累计非重复计数

Question

我正在查询以获取每日 uid 的累计不同计数。

示例：假设有 2 个 uid (100,200) 出现在日期 2016-11-01，并且它们也在第二天出现，新的 uid 300 (100,200,300) 出现在 2016-11-02 此时我希望商店累积计数为 3 而不是 5，因为（用户 ID 100 和 200 已在过去一天出现）。

Input table:

    date            uid         
2016-11-01          100
2016-11-01          200
2016-11-01          300
2016-11-01          400         
2016-11-02          100
2016-11-02          200                 
2016-11-03          300
2016-11-03          400
2016-11-03          500
2016-11-03          600
2016-11-04          700

Expected query result:

date            daily_cumulative_count
2016-11-01              4   
2016-11-02              4
2016-11-03              6
2016-11-04              7

到目前为止，我能够获得每天的累计不同计数，但它也包括前一天的不同 uid。

SELECT 
  date, 
  SUM(count) OVER (
    ORDER BY date ASC 
    ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW
  )
FROM (
  SELECT 
    date, 
    COUNT(DISTINCT uid) AS count
  FROM sample_table
  GROUP by 1
)ORDER BY date DESC;

任何形式的帮助将不胜感激。

Answer 1

最简单的方法：

SELECT *, count(*) over (order by fst_date ) cum_uids
  FROM (
SELECT uid, min(date) fst_date FROM t GROUP BY uid
 ) t

或者类似的东西

Answer 2

WITH firstseen AS (
  SELECT uid, MIN(date) date
  FROM sample_table
  GROUP BY 1
)
SELECT DISTINCT date, COUNT(uid) OVER (ORDER BY date) daily_cumulative_count 
FROM firstseen
ORDER BY 1

使用

SELECT DISTINCT

因为

(date, COUNT(uid))

会重复很多次。

说明：对于每个日期

dt

，它会计算从最早日期到

dt

的uid，因为我们指定了

ORDER BY date

，并且默认为

BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW

。

Answer 3

您可以使用

exists

检查之前的任何日期是否存在 ID。然后获取运行总和并找到每个组的最大值，这将为您提供每日不同的累积计数。

select dt, max(col) as daily_cumulative_count
from (select t1.*, 
      sum(case when not exists (select 1 from t where t1.dt > dt and id = t1.uid) then 1 else 0 end) over(order by dt) col
      from t t1) x 
group by dt

Answer 4

请尝试以下操作...

SELECT date AS date
       COUNT( uid ) AS daily_cumulative_count
FROM ( SELECT leftTable.date AS date,
              rightTable.uid AS uid
       FROM sample_table AS leftTable
       JOIN sample_table AS rightTable ON leftTable.date >= rightTable.date
       GROUP BY leftTable.date,
                rightTable.uid
     ) AS allUIDSForDateFinder
GROUP BY date;

此语句首先将

sample_table

的一个实例连接到另一个实例，使得

leftTable

中的每条记录都与

rightTable

中具有较早或相等

date

值的每条记录的副本相关联。这有效地将一个列表附加到所有

date

值中的每个

uid

，直到并包括该

date

值。

通过使用

date

，将生成的数据集细化为独特的

uid

和

GROUP BY

组合。

来自子查询

allUIDSForDateFinder

的精炼数据集随后按查询主体按

date

进行分组，并执行与每个组关联的

COUNT()

值。如果您有任何疑问或意见，请随时发表评论。

Answer 5

版本：

uid

版本选择：

WITH t as (
    SELECT uid
           , min(dt) fst_date 
    FROM input_table 
    GROUP BY uid
)
SELECT DISTINCT fst_date
                , count(uid) over (order by fst_date ) daily_cumulative_count
FROM t

Answer 6

SELECT DISTINCT fst_date
                , count(uid) over (order by fst_date ) daily_cumulative_count
FROM (
    SELECT uid
           , min(dt) fst_date 
    FROM input_table 
    GROUP BY uid
     ) t

解决方案：

CREATE TABLE MyTable
(
    fecha   VARCHAR(512),
    uid     INT
);

INSERT INTO MyTable (fecha, uid ) VALUES ('1/11/2016', '100');
INSERT INTO MyTable (fecha, uid ) VALUES ('1/11/2016', '200');
INSERT INTO MyTable (fecha, uid ) VALUES ('1/11/2016', '300');
INSERT INTO MyTable (fecha, uid ) VALUES ('1/11/2016', '400');
INSERT INTO MyTable (fecha, uid ) VALUES ('2/11/2016', '100');
INSERT INTO MyTable (fecha, uid ) VALUES ('2/11/2016', '200');
INSERT INTO MyTable (fecha, uid ) VALUES ('3/11/2016', '300');
INSERT INTO MyTable (fecha, uid ) VALUES ('3/11/2016', '400');
INSERT INTO MyTable (fecha, uid ) VALUES ('3/11/2016', '500');
INSERT INTO MyTable (fecha, uid ) VALUES ('3/11/2016', '600');
INSERT INTO MyTable (fecha, uid ) VALUES ('4/11/2016', '700');
INSERT INTO MyTable (fecha, uid ) VALUES ('5/11/2016', '700');
INSERT INTO MyTable (fecha, uid ) VALUES ('6/11/2016', '700');
INSERT INTO MyTable (fecha, uid ) VALUES ('7/11/2016', '700');
INSERT INTO MyTable (fecha, uid ) VALUES ('8/11/2016', '700');
INSERT INTO MyTable (fecha, uid ) VALUES ('8/11/2016', '900');

您可以在这里快速测试一下

Answer 7

SELECT 
    t1.fecha, 
    COUNT(DISTINCT t2.uid) as daily_cumulative_count
FROM 
    MyTable t1
INNER JOIN 
    MyTable t2
ON 
    t1.fecha >= t2.fecha
GROUP BY 
    t1.fecha
ORDER BY 
    t1.fecha