SQL为具有类别总计的不同行加上非不同的权重

Question

我想在个人可以为多个单元格做出贡献的情况下对一些加权调查数据进行交叉制表。 面临的挑战是确保小计和总计不会重复计算。

我可以使用类似于如何SUM DISTINCT行的解决方案的方法来获取单个单元格值而不是总计？ 或与其他列不同 。 我正在尝试使用Oracle CUBE语句以很好的方式获取总计。

这是一个婴儿的例子。 假设我们根据他们拥有的宠物和他们的爱好来计算人数。 问题是一个人可能有一个以上的宠物，或一个以上的爱好。 我们需要转换这组单位记录：

person_id, weight
1, 10
2, 10
3, 12

person_id, pet
1, "cat"
1, "dog"
2, "cat"
3, "cat"

person_id, hobby
1, "chess"
2, "chess"
2, "skydiving"
3, "skydiving"

进入这对表：

    Unweighted count

      | chess | skydiving | total
------+-------+-----------+--------
cat   |  2    |  2        | 3
------+-------+-----------+--------
dog   |  1    |  0        | 1
------+-------+-----------+--------
total |  2    |  2        | 3      


Weighted count

      | chess | skydiving | total
------+-------+-----------+--------
cat   |  20   |  22       | 32
------+-------+-----------+--------
dog   |  10   |  0        | 10
------+-------+-----------+--------
total |  20   |  22       | 32     

请注意，“cat”行的未加权总计为3，而不是2 + 2 = 4，因为人数2在两个不同的位置计算。 只有三个不同的人参与这一行。 其他总数也是如此。

请注意，“猫，象棋”的加权总数为20 = 10 + 10，因为两个不同的人每人对该单元格贡献10。

请注意，加权表的总计为32.这来自1和2人，每人贡献10人，3人贡献12人。总计不仅仅是所有单个单元格的总和！

对于未加权的计数，我可以通过以下方式获得所有细胞计数和总计：

CREATE TABLE weights(person_id INTEGER, weight INTEGER);
INSERT INTO weights(person_id,weight) VALUES (1,10);
INSERT INTO weights(person_id,weight) VALUES (2,10);
INSERT INTO weights(person_id,weight) VALUES (3,12);

CREATE TABLE pets(person_id INTEGER, pet VARCHAR(3));
INSERT INTO pets(person_id,pet) VALUES (1,'cat');
INSERT INTO pets(person_id,pet) VALUES (1,'dog');
INSERT INTO pets(person_id,pet) VALUES (2,'cat');
INSERT INTO pets(person_id,pet) VALUES (3,'cat');

CREATE TABLE hobbies(person_id INTEGER, hobby VARCHAR(9));
INSERT INTO hobbies(person_id,hobby) VALUES (1,'chess');
INSERT INTO hobbies(person_id,hobby) VALUES (2,'chess');
INSERT INTO hobbies(person_id,hobby) VALUES (2,'skydiving');
INSERT INTO hobbies(person_id,hobby) VALUES (3,'skydiving');

SELECT pet, hobby, COUNT(DISTINCT weights.person_id)
FROM weights JOIN pets on weights.person_id=pets.person_ID
JOIN hobbies on weights.person_id=hobbies.person_id
GROUP BY CUBE(pet, hobby);

COUNT(DISTINCT ...)和CUBE给出了正确的总数。

对于加权计数，如果我尝试相同的想法：

SELECT pet, hobby, SUM(DISTINCT weight)
FROM weights JOIN pets on weights.person_id=pets.person_ID
JOIN hobbies on weights.person_id=hobbies.person_id
GROUP BY CUBE(pet, hobby);

“猫，棋”细胞来到10而不是20，因为人1和2都具有相同的重量。 删除“不同”关键字意味着单个细胞计数是正确的，但总数是错误的（它产生总计52，其应该是32，因为人1和2在总数中被重复计算）。

有什么建议？

Answer 1

试试这个，下面给出了正确的结果，但它是最简化的一个

SELECT pet, hobby, SUM(weight)
FROM weights JOIN pets on weights.person_id=pets.person_ID
JOIN hobbies on weights.person_id=hobbies.person_id
GROUP BY pet, hobby
UNION
SELECT pet, NULL, SUM(weight)
FROM weights JOIN pets on weights.person_id=pets.person_ID
GROUP BY pet
UNION
SELECT NULL, hobby, SUM(weight)
FROM weights JOIN hobbies on weights.person_id=hobbies.person_id
GROUP BY hobby
UNION
SELECT SUM(weight)
FROM weights

