Categorize website visitors starting from the first occasion, based on if condition
0
Could you please help me with sql statement, preferreby it should work in big query. I have 3 columns userid, date, hostname. I need to create additional column - client_type on the following condition: when userid first time comes to hostname = "online-store.com" then from this date on client_type for this particular userid will be always "current_client" else "visitor".
For example, in the image (link attached) we have userid = 1 and 4 who had become "current client". User 4 was just a visitor, but after visiting hostname = "online-store.com" he will be always classified as "current client".enter image description here
google-bigquery intervals between dateadd asked Nov 18 at 18:39
Andrey
12
|
show 1 more comment
0
Could you please help me with sql statement, preferreby it should work in big query. I have 3 columns userid, date, hostname. I need to create additional column - client_type on the following condition: when userid first time comes to hostname = "online-store.com" then from this date on client_type for this particular userid will be always "current_client" else "visitor".
For example, in the image (link attached) we have userid = 1 and 4 who had become "current client". User 4 was just a visitor, but after visiting hostname = "online-store.com" he will be always classified as "current client".enter image description here
google-bigquery intervals between dateadd asked Nov 18 at 18:39
Andrey
12
Please edit your question to show a Minimal, Complete, and Verifiable example of the code and most importantly data that you are having problems with, then we can try to help with the your problem. You can also read How to Ask.
– Mikhail Berlyant
Nov 21 at 16:14
take a closer look especially to that part -Not all questions benefit from including code. But if your problem is with code you've written, you should include some. But don't just copy in your entire code! ..., it likely includes a lot of irrelevant details that readers will need to ignore when trying to reproduce the problem. Here are some guidelines: ...
– Mikhail Berlyant
Nov 21 at 16:16
Mikhail, thanks a lot for advice, I have rewritten my question and added a picture. Also my question has been modified a bit.
– Andrey
Nov 22 at 8:03
see the answer. please in your next/new questions - avoid using images and rather provide data examples as plain text so we can use it while helping you :o)
– Mikhail Berlyant
Nov 22 at 20:20
ok, =) understand
– Andrey
Nov 23 at 8:42
|
show 1 more comment
0
0
0
Could you please help me with sql statement, preferreby it should work in big query. I have 3 columns userid, date, hostname. I need to create additional column - client_type on the following condition: when userid first time comes to hostname = "online-store.com" then from this date on client_type for this particular userid will be always "current_client" else "visitor".
For example, in the image (link attached) we have userid = 1 and 4 who had become "current client". User 4 was just a visitor, but after visiting hostname = "online-store.com" he will be always classified as "current client".enter image description here
google-bigquery intervals between dateadd asked Nov 18 at 18:39
Andrey
12
Could you please help me with sql statement, preferreby it should work in big query. I have 3 columns userid, date, hostname. I need to create additional column - client_type on the following condition: when userid first time comes to hostname = "online-store.com" then from this date on client_type for this particular userid will be always "current_client" else "visitor".
For example, in the image (link attached) we have userid = 1 and 4 who had become "current client". User 4 was just a visitor, but after visiting hostname = "online-store.com" he will be always classified as "current client".enter image description here
google-bigquery intervals between dateadd google-bigquery intervals between dateadd asked Nov 18 at 18:39
Andrey
12
asked Nov 18 at 18:39
Andrey
12
edited Nov 22 at 8:26
asked Nov 18 at 18:39
Andrey
12
asked Nov 18 at 18:39
Andrey
12
asked Nov 18 at 18:39
Andrey
12
12
Please edit your question to show a Minimal, Complete, and Verifiable example of the code and most importantly data that you are having problems with, then we can try to help with the your problem. You can also read How to Ask.
– Mikhail Berlyant
Nov 21 at 16:14
take a closer look especially to that part -Not all questions benefit from including code. But if your problem is with code you've written, you should include some. But don't just copy in your entire code! ..., it likely includes a lot of irrelevant details that readers will need to ignore when trying to reproduce the problem. Here are some guidelines: ...
– Mikhail Berlyant
Nov 21 at 16:16
Mikhail, thanks a lot for advice, I have rewritten my question and added a picture. Also my question has been modified a bit.
– Andrey
Nov 22 at 8:03
see the answer. please in your next/new questions - avoid using images and rather provide data examples as plain text so we can use it while helping you :o)
– Mikhail Berlyant
Nov 22 at 20:20
ok, =) understand
– Andrey
Nov 23 at 8:42
|
show 1 more comment
Please edit your question to show a Minimal, Complete, and Verifiable example of the code and most importantly data that you are having problems with, then we can try to help with the your problem. You can also read How to Ask.
