I know it's not possible to nest aggregate functions. But I want to achieve something like this and quite confused about how to do this compromising performance.
SELECT
date,
count(CASE WHEN SUM(active_time) > 5 THEN user_id END) AS total_active_users,
count(CASE WHEN SUM(active_time) > 5 AND is_admin = true THEN user_id END) AS total_active_admin_users
FROM
(
SELECT date, user_id, user_name, active_time, is_admin FROM users
)
GROUP BY date
It's really appreciated if someone could suggest a way to achieve this.
Perhaps you want something like this:
select date,
sum(case when sum_active_time > 5 then 1 else 0 end) as total_active_users,
sum(case when sum_active_time > 5 and is_admin then 1 else 0 end) as total_active_admin_users
from (select u.*, sum(active_time) over (partition by user_id) as sum_active_time
from users
) u
group by date;
However, I would expect user_id to be unique in a table called users. That makes me wonder why you need to do a count or sum at all. So, you might want:
select date,
sum(case when active_time > 5 then 1 else 0 end) as total_active_users,
sum(case when active_time > 5 and is_admin then 1 else 0 end) as total_active_admin_users
from users
group by date;
SELECT date,
COUNT(user_id) as total_active_users,
COUNT(CASE WHEN is_admin = 1 THEN user_id END ) as total_active_admin_users
FROM (
SELECT date, is_admin, user_id
FROM users
GROUP BY date, is_admin, user_id
HAVING SUM(active_time) > 5
) t
GROUP BY date
Related
I have a code as below where I want to count number of first purchases for a given period of time. I have a column in my sales table where if the buyer is not a first time buyer, then is_first_purchase = 0
For example:
buyer_id = 456391 is already an existing buyer who made purchases on 2 different dates.
Hence is_first_purchase column will show as 0 as per below.
If i do a count() on is_first_purchase for this buyer_id = 456391 then it should return 0 instead of 2.
My query is as follows:
with first_purchases as
(select *,
case when is_first_purchase = 1 then 'Yes' else 'No' end as first_purchase
from sales)
select
count(case when first_purchase = 'Yes' then 1 else 0 end) as no_of_first_purchases
from first_purchases
where buyer_id = 456391
and date_id between '2021-02-01' and '2021-03-01'
order by 1 desc;
It returned the below which is not an intended output
Appreciate if someone can help explain how to exclude is_first_purchase = 0 from the count, thanks.
Because COUNT function count when the value isn't NULL (include 0), if you don't want to count, need to let CASE WHEN return NULL
There are two ways you can count as your expectation, one is SUM other is COUNT but remove the part of else 0
SUM(case when first_purchase = 'Yes' then 1 else 0 end) as no_of_first_purchases
COUNT(case when first_purchase = 'Yes' then 1 end) as no_of_first_purchases
From your question, I would combine CTE and main query as below
select
COUNT(case when is_first_purchase = 1 then 1 end) as no_of_first_purchases
from sales
where buyer_id = 456391
and date_id between '2021-02-01' and '2021-03-01'
order by 1 desc;
I think that you are using COUNT() when you want SUM().
with first_purchases as
(select *,
case when is_first_purchase = 1 then 'Yes' else 'No' end as first_purchase
from sales)
select
SUM(case when first_purchase = 'Yes' then 1 else 0 end) as no_of_first_purchases
from first_purchases
where buyer_id = 456391
and date_id between '2021-02-01' and '2021-03-01'
order by 1 desc;
You could simplify your query as:
SELECT COUNT(*) AS
FROM sales no_of_first_purchases
WHERE is_first_purchase = 1
AND buyer_id = 456391
AND date_id BETWEEN '2021-02-01' AND '2021-03-01'
ORDER BY 1 DESC;
It is better to avoid the use of functions like IF and CASE when it can be done with WHERE.
