因此，我写了一个查询来计算保留率，新生和回国学生的增长率。下面的代码返回类似于此的结果。
Row visit_month student_type numberofstd growth
1 2013 new 574 null
2 2014 new 220 -62%
3 2014 retained 442 245%
4 2015 new 199 -10%
5 2015 retained 533 21%
6 2016 new 214 8%
7 2016 retained 590 11%
8 2016 returning 1 -100%
我尝试过的查询。
with visit_log AS (
SELECT studentid,
cast(substr(session, 1, 4) as numeric) as visit_month,
FROM abaresult
GROUP BY 1,
2
ORDER BY 1,
2),
time_lapse_2 AS (
SELECT studentid,
Visit_month,
lag(visit_month, 1) over (partition BY studentid ORDER BY studentid, visit_month) lag
FROM visit_log),
time_diff_calculated_2 AS (
SELECT studentid,
visit_month,
lag,
visit_month - lag AS time_diff
FROM time_lapse_2),
student_categorized AS (
SELECT studentid,
visit_month,
CASE
WHEN time_diff=1 THEN 'retained'
WHEN time_diff>1 THEN 'returning'
WHEN time_diff IS NULL THEN 'new'
END AS student_type,
FROM time_diff_calculated_2)
SELECT visit_month,
student_type,
Count(distinct studentid) as numberofstd,
ROUND(100 * (COUNT(student_type) - LAG(COUNT(student_type), 1) OVER (ORDER BY student_type)) / LAG(COUNT(student_type), 1) OVER (ORDER BY student_type),0) || '%' AS growth
FROM student_categorized
group by 1,2
order by 1,2
上面的查询根据上次会话student_type类别的数字计算保留率，新增率和返回率。
我正在寻找一种方法，而不是根据每个类别的访问量/每月访问量来计算这些数字。有没有办法可以做到这一点？
我正在尝试获取与此类似的表
Row visit_month student_type totalstd numberofstd growth
1 2013 new 574 574 null
2 2014 new 662 220 62%
3 2014 retained 662 442 22%
4 2015 new 732 199 10%
5 2015 retained 732 533 21%
6 2016 new 804 214 8%
7 2016 retained 804 590 11%
8 2016 returning 804 1 100%
注意：
totalstd是每节课的学生总数，由new + retention + returning获得。
假设增长计算。
请帮助！谢谢。
虽然我没有您的源数据，但是我依靠您共享的查询和输出结果。
我创建了一些额外的代码以输出所需的结果。我想指出，我没有访问BigQuery的编译的权限，因为我没有数据。因此，我试图防止自己查询的任何可能的错误。此外，**之间的查询保持不变，并从您的代码中复制。下面是代码（它是您的代码和我创建的额外位的组合）：
#*****************************************************************
with visit_log AS (
SELECT studentid,
cast(substr(session, 1, 4) as numeric) as visit_month,
FROM abaresult
GROUP BY 1,
2
ORDER BY 1,
2),
time_lapse_2 AS (
SELECT studentid,
Visit_month,
lag(visit_month, 1) over (partition BY studentid ORDER BY studentid, visit_month) lag
FROM visit_log),
time_diff_calculated_2 AS (
SELECT studentid,
visit_month,
lag,
visit_month - lag AS time_diff
FROM time_lapse_2),
student_categorized AS (
SELECT studentid,
visit_month,
CASE
WHEN time_diff=1 THEN 'retained'
WHEN time_diff>1 THEN 'returning'
WHEN time_diff IS NULL THEN 'new'
END AS student_type,
FROM time_diff_calculated_2)
#**************************************************************
#Code I added
#each unique visit_month will have its count
WITH total_stud AS (
SELECT visit_month, count(distinct studentid) as totalstd FROM visit_log
GROUP BY 1
ORDER BY visit_month
),
#After you have your student_categorized temp table, create another one
#It will have the count of the number of students per visit_month per student_type
number_std_monthType AS (
SELECT visit_month,student_type, Count(distinct studentid) as numberofstd from student_categorized
GROUP BY 1, 2
),
#You will have one row per combination of visit_month and student_type
student_categorized2 AS(
SELECT DISTINCT visit_month,student_type FROM student_categorized2
GROUP BY 1,2
),
#before calculation, create the table with all the necessary data
#you have the desired table without the growth
#notice that I used two keys to join t1 and t3 so the results are correct
final_t AS (
SELECT t1.visit_month,
t1.student_type,
t2.totalstd as totalstd,
t3.numberofstd
FROM student_categorized2 t1
LEFT JOIN total_stud AS t2 ON t1.visit_month = t2.visit_month
LEFT JOIN number_std_monthType t3 ON (t1.visit_month = t3.visit_month and t1.student_type = t3.student_type)
ORDER BY
)
#Now all the necessary values to calculate growth are in the temp table final_t
SELECT visit_month, student_type, totalstd, numberofstd,
ROUND(100 * (totalstd - LAG(totalstd) OVER (PARTITION BY visit_month ORDER BY visit_month ASC) /LAG(totalstd) OVER (PARTITION BY visit_month ORDER BY visit_month ASC) || '%' AS growth
FROM final_t
注意，一旦在不同的临时表]中计算了每个计数，我便会在最终表中使用LEFT JOIN进行计数。另外，我没有使用您的最终SELECT声明。
如果您对代码有任何疑问，请随时提出。