– Mikhail Berlyant
Nov 21 at 16:14
take a closer look especially to that part -Not all questions benefit from including code. But if your problem is with code you've written, you should include some. But don't just copy in your entire code! ..., it likely includes a lot of irrelevant details that readers will need to ignore when trying to reproduce the problem. Here are some guidelines: ...
– Mikhail Berlyant
Nov 21 at 16:16
Mikhail, thanks a lot for advice, I have rewritten my question and added a picture. Also my question has been modified a bit.
– Andrey
Nov 22 at 8:03
see the answer. please in your next/new questions - avoid using images and rather provide data examples as plain text so we can use it while helping you :o)
– Mikhail Berlyant
Nov 22 at 20:20
ok, =) understand
– Andrey
Nov 23 at 8:42
Please edit your question to show a Minimal, Complete, and Verifiable example of the code and most importantly data that you are having problems with, then we can try to help with the your problem. You can also read How to Ask.
– Mikhail Berlyant
Nov 21 at 16:14
Please edit your question to show a Minimal, Complete, and Verifiable example of the code and most importantly data that you are having problems with, then we can try to help with the your problem. You can also read How to Ask.
– Mikhail Berlyant
Nov 21 at 16:14
take a closer look especially to that part -
Not all questions benefit from including code. But if your problem is with code you've written, you should include some. But don't just copy in your entire code! ..., it likely includes a lot of irrelevant details that readers will need to ignore when trying to reproduce the problem. Here are some guidelines: ...– Mikhail Berlyant
Nov 21 at 16:16
take a closer look especially to that part -
Not all questions benefit from including code. But if your problem is with code you've written, you should include some. But don't just copy in your entire code! ..., it likely includes a lot of irrelevant details that readers will need to ignore when trying to reproduce the problem. Here are some guidelines: ...– Mikhail Berlyant
Nov 21 at 16:16
Mikhail, thanks a lot for advice, I have rewritten my question and added a picture. Also my question has been modified a bit.
– Andrey
Nov 22 at 8:03
Mikhail, thanks a lot for advice, I have rewritten my question and added a picture. Also my question has been modified a bit.
– Andrey
Nov 22 at 8:03
see the answer. please in your next/new questions - avoid using images and rather provide data examples as plain text so we can use it while helping you :o)
– Mikhail Berlyant
Nov 22 at 20:20
see the answer. please in your next/new questions - avoid using images and rather provide data examples as plain text so we can use it while helping you :o)
– Mikhail Berlyant
Nov 22 at 20:20
ok, =) understand
– Andrey
Nov 23 at 8:42
ok, =) understand
– Andrey
Nov 23 at 8:42
|
show 1 more comment
2 Answers
2
active
oldest
votes
0
Below is for BigQuery Standard SQL
#standardSQL
SELECT
userid, date, hostname,
IF(0 = COUNTIF(hostname = 'online-store.com') OVER(
PARTITION BY userid ORDER BY date ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW
), 'visitor', 'current_client') client_type
FROM `project.dataset.table`
You can test, play with above using dummy data you provided in your question
#standardSQL
WITH `project.dataset.table` AS (
SELECT 1 userid, DATE '2018-02-01' date, 'online-store.com' hostname UNION ALL
SELECT 2, '2018-02-01', 'other' UNION ALL
SELECT 3, '2018-02-01', 'other' UNION ALL
SELECT 4, '2018-02-01', 'other' UNION ALL
SELECT 1, '2018-02-01', 'other' UNION ALL
SELECT 1, '2018-04-07', 'other' UNION ALL
SELECT 4, '2018-04-08', 'online-store.com' UNION ALL
SELECT 5, '2018-04-08', 'other' UNION ALL
SELECT 6, '2018-04-08', 'other' UNION ALL
SELECT 4, '2018-04-08', 'other' UNION ALL
SELECT 8, '2018-04-08', 'other' UNION ALL
SELECT 1, '2018-07-07', 'other' UNION ALL
SELECT 1, '2018-11-22', 'online-store.com'
)
SELECT
userid, date, hostname,
