Question

我们说我的收藏品有用户名和唯一的提交ID。

{"name" : "a","commit_id": "078d40cc537"},
{"name" : "b","commit_id": "078d40cc538"},
{"name" : "c","commit_id": "078d40cc539"},
{"name" : "a","commit_id": "078d40cc540"},
{"name" : "c","commit_id": "078d40cc541"},
{"name" : "c","commit_id": "078d40cc542"},
{"name" : "d","commit_id": "078d40cc543"}

我想生成按提交数量分组的输出。任何帮助将不胜感激。

1 commit  = 2  #user b,d
2 commits = 1  #user a
3 commits = 1  #user c

Answer 1

不确定。聚合“管道”正是如此，因为您可以将“链”或“管道”阶段组合在一起。要获得结果，您基本上需要连续两个$group阶段：

db.commits.aggregate([
    { "$group": {
        "_id": "$name",
        "commits": { "$sum": 1 }  
    }},
    { "$group": {
        "_id": "$commits",
        "users": { "$push": "$_id" },
        "howMany": { "$sum": 1 }
    }},
    { "$sort": { "_id": -1 } }
])

所以每个“用户”的第一个总数，第二个按“计数”收集它们。可选择按以下形式排序：

{ "_id" : 3, "users" : [ "c" ], "howMany": 1 }
{ "_id" : 2, "users" : [ "a" ], "howMany": 1 }
{ "_id" : 1, "users" : [ "d", "b" ], "howMany": 2 }

对舞台出现的次数没有限制（在BSON大小限制内），因此您不仅限于拥有单个$ group或其他管道阶段。

Mongo聚合按$ sum分组？

1 个答案: