python - Pymongo 和聚合框架查询

Question

我尝试在 pymongo 中编写聚合查询，但我自己做不到，我尝试过文档，尝试过谷歌，尝试过 Stack Overflow，但我只是找不到答案。我的数据样本:

{"_id": id, "site": "site A", "weekday": 1, "value": 1}
{"_id": id, "site": "site B", "weekday": 2, "value": 0}
{"_id": id, "site": "site C", "weekday": 3, "value": 1}
{"_id": id, "site": "site A", "weekday": 2, "value": 0}
{"_id": id, "site": "site B", "weekday": 3, "value": -1}
{"_id": id, "site": "site C", "weekday": 2, "value": 1}
{"_id": id, "site": "site A", "weekday": 1, "value": -1}
{"_id": id, "site": "site B", "weekday": 3, "value": 1}

我需要的是:

对于单个站点，假设“站点 A”，我需要每个工作日的字典列表(总共 7 个)，其中“值”的数量大于 0、等于 0 和小于 0。全部按工作日排序。

所以我的输出应该是这样的:

{"weekday": 1, "greaterCount": x, "lesserCount": y, "zeroCount": z}
{"weekday": 2, "greaterCount": x, "lesserCount": y, "zeroCount": z}
{"weekday": 3, "greaterCount": x, "lesserCount": y, "zeroCount": z}
{"weekday": 4, "greaterCount": x, "lesserCount": y, "zeroCount": z}
{"weekday": 5, "greaterCount": x, "lesserCount": y, "zeroCount": z}
{"weekday": 6, "greaterCount": x, "lesserCount": y, "zeroCount": z}
{"weekday": 7, "greaterCount": x, "lesserCount": y, "zeroCount": z}

当然，不同工作日的greaterCount、lesserCount和zeroCount的值会有所不同，因为我很懒，所以我的示例输出中的每个字典中都有x、y和z。

Answer 1

您在这里基本上要寻找的是 $cond运算符(operator)。这是一个“三元”条件，其计算结果是从逻辑条件的 true/false 返回一个值。

在这种情况下，每个“逻辑”测试都会查看当前的“值”字段，并确定测试的 true 位置，例如 $gt是否向 $sum 返回正值或改为 0 值:

db.collection.aggregate([ 
    { "$group": {
        "_id": "$weekday",
        "greaterCount": {
            "$sum": {
                "$cond": [
                    { "$gt": [ "$value", 0 ] },
                    1,
                    0
                ]
            }
        },
        "lesserCount": {
            "$sum": {
                "$cond": [
                    { "$lt": [ "$value", 0 ] },
                    1,
                    0
                ]
            }
        },
        "zeroCount": {
            "$sum": {
                "$cond": [
                    { "$eq": [ "$value", 0 ] },
                    1,
                    0
                ]
            }
        }

    }}
])

sample 产生的结果:

{ "_id" : 3, "greaterCount" : 2, "lesserCount" : 1, "zeroCount" : 0 }
{ "_id" : 2, "greaterCount" : 1, "lesserCount" : 0, "zeroCount" : 2 }
{ "_id" : 1, "greaterCount" : 1, "lesserCount" : 1, "zeroCount" : 0 }