重新考虑数据库组并合并到带有子阵列的单个文档中

Question

这是将每个缩减分组并合并到带有子数组的单个文档中的最佳方法吗？

r.expr([
  {id: 1, foo: 1, bar: 2, date: r.time(2016, 1, 1, 'Z')},
  {id: 1, foo: 4, bar: 1, date: r.time(2016, 1, 3, 'Z')},
  {id: 1, foo: 10, bar: 0, date: r.time(2016, 1, 2, 'Z')},
  {id: 2, foo: 5, bar: 3, date: r.time(2016, 1, 1, 'Z')},
  {id: 2, foo: 3, bar: 6, date: r.time(2016, 1, 2, 'Z')}
]).group('id').orderBy('date').map(function(d){
   return d
     .without('foo', 'bar', 'date')
     .merge({stats: [d.pluck('foo', 'bar', 'date')]})
 }).reduce(function(left, right){
   return left .without('stats').merge({
       stats: left('stats').append(right('stats')(0))
     })
 }).ungroup().map(function(g){
   return g('reduction')
 })

输出：

[
  {
    "id": 1 ,
    "stats": [
      {
        "foo": 1, "bar": 2 , "date": Fri Jan 01 2016 00:00:00 GMT+00:00 
      },
      {
        "foo": 10, "bar": 0 , "date": Sat Jan 02 2016 00:00:00 GMT+00:00 
      } ,
      {
        "foo": 4, "bar": 1 , "date": Sun Jan 03 2016 00:00:00 GMT+00:00 
      }
    ]
  },
  {
    "id": 2 ,
    "stats": [
      {
        "foo": 5, "bar": 3, "date": Fri Jan 01 2016 00:00:00 GMT+00:00
      } ,
      {
        "foo": 3, "bar": 6, "date": Sat Jan 02 2016 00:00:00 GMT+00:00
      }
    ]
  }
]

Answer 1

这应该有效：

r.expr([
  {id: 1, foo: 1, bar: 2, date: r.time(2016, 1, 1, 'Z')},
  {id: 1, foo: 4, bar: 1, date: r.time(2016, 1, 3, 'Z')},
  {id: 1, foo: 10, bar: 0, date: r.time(2016, 1, 2, 'Z')},
  {id: 2, foo: 5, bar: 3, date: r.time(2016, 1, 1, 'Z')},
  {id: 2, foo: 3, bar: 6, date: r.time(2016, 1, 2, 'Z')}
]).group('id')
  .orderBy('date')
  .without('id')
  .ungroup()
  .map(rec => { return { id : rec('group'), stats : rec('reduction') }; } )