IF(0 = COUNTIF(hostname = 'online-store.com') OVER(
PARTITION BY userid ORDER BY date ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW
), 'visitor', 'current_client') client_type
FROM `project.dataset.table`
ORDER BY date
with result
Row userid date hostname client_type
1 1 2018-02-01 online-store.com current_client
2 1 2018-02-01 other current_client
3 2 2018-02-01 other visitor
4 3 2018-02-01 other visitor
5 4 2018-02-01 other visitor
6 1 2018-04-07 other current_client
7 4 2018-04-08 online-store.com current_client
8 4 2018-04-08 other current_client
9 5 2018-04-08 other visitor
10 6 2018-04-08 other visitor
11 8 2018-04-08 other visitor
12 1 2018-07-07 other current_client
13 1 2018-11-22 online-store.com current_client
answered Nov 22 at 20:19
Mikhail Berlyant
55.1k43368
add a comment |
0
This should be good:
#standardSQL
with userdates as (
select userid, hostname, min(date) as mindate from `dataset.table` where hostname = 'online-store.com' group by userid, hostname
)
select u.userid, u.date, u.hostname, case when u.date >= ud.mindate then 'current_user' else 'visitor' end as client_type
from `dataset.table` u
left outer join userdates ud on u.userid = ud.userid
order by 1, 2
answered Nov 22 at 23:33
khan
1,85383052
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
oldest
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active
oldest
votes
0
Below is for BigQuery Standard SQL
#standardSQL
SELECT
userid, date, hostname,
IF(0 = COUNTIF(hostname = 'online-store.com') OVER(
PARTITION BY userid ORDER BY date ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW
), 'visitor', 'current_client') client_type
FROM `project.dataset.table`
You can test, play with above using dummy data you provided in your question
#standardSQL
WITH `project.dataset.table` AS (
SELECT 1 userid, DATE '2018-02-01' date, 'online-store.com' hostname UNION ALL
SELECT 2, '2018-02-01', 'other' UNION ALL
SELECT 3, '2018-02-01', 'other' UNION ALL
SELECT 4, '2018-02-01', 'other' UNION ALL
SELECT 1, '2018-02-01', 'other' UNION ALL
SELECT 1, '2018-04-07', 'other' UNION ALL
SELECT 4, '2018-04-08', 'online-store.com' UNION ALL
SELECT 5, '2018-04-08', 'other' UNION ALL
SELECT 6, '2018-04-08', 'other' UNION ALL
SELECT 4, '2018-04-08', 'other' UNION ALL
SELECT 8, '2018-04-08', 'other' UNION ALL
SELECT 1, '2018-07-07', 'other' UNION ALL
SELECT 1, '2018-11-22', 'online-store.com'
)
SELECT
userid, date, hostname,
IF(0 = COUNTIF(hostname = 'online-store.com') OVER(
PARTITION BY userid ORDER BY date ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW
), 'visitor', 'current_client') client_type
FROM `project.dataset.table`
ORDER BY date
with result
Row userid date hostname client_type
1 1 2018-02-01 online-store.com current_client
2 1 2018-02-01 other current_client
3 2 2018-02-01 other visitor
4 3 2018-02-01 other visitor
5 4 2018-02-01 other visitor
6 1 2018-04-07 other current_client
7 4 2018-04-08 online-store.com current_client
8 4 2018-04-08 other current_client
9 5 2018-04-08 other visitor
10 6 2018-04-08 other visitor
11 8 2018-04-08 other visitor
12 1 2018-07-07 other current_client
13 1 2018-11-22 online-store.com current_client
answered Nov 22 at 20:19
Mikhail Berlyant
55.1k43368
add a comment |
0
Below is for BigQuery Standard SQL
#standardSQL
SELECT
userid, date, hostname,
IF(0 = COUNTIF(hostname = 'online-store.com') OVER(
PARTITION BY userid ORDER BY date ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW
), 'visitor', 'current_client') client_type
FROM `project.dataset.table`
You can test, play with above using dummy data you provided in your question
#standardSQL
WITH `project.dataset.table` AS (
SELECT 1 userid, DATE '2018-02-01' date, 'online-store.com' hostname UNION ALL
SELECT 2, '2018-02-01', 'other' UNION ALL
SELECT 3, '2018-02-01', 'other' UNION ALL
SELECT 4, '2018-02-01', 'other' UNION ALL
SELECT 1, '2018-02-01', 'other' UNION ALL
SELECT 1, '2018-04-07', 'other' UNION ALL
SELECT 4, '2018-04-08', 'online-store.com' UNION ALL