The simplest approach for Trino (f.k.a. Presto SQL) is to use an aggregate with a filter:
count(name) FILTER (WHERE first_purchase = 'Yes') AS no_of_first_purchases
I'm trying to figure out if what I'm trying to do is possible. Instead of resorting to multiple queries on a table, I wanted to group the records by business date and id then group by the id and select one date for a field and another date for the other field.
SELECT
*
{AMOUNT FROM DATE}
{AMOUNT FROM OTHER DATE}
FROM (
SELECT
date,
id,
SUM(amount) AS amount
FROM
table
GROUP BY id, date
AS subquery
GROUP BY id
It seems that you're looking to do a pivot query. I usually use cross tabs for this. Based on the query you posted, it could look like:
SELECT
id,
SUM(CASE WHEN date = '20190901' THEN amount ELSE 0 END) AmountFromSept01,
SUM(CASE WHEN date = '20191001' THEN amount ELSE 0 END) AmountFromOct01
FROM (
SELECT
date,
id,
SUM(amount) AS amount
FROM
table
GROUP BY id, date
)AS subquery
GROUP BY id;
You could also use a CTE.
WITH CTE AS(
SELECT
date,
id,
SUM(amount) AS amount
FROM
table
GROUP BY id, date
)
SELECT
id,
SUM(CASE WHEN date = '20190901' THEN amount ELSE 0 END) AmountFromSept01,
SUM(CASE WHEN date = '20191001' THEN amount ELSE 0 END) AmountFromOct01
FROM CTE
GROUP BY id;
Or even be a rebel and do the operation directly.
SELECT
id,
SUM(CASE WHEN date = '20190901' THEN amount ELSE 0 END) AmountFromSept01,
SUM(CASE WHEN date = '20191001' THEN amount ELSE 0 END) AmountFromOct01
FROM CTE
GROUP BY id;
However, some people have tested for performance and found that pre-aggregating can improve performance.
If I understand you correctly, then you're just trying to pivot, but only with two particular dates:
select id,
date1 = sum(iif(date = '2000-01-01', amount, null)),
date2 = sum(iif(date = '2000-01-02', amount, null))
from [table]
group by id
I have the table in PostgreSQL DB
Need to calculate SUM of counts for each event_type (example for 4 and 1)
When I use query like this
SELECT account_id, date,
CASE
WHEN event_type = 1 THEN SUM(count)
ELSE null
END AS shows,
CASE
WHEN event_type = 4 THEN SUM(count)
ELSE null
END AS clicks
FROM widgetstatdaily WHERE account_id = 272 AND event_type = 1 OR event_type = 4 GROUP BY account_id, date, event_type ORDER BY date
I receive this table
With <null> fields. It's because I have event_type in select and I need to GROUP BY on it.
How I can make query to receive grouped by account_id and date result without null's in cells? Like (first row)
272 2018-03-28 00:00:00.000000 57 2
May be I can group it after receiving result
You need conditional aggregation and some other fixes. Try this:
SELECT account_id, date,
SUM(CASE WHEN event_type = 1 THEN count END) as shows,
SUM(CASE WHEN event_type = 4 THEN count END) as clicks
FROM widgetstatdaily
WHERE account_id = 272 AND
event_type IN (1, 4)
GROUP BY account_id, date
ORDER BY date;
Notes:
The CASE expression should be an argument to the SUM().
The ELSE NULL is redundant. The default without an ELSE is NULL.
The logic in the WHERE clause is probably not what you intend. That is fixed using IN.
try its
SELECT account_id, date,
SUM(CASE WHEN event_type = 1 THEN count else 0 END) as shows,
SUM(CASE WHEN event_type = 4 THEN count else 0 END) as clicks
FROM widgetstatdaily
WHERE account_id = 272 AND
event_type IN (1, 4)
GROUP BY account_id, date
ORDER BY date;
I'm trying to analyze a funnel using event data in Redshift and have difficulties finding an efficient query to extract that data.