仍然在单一选择

Answer 2

您可以使用嵌套查询执行此操作，其中内部查询指定从行到表单元格的映射（即哪些记录在每个表格单元格的范围内），外部查询指定要应用的汇总函数：

SELECT pet, hobby, COUNT(1), SUM(weight) FROM
(SELECT pet, hobby, weights.person_ID, weight
FROM weights JOIN pets on weights.person_id=pets.person_ID
JOIN hobbies on weights.person_id=hobbies.person_id
GROUP BY CUBE(pet, hobby), weights.person_ID, weight)
GROUP BY pet, hobby;

结果

旁白：你也可以不使用CUBE运算符编写内部查询，但它更麻烦：

WITH
    pet_cube_map as (SELECT DISTINCT pet, NULL as pet_cubed FROM pets UNION ALL SELECT DISTINCT pet, pet as pet_cubed FROM pets),
    hobby_cube_map as (SELECT DISTINCT hobby, NULL as hobby_cubed FROM hobbies UNION ALL SELECT DISTINCT hobby, hobby as hobby_cubed FROM hobbies)
SELECT DISTINCT pet_cubed as pet, hobby_cubed as hobby, weights.person_ID, weight
FROM weights
    JOIN pets on weights.person_ID=pets.person_ID
    JOIN pet_cube_map on pets.pet=pet_cube_map.pet
    JOIN hobbies on weights.person_ID=hobbies.person_ID
    JOIN hobby_cube_map on hobbies.hobby=hobby_cube_map.hobby
;

Answer 3

我想你需要像这样做一些数学：

;WITH t AS (
    SELECT 
        p.pet,
        SUM(DISTINCT CASE WHEN h.hobby = 'chess' THEN POWER(2,h.person_id) ELSE 0 END) chess,
        SUM(DISTINCT CASE WHEN h.hobby = 'skydiving' THEN POWER(2,h.person_id) ELSE 0 END) skydiving,
        SUM(DISTINCT POWER(2,h.person_id)) total
    FROM 
        hobbies h
        LEFT JOIN
        pets p ON h.person_id = p.person_id
    GROUP BY
        p.pet
    UNION ALL 
    SELECT 
        'total',
        SUM(DISTINCT CASE WHEN h.hobby = 'chess' THEN POWER(2,h.person_id) ELSE 0 END),
        SUM(DISTINCT CASE WHEN h.hobby = 'skydiving' THEN POWER(2,h.person_id) ELSE 0 END),
        SUM(DISTINCT POWER(2,h.person_id))
    FROM 
        hobbies h
), w(person_id, weight) as (
    SELECT POWER(2,person_id), weight
    FROM weights
), cte(person_id, weight) AS (
    SELECT *
    FROM w
    UNION ALL
    SELECT w1.person_id + w2.person_id, w1.weight + w2.weight
    FROM cte w1 JOIN w w2 ON w2.person_id > w1.person_id
)
SELECT 
    pet,
    COALESCE((SELECT cte.weight FROM cte WHERE cte.person_id = t.chess), 0) AS chess,
    COALESCE((SELECT cte.weight FROM cte WHERE cte.person_id = t.skydiving), 0) AS skydiving,
    COALESCE((SELECT cte.weight FROM cte WHERE cte.person_id = t.total), 0) AS total
FROM t;

没有立方体，静态的方式，有点脏。 但我只是在SQL Server中测试它;）。

---

这可以是立方体版本（未经测试）：

;With t as (
SELECT  h.hobby, p.pet, POWER(2,h.person_id) weight
FROM    hobbies h 
JOIN    pets p
ON      h.person_id = p.person_id
JOIN    weights w
ON      h.person_id = w.person_id
), w(person_id, weight) as (
    SELECT POWER(2,person_id), weight
    FROM weights
), cte(person_id, weight) AS (
    SELECT *
    FROM w
    UNION ALL
    SELECT w1.person_id + w2.person_id, w1.weight + w2.weight
    FROM cte w1 JOIN w w2 ON w2.person_id > w1.person_id
), c as (
SELECT  
    hobby, pet, SUM(DISTINCT weight) person_id
FROM    t
GROUP BY CUBE(hobby, pet)
)
SELECT c.hobby, c.pet, cte.weight
FROM c JOIN
    cte ON c.person_id = cte.person_id;