SELECT 5, '2018-04-08', 'other' UNION ALL
SELECT 6, '2018-04-08', 'other' UNION ALL
SELECT 4, '2018-04-08', 'other' UNION ALL
SELECT 8, '2018-04-08', 'other' UNION ALL
SELECT 1, '2018-07-07', 'other' UNION ALL
SELECT 1, '2018-11-22', 'online-store.com'
)
SELECT
userid, date, hostname,
IF(0 = COUNTIF(hostname = 'online-store.com') OVER(
PARTITION BY userid ORDER BY date ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW
), 'visitor', 'current_client') client_type
FROM `project.dataset.table`
ORDER BY date
with result
Row userid date hostname client_type
1 1 2018-02-01 online-store.com current_client
2 1 2018-02-01 other current_client
3 2 2018-02-01 other visitor
4 3 2018-02-01 other visitor
5 4 2018-02-01 other visitor
6 1 2018-04-07 other current_client
7 4 2018-04-08 online-store.com current_client
8 4 2018-04-08 other current_client
9 5 2018-04-08 other visitor
10 6 2018-04-08 other visitor
11 8 2018-04-08 other visitor
12 1 2018-07-07 other current_client
13 1 2018-11-22 online-store.com current_client
answered Nov 22 at 20:19
Mikhail Berlyant
55.1k43368
add a comment |
0
0
0
Below is for BigQuery Standard SQL
#standardSQL
SELECT
userid, date, hostname,
IF(0 = COUNTIF(hostname = 'online-store.com') OVER(
PARTITION BY userid ORDER BY date ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW
), 'visitor', 'current_client') client_type
FROM `project.dataset.table`
You can test, play with above using dummy data you provided in your question
#standardSQL
WITH `project.dataset.table` AS (
SELECT 1 userid, DATE '2018-02-01' date, 'online-store.com' hostname UNION ALL
SELECT 2, '2018-02-01', 'other' UNION ALL
SELECT 3, '2018-02-01', 'other' UNION ALL
SELECT 4, '2018-02-01', 'other' UNION ALL
SELECT 1, '2018-02-01', 'other' UNION ALL
SELECT 1, '2018-04-07', 'other' UNION ALL
SELECT 4, '2018-04-08', 'online-store.com' UNION ALL
SELECT 5, '2018-04-08', 'other' UNION ALL
SELECT 6, '2018-04-08', 'other' UNION ALL
SELECT 4, '2018-04-08', 'other' UNION ALL
SELECT 8, '2018-04-08', 'other' UNION ALL
SELECT 1, '2018-07-07', 'other' UNION ALL
SELECT 1, '2018-11-22', 'online-store.com'
)
SELECT
userid, date, hostname,
IF(0 = COUNTIF(hostname = 'online-store.com') OVER(
PARTITION BY userid ORDER BY date ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW
), 'visitor', 'current_client') client_type
FROM `project.dataset.table`
ORDER BY date
with result
Row userid date hostname client_type
1 1 2018-02-01 online-store.com current_client
2 1 2018-02-01 other current_client
3 2 2018-02-01 other visitor
4 3 2018-02-01 other visitor
5 4 2018-02-01 other visitor
6 1 2018-04-07 other current_client
7 4 2018-04-08 online-store.com current_client
8 4 2018-04-08 other current_client
9 5 2018-04-08 other visitor
10 6 2018-04-08 other visitor
11 8 2018-04-08 other visitor
12 1 2018-07-07 other current_client
13 1 2018-11-22 online-store.com current_client
answered Nov 22 at 20:19
Mikhail Berlyant
55.1k43368
Below is for BigQuery Standard SQL
#standardSQL
SELECT
userid, date, hostname,
IF(0 = COUNTIF(hostname = 'online-store.com') OVER(
PARTITION BY userid ORDER BY date ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW
), 'visitor', 'current_client') client_type
FROM `project.dataset.table`
You can test, play with above using dummy data you provided in your question
#standardSQL
WITH `project.dataset.table` AS (
SELECT 1 userid, DATE '2018-02-01' date, 'online-store.com' hostname UNION ALL
SELECT 2, '2018-02-01', 'other' UNION ALL
SELECT 3, '2018-02-01', 'other' UNION ALL
SELECT 4, '2018-02-01', 'other' UNION ALL
SELECT 1, '2018-02-01', 'other' UNION ALL
SELECT 1, '2018-04-07', 'other' UNION ALL
SELECT 4, '2018-04-08', 'online-store.com' UNION ALL
SELECT 5, '2018-04-08', 'other' UNION ALL
SELECT 6, '2018-04-08', 'other' UNION ALL
SELECT 4, '2018-04-08', 'other' UNION ALL
SELECT 8, '2018-04-08', 'other' UNION ALL
SELECT 1, '2018-07-07', 'other' UNION ALL
SELECT 1, '2018-11-22', 'online-store.com'
)
SELECT
userid, date, hostname,
IF(0 = COUNTIF(hostname = 'online-store.com') OVER(