For example, in Redshift I have:
timestamp action user id
--------- ------ -------
2015-05-05 12:00 homepage 1
2015-05-05 12:01 product page 1
2015-05-05 12:02 homepage 2
2015-05-05 12:03 checkout 1
I would like to extract the funnel statistics. For example:
homepage_count product_page_count checkout_count
-------------- ------------------ --------------
100 50 25
Where homepage_count represent the distinct number of users who visited the homepage, product_page_count represents the distinct numbers of users who visited the homepage after visiting the homepage, and checkout_count represents the number of users who checked out after visiting the homepage and the product page.
What would be the best query to achieve that with Amazon Redshift? Is it possible to do with a single query?
I think the best method might be to add flags to the data for the first visit of each type for each user and then use these for aggregation logic:
select sum(case when ts_homepage is not null then 1 else 0 end) as homepage_count,
sum(case when ts_productpage > ts_homepage then 1 else 0 end) as productpage_count,
sum(case when ts_checkout > ts.productpage and ts.productpage > ts.homepage then 1 else 0 end) as checkout_count
from (select userid,
min(case when action = 'homepage' then timestamp end) as ts_homepage,
min(case when action = 'product page' then timestamp end) as ts_productpage,
min(case when action = 'checkout' then timestamp end) as ts_checkout
from table t
group by userid
) t
The above answer is very much correct . I have modified it for people using it for AWS Mobile Analytics and Redshift.
select sum(case when ts_homepage is not null then 1 else 0 end) as homepage_count,
sum(case when ts_productpage > ts_homepage then 1 else 0 end) as productpage_count,
sum(case when ts_checkout > ts_productpage and ts_productpage > ts_homepage then 1 else 0 end) as checkout_count
from (select client_id,
min(case when event_type = 'App Launch' then event_timestamp end) as ts_homepage,
min(case when event_type = 'SignUp Success' then event_timestamp end) as ts_productpage,
min(case when event_type = 'Start Quiz' then event_timestamp end) as ts_checkout
from awsma.v_event
group by client_id
) ts;
Just in case more precise model required: when product page can be opened twice. First time before home page and second one after. This case usually should be considered as conversion as well.
Redshift SQL query:
SELECT
COUNT(
DISTINCT CASE WHEN cur_homepage_time IS NOT NULL
THEN user_id END
) Step1,
COUNT(
DISTINCT CASE WHEN cur_homepage_time IS NOT NULL AND cur_productpage_time IS NOT NULL
THEN user_id END
) Step2,
COUNT(
DISTINCT CASE WHEN
cur_homepage_time IS NOT NULL AND cur_productpage_time IS NOT NULL AND cur_checkout_time IS NOT NULL
THEN user_id END
) Step3
FROM (
SELECT
user_id,
timestamp,
COALESCE(homepage_time,
LAG(homepage_time) IGNORE NULLS OVER(PARTITION BY user_id
ORDER BY time)
) cur_homepage_time,
COALESCE(productpage_time,
LAG(productpage_time) IGNORE NULLS OVER(PARTITION BY distinct_id
ORDER BY time)
) cur_productpage_time,
COALESCE(checkout_time,
LAG(checkout_time) IGNORE NULLS OVER(PARTITION BY distinct_id
ORDER BY time)
) cur_checkout_time
FROM
(
SELECT
timestamp,
user_id,
(CASE WHEN event = 'homepage'
THEN timestamp END) homepage_time,
(CASE WHEN event = 'product page'
THEN timestamp END) productpage_time,
(CASE WHEN event = 'checkout'
THEN timestamp END) checkout_time
FROM events
WHERE timestamp > '2016-05-01' AND timestamp < '2017-01-01'
ORDER BY user_id, timestamp
) event_times
ORDER BY user_id, timestamp
) event_windows
This query fills each row's cur_homepage_time, cur_productpage_time and cur_checkout_time with recent timestamp of event occurrences. So in case for some specific time (read row) event occured then particular column is not NULL.
