Mongo 不同的聚合

Mongo Distinct Aggegation(Mongo 不同的聚合)

问题描述

我正在尝试使用聚合框架在 mongo 中执行组计数，但结果并不完全符合预期.

I'm trying to use the aggregation framework to perform group counts in mongo but the results are not exactly as expected.

考虑下面的集合

 $people->insert(array("user_id" => "1", "day" => "Monday", 'age' => 18));
 $people->insert(array("user_id" => "3", "day" => "Monday", 'age' => 24));
 $people->insert(array("user_id" => "1", "day" => "Monday", 'age' => 18));
 $people->insert(array("user_id" => "1", "day" => "Monday", 'age' => 18));
 $people->insert(array("user_id" => "2", "day" => "Monday", 'age' => 25));
 $people->insert(array("user_id" => "4", "day" => "Monday", 'age' => 33));
 $people->insert(array("user_id" => "1", "day" => "Tuesday", 'age' => 18));
 $people->insert(array("user_id" => "2", "day" => "Tuesday", 'age' => 25));
 $people->insert(array("user_id" => "1", "day" => "Wednesday", 'age' => 18));
 $people->insert(array("user_id" => "2", "day" => "Thursday", 'age' => 25));
 $people->insert(array("user_id" => "1", "day" => "Friday", 'age' => 18));

我使用下面的查询尝试计算一周中每一天的不同条目(user_id)的数量.

I use the query below try count the number of distinct entries (user_id) for each day of the week.

$query = array(
        array(
            '$project' => array(
                'user_id' =>1,
                'day' =>1,
            ),
        ),
        array(
            '$group' => array(
                '_id'  => array(
                    'user_id' => '$user_id',
                 'day' => '$day'),
                'count' => array('$sum' => 1),
            )
        ));

所以对于上面的集合，结果应该是

So for the collection above the results should be

Monday = 3     Tues = 2,     Wed = 1,     Thur = 1 and    Friday = 1

但它不会将所有 DISTINCT users_id 的总数归为一天，而是每天为我提供每个现有 user_id 的总数.`

but it does not group the totals of all DISTINCT users_id under a day, and instead for each day it gives me a total for each existing user_id.`

结果(不完整)

     [result] => Array
    (
        [0] => Array
            (
                [_id] => Array
                    (
                        [user_id] => 1
                        [day] => Friday
                    )

                [count] => 1
            )

        [1] => Array
            (
                [_id] => Array
                    (
                        [user_id] => 1
                        [day] => Wednesday
                    )

                [count] => 1
            )

        [2] => Array
            (
                [_id] => Array
                    (
                        [user_id] => 2
                        [day] => Tuesday
                    )

                [count] => 1
            )
... ... ...

有人可以帮我过滤每日总计，使其仅包括每天不同的总计

Can someone help me to filter the daily totals so that it only includes distinct totals per day

我查看了 $unwind 但不能真的让我头疼.`

I have looked at the $unwind but couldn't really get my head around it. `

推荐答案

如果我对问题的理解是正确的，那么您要解决的问题是

If I'm understanding the question right, what you're trying to get at is

 totals of all DISTINCT users_id under a day

或者据我了解:每天唯一用户 ID 的计数.

Or as I understand it: Count of unique user_ids per day.

为此，您可以使用已有的组并删除计数，以便您只有一个唯一的 _id.user_id 和 _id.day 值:

For that, you could take the group you already have and cut out the count so that you just have a unique _id.user_id and _id.day value:

'$group' => array(
            '_id'  => array(
                'user_id' => '$user_id',
                'day' => '$day'
            )
        )

然后将其传送到另一个 $group 语句，该语句计算每天的文档数量，因为每个唯一的 user_id/day组合:

'$group' => array(
            '_id'  => '$_id.day',
            'count' => array('$sum' => 1)
        )