如何在MongoDB中按字段获得聚集组旁边的计数字段？

Question

我正在使用餐厅样本数据集，其中的数据如下：

{
    "_id" : ObjectId("5728c18870b1f4f542bd7c52"),
    "borough" : "Brooklyn",
    "cuisine" : "American",
    "name" : "Riviera Caterer",
    "restaurant_id" : "40356018"
}

{
    "_id" : ObjectId("5728c18870b1f4f542bd7c55"),
    "borough" : "Queens",
    "cuisine" : "Jewish/Kosher",
    "name" : "Tov Kosher Kitchen",
    "restaurant_id" : "40356068"
}

(跳过一些不需要的字段)

我想做的基本上是将数据按菜系分组，计数和菜式相邻，很像：

select count(1) as Count, cuisine 
from restaurants 
group by cuisine

会屈服

 Count | Cuisine
 --------------
   10  | American 
    5  | Indian

试炼

db.restaurants.aggregate([
    {"$group" : {_id:"$cuisine", 
                 count_1:{$sum:1}}}
])

结果如下：

{ "_id" : "American ", "count_1" : 6183 }
{ "_id" : "Bakery", "count_1" : 691 }
{ "_id" : "Irish", "count_1" : 190 }
{ "_id" : "Chicken", "count_1" : 410 }
{ "_id" : "Ice Cream, Gelato, Yogurt, Ices", "count_1" : 348 }
{ "_id" : "Egyptian", "count_1" : 14 }
{ "_id" : "Jewish/Kosher", "count_1" : 316 }
{ "_id" : "Delicatessen", "count_1" : 321 }
{ "_id" : "Indonesian", "count_1" : 8 }
{ "_id" : "Russian", "count_1" : 88 }
{ "_id" : "Spanish", "count_1" : 637 }
{ "_id" : "Chinese", "count_1" : 2418 }
{ "_id" : "Hamburgers", "count_1" : 433 }
{ "_id" : "Hotdogs", "count_1" : 34 }
{ "_id" : "Sandwiches/Salads/Mixed Buffet", "count_1" : 255 }
{ "_id" : "Pancakes/Waffles", "count_1" : 16 }
{ "_id" : "Turkish", "count_1" : 70 }
{ "_id" : "Caribbean", "count_1" : 657 }
{ "_id" : "Donuts", "count_1" : 479 }
{ "_id" : "Bagels/Pretzels", "count_1" : 168 }

是否有可能得到这样的数据：

{ "American" : 6183 }
{ "Bakery"   : 691 }
{ "Irish"    : 190 }
{ "Chicken"  : 410 }
{ "Ice Cream, Gelato, Yogurt, Ices" : 348 }
{ "Egyptian" : 14 }
{ "Jewish/Kosher" : 316 }
{ "Delicatessen": 321 }

诸若此类?

尝试使用像这样的投影：

db.restaurants.aggregate(
    { $group: { "_id": {cuisine:"$cuisine"},
                "Count":  {"$sum":1}
              }
    },
    { "$project":
       { "_id": 0,
         "Cuisine": "$_id.cuisine",
         "Total":"$Count",
        }
     }
   ) 

但并没有完全理解：

{ "Cuisine" : "American ", "Total" : 6183 }
{ "Cuisine" : "Bakery", "Total" : 691 }
{ "Cuisine" : "Irish", "Total" : 190 }
{ "Cuisine" : "Chicken", "Total" : 410 }
{ "Cuisine" : "Ice Cream, Gelato, Yogurt, Ices", "Total" : 348 }
{ "Cuisine" : "Egyptian", "Total" : 14 }
{ "Cuisine" : "Jewish/Kosher", "Total" : 316 }
{ "Cuisine" : "Delicatessen", "Total" : 321 }
{ "Cuisine" : "Indonesian", "Total" : 8 }
{ "Cuisine" : "Russian", "Total" : 88 }
{ "Cuisine" : "Spanish", "Total" : 637 }
{ "Cuisine" : "Chinese", "Total" : 2418 }
{ "Cuisine" : "Hamburgers", "Total" : 433 }
{ "Cuisine" : "Hotdogs", "Total" : 34 }
{ "Cuisine" : "Sandwiches/Salads/Mixed Buffet", "Total" : 255 }
{ "Cuisine" : "Pancakes/Waffles", "Total" : 16 }
{ "Cuisine" : "Turkish", "Total" : 70 }
{ "Cuisine" : "Caribbean", "Total" : 657 }
{ "Cuisine" : "Donuts", "Total" : 479 }
{ "Cuisine" : "Bagels/Pretzels", "Total" : 168 }

基本上，我希望下面的计数按字段分组，但在MongoDB中。

Answer 1

尝尝这个

db.restaurants.aggregate([
  {
    "$group" : {_id:"$cuisine", count_1:{$sum:1}}
  }
]).
result.
map(function(o) {
  var r = {};
  r[o._id] = o.count_1;
  return r;
});

Answer 2

这将给出预期的输出。

db.restaurants.aggregate([
      {
        $group : {_id:"$cuisine", mycount:{$sum:1}}
      }]).map(function(o) {
      var r = {};
      r[o._id] = o.mycount;
      return r;
    });