More info here.
how to calculate count based on rows?
SOURCE TABLE
each employee can take 2 days off
Employee-----First_Day_Off-----Second_Day_Off
1------------10/21/2009--------12/6/2009
2------------09/3/2009--------12/6/2009
3------------09/3/2009--------NULL
4
5
.
.
.
Now i need a table that shows the dates and number of people taking off on that day
Date---------First_Day_Off-------Second_Day_Off
10/21/2009---1-------------------0
12/06/2009---1--------------------1
09/3/2009----2--------------------0
Any ideas?
Oracle 9i+, using Subquery Factoring (WITH):
WITH sample AS (
SELECT a.employee,
a.first_day_off AS day_off,
1 AS day_number
FROM YOUR_TABLE a
WHERE a.first_day_off IS NOT NULL
UNION ALL
SELECT b.employee,
b.second_day_off,
2 AS day_number
FROM YOUR_TABLE b
WHERE b.second_day_off IS NOT NULL)
SELECT s.day_off AS date,
SUM(CASE WHEN s.day_number = 1 THEN 1 ELSE 0 END) AS first_day_off,
SUM(CASE WHEN s.day_number = 2 THEN 1 ELSE 0 END) AS second_day_off
FROM sample s
GROUP BY s.day_off
Non Subquery Version
SELECT s.day_off AS date,
SUM(CASE WHEN s.day_number = 1 THEN 1 ELSE 0 END) AS first_day_off,
SUM(CASE WHEN s.day_number = 2 THEN 1 ELSE 0 END) AS second_day_off
FROM (SELECT a.employee,
a.first_day_off AS day_off,
1 AS day_number
FROM YOUR_TABLE a
WHERE a.first_day_off IS NOT NULL
UNION ALL
SELECT b.employee,
b.second_day_off,
2 AS day_number
FROM YOUR_TABLE b
WHERE b.second_day_off IS NOT NULL) s
GROUP BY s.day_off
It is a bit awkward to handle these queries, since you have days off stored in different columns. A better layout would be to have something like
EMPLOYEE_ID DAY_OFF
Then you would have multiple rows if an employee took multiple days off
EMPLOYEE_ID DAY_OFF
1 10/21/2009
1 12/6/2009
2 09/3/2009
2 12/6/2009
3 09/3/2009
...
In that case, you could find out how many days off each person took by using the following query:
SELECT EMPLOYEE_ID, COUNT(*) AS NUM_DAYS_OFF FROM DAYS_OFF_TABLE GROUP BY EMPLOYEE_ID
And the number of people who took days off on each date like this:
SELECT DAY_OFF, COUNT(*) AS NUM_PEOPLE FROM DAYS_OFF_TABLE GROUP BY DAY_OFF
But I digress...
You can try to use an SQL CASE statement to help with this:
SELECT Employee, CASE
WHEN First_Day_Off is NULL AND Second_Day_Off is NULL THEN 0
WHEN First_Day_Off is NOT NULL AND Second_Day_Off is NULL THEN 1
WHEN First_Day_Off is NULL AND Second_Day_Off is NOT NULL THEN 1
ELSE 2
END AS NUM_DAYS_OFF
FROM DAYS_OFF_TABLE
(note that you may need to change around the syntax slightly depending on your database.
Getting dates and number of people who took off on that day might be more complicated.
I don't know if this would work, but you can try it:
SELECT
Date_Off,
COUNT(*) AS Num_People
FROM
(SELECT
First_Day_Off, COUNT(*) AS Num_People FROM DAYS_OFF_TABLE WHERE First_Day_Off IS NOT NULL GROUP BY First_Day_Off
UNION
SELECT Second_Day_Off, COUNT(*) AS Num_People FROM DAYS_OFF_TABLE WHERE Second_Day_Off IS NOT NULL GROUP BY Second_Day_Off)
GROUP BY
Num_People