PARTITION BY userid ORDER BY date ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW
), 'visitor', 'current_client') client_type
FROM `project.dataset.table`
ORDER BY date
with result
Row userid date hostname client_type
1 1 2018-02-01 online-store.com current_client
2 1 2018-02-01 other current_client
3 2 2018-02-01 other visitor
4 3 2018-02-01 other visitor
5 4 2018-02-01 other visitor
6 1 2018-04-07 other current_client
7 4 2018-04-08 online-store.com current_client
8 4 2018-04-08 other current_client
9 5 2018-04-08 other visitor
10 6 2018-04-08 other visitor
11 8 2018-04-08 other visitor
12 1 2018-07-07 other current_client
13 1 2018-11-22 online-store.com current_client
answered Nov 22 at 20:19
Mikhail Berlyant
55.1k43368
answered Nov 22 at 20:19
Mikhail Berlyant
55.1k43368
answered Nov 22 at 20:19
Mikhail Berlyant
55.1k43368
answered Nov 22 at 20:19
Mikhail Berlyant
55.1k43368
55.1k43368
add a comment |
add a comment |
0
This should be good:
#standardSQL
with userdates as (
select userid, hostname, min(date) as mindate from `dataset.table` where hostname = 'online-store.com' group by userid, hostname
)
select u.userid, u.date, u.hostname, case when u.date >= ud.mindate then 'current_user' else 'visitor' end as client_type
from `dataset.table` u
left outer join userdates ud on u.userid = ud.userid
order by 1, 2
answered Nov 22 at 23:33
khan
1,85383052
add a comment |
0
This should be good:
#standardSQL
with userdates as (
select userid, hostname, min(date) as mindate from `dataset.table` where hostname = 'online-store.com' group by userid, hostname
)
select u.userid, u.date, u.hostname, case when u.date >= ud.mindate then 'current_user' else 'visitor' end as client_type
from `dataset.table` u
left outer join userdates ud on u.userid = ud.userid
order by 1, 2
answered Nov 22 at 23:33
khan
1,85383052
add a comment |
0
0
0
This should be good:
#standardSQL
with userdates as (
select userid, hostname, min(date) as mindate from `dataset.table` where hostname = 'online-store.com' group by userid, hostname
)
select u.userid, u.date, u.hostname, case when u.date >= ud.mindate then 'current_user' else 'visitor' end as client_type
from `dataset.table` u
left outer join userdates ud on u.userid = ud.userid
order by 1, 2
answered Nov 22 at 23:33
khan
1,85383052
This should be good:
#standardSQL
with userdates as (
select userid, hostname, min(date) as mindate from `dataset.table` where hostname = 'online-store.com' group by userid, hostname
)
select u.userid, u.date, u.hostname, case when u.date >= ud.mindate then 'current_user' else 'visitor' end as client_type
from `dataset.table` u
left outer join userdates ud on u.userid = ud.userid
order by 1, 2
answered Nov 22 at 23:33
khan
1,85383052
answered Nov 22 at 23:33
khan
1,85383052
answered Nov 22 at 23:33
khan
1,85383052
answered Nov 22 at 23:33
khan
1,85383052
1,85383052
add a comment |
add a comment |
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Please edit your question to show a Minimal, Complete, and Verifiable example of the code and most importantly data that you are having problems with, then we can try to help with the your problem. You can also read How to Ask.
– Mikhail Berlyant
Nov 21 at 16:14
take a closer look especially to that part -
Not all questions benefit from including code. But if your problem is with code you've written, you should include some. But don't just copy in your entire code! ..., it likely includes a lot of irrelevant details that readers will need to ignore when trying to reproduce the problem. Here are some guidelines: ...– Mikhail Berlyant
Nov 21 at 16:16
Mikhail, thanks a lot for advice, I have rewritten my question and added a picture. Also my question has been modified a bit.
– Andrey
Nov 22 at 8:03
see the answer. please in your next/new questions - avoid using images and rather provide data examples as plain text so we can use it while helping you :o)
– Mikhail Berlyant
Nov 22 at 20:20
ok, =) understand
– Andrey
Nov 23 at 8:42