。最后的结果是我不想有“id”属性
文件数组：

const file = [{
  id: "xxx",
  folder: "A"
},{
  id: "bbb",
  folder: "B"
},{
  id: "ccc",
  folder: "B"
},{
  id: "ddd",
  folder: "C"
},{
  id: "aaa",
  folder: "C"
},{
  id: "ggg",
  folder: "A"
}]

计数数组：

const counts = [{
  id: "bbb",
  actions: {
    download: 5,
    seen: 10,
    shared: 3
  }
},{
  id: "xxx",
  actions: {
    download: 2,
    seen: 5,
    shared: 8
  }
},{
  id: "ccc",
  actions: {
    download: 7,
    seen: 9,
    shared: 5
  }
},{
  id: "ggg",
  actions: {
    download: 0,
    seen: 2,
    shared: 3
  }
},{
  id: "eee",
  actions: {
    download: 50,
    seen: 55,
    shared: 3
  }
},{
  id: "fff",
  actions: {
    download: 5,
    seen: 4,
    shared: 3
  }
}]

结果:

const result = [{
  folder: "A",
  actions: {
    download: 2,
    seen: 7,
    shared: 11
  }
},{
  folder: "B",
  actions: {
    download: 12,
    seen: 19,
    shared: 8
  }
},{
  folder: "C"
}]

我的代码：

const join = files.reduce((arr, e) => {
  arr.push(Object.assign({}, e, counts.find(a => a.id == e.id)))
  return arr;
}, [])

join.forEach(function(v){ delete v.id });

const result = Object.values(
                 join.reduce((a, c) => (
                   a[c.folder] = a[c.folder] && a[c.folder].actions?
                   (a[c.folder].actions.download += c.actions.download,
                   a[c.folder].actions.seen += c.actions.seen,
                   a[c.folder].actions.shared += c.actions.shared , 
                   a[c.folder]) :
                   c, a), {}
                 )
               )

有没有一种合适的方法可以在更少的线路上进行优化？
事先非常感谢。