i have the following code,
SELECT
years_month_count.day_date,
years_month_count.year_date,
years_month_count.month_date,
years_month_count.no_of_customers_day,
sum(years_month_count.no_of_customers_day) OVER (PARTITION BY year_date ORDER BY day_date) AS no_of_customers_ytd
FROM (
SELECT
DATE(date) as day_date,
DATE_PART('year',date) as year_date,
DATE_PART('month',date) as month_date,
count(prepare_first_buyer.person_id) as no_of_customers_day
FROM (
SELECT
DATE(bestelldatum),
person_id,
ROW_NUMBER() OVER (PARTITION BY person_id ORDER BY person_id)
FROM ani.bestellung
) prepare_first_buyer
WHERE row_number=1
GROUP BY DATE(date), DATE_PART('year',date),DATE_PART('month',date)
ORDER BY DATE(date), DATE_PART('year',date),DATE_PART('month',date)
) years_month_count
the Output looks like this:
day_date
year_date
month_date
no_of_customers_day
no_of_Customers_ytd
2017-04-04
2017
4
6
6
2017-04-05
2017
4
4
10
...
...
...
...
...
...
...
...
...
...
and so on.
The no_of_customers_ytd will be set to zero at the beginning of every new year (1.January).
But what i need is to set it to zero on a special date, lets say 1.June every year.
So i need a sum from everything between the 1.June till 30.March for every Year.
Thx for the help.
step-by-step demo:db<>fiddle
SELECT
*,
SUM(value) OVER (PARTITION BY -- 4
date_part('year', -- 3
the_date - interval '5 months' -- 2
)
)
FROM t
WHERE date_part('month', the_date)::int NOT BETWEEN 4 AND 5 -- 1
Filter all dates you are not requiring. In your example all dates with months 4 and 5
Shift your date range start to the beginning of the year. In your example you have to shift: year-06-01 to year-01-01, so you need to subtract 5 months. Because your date range never exceeds a year, all your relevant data now has the same year, which makes a great group criterion
Extract the year part to use it as group/partition criterion
Do your calculation on this criterion
Related
I am trying to optimize the below query to help fetch all customers in the last three months who have a monthly order frequency +4 for the past three months.
Customer ID
Feb
Mar
Apr
0001
4
5
6
0002
3
2
4
0003
4
2
3
In the above table, the customer with Customer ID 0001 should only be picked, as he consistently has 4 or more orders in a month.
Below is a query I have written, which pulls all customers with an average purchase frequency of 4 in the last 90 days, but not considering there is a consistent purchase of 4 or more last three months.
Query:
SELECT distinct lines.customer_id Customer_ID, (COUNT(lines.order_id)/90) PurchaseFrequency
from fct_customer_order_lines lines
LEFT JOIN product_table product
ON lines.entity_id= product.entity_id
AND lines.vendor_id= product.vendor_id
WHERE LOWER(product.country_code)= "IN"
AND lines.date >= DATE_SUB(CURRENT_DATE() , INTERVAL 90 DAY )
AND lines.date < CURRENT_DATE()
GROUP BY Customer_ID
HAVING PurchaseFrequency >=4;
I tried to use window functions, however not sure if it needs to be used in this case.
I would sum the orders per month instead of computing the avg and then retrieve those who have that sum greater than 4 in the last three months.
Also I think you should select your interval using "month(CURRENT_DATE()) - 3" instead of using a window of 90 days. Of course if needed you should handle the case of when current_date is jan-feb-mar and in that case go back to oct-nov-dec of the previous year.
I'm not familiar with Google BigQuery so I can't write your query but I hope this helps.
So I've found the solution to this using WITH operator as below:
WITH filtered_orders AS (
select
distinct customer_id ID,
extract(MONTH from date) Order_Month,
count(order_id) CountofOrders
from customer_order_lines` lines
where EXTRACT(YEAR FROM date) = 2022 AND EXTRACT(MONTH FROM date) IN (2,3,4)
group by ID, Order_Month
having CountofOrders>=4)
select distinct ID
from filtered_orders
group by ID
having count(Order_Month) =3;
Hope this helps!
An option could be first count the orders by month and then filter users which have purchases on all months above your threshold:
WITH ORDERS_BY_MONTH AS (
SELECT
DATE_TRUNC(lines.date, MONTH) PurchaseMonth,
lines.customer_id Customer_ID,
COUNT(lines.order_id) PurchaseFrequency
FROM fct_customer_order_lines lines
LEFT JOIN product_table product
ON lines.entity_id= product.entity_id
AND lines.vendor_id= product.vendor_id
WHERE LOWER(product.country_code)= "IN"
AND lines.date >= DATE_SUB(CURRENT_DATE() , INTERVAL 90 DAY )
AND lines.date < CURRENT_DATE()
GROUP BY PurchaseMonth, Customer_ID
)
SELECT
Customer_ID,
AVG(PurchaseFrequency) AvgPurchaseFrequency
FROM ORDERS_BY_MONTH
GROUP BY Customer_ID
HAVING COUNT(1) = COUNTIF(PurchaseFrequency >= 4)
#EDIT - Following the comments, I rephrase my question
I have a BigQuery table that i want to use to get some KPI of my application.
In this table, I save each create or update as a new line in order to keep a better history.
So I have several times the same data with a different state.
Example of the table :
uuid |status |date
––––––|–––––––––––|––––––––––
3 |'inactive' |2018-05-12
1 |'active' |2018-05-10
1 |'inactive' |2018-05-08
2 |'active' |2018-05-08
3 |'active' |2018-05-04
2 |'inactive' |2018-04-22
3 |'inactive' |2018-04-18
We can see that we have multiple value of each data.
What I would like to get:
I would like to have the number of current 'active' entry (So there must be no 'inactive' entry with the same uuid after). And to complicate everything, I need this total per day.
So for each day, the amount of 'active' entries, including those from previous days.
So with this example I should have this result :
date | actives
____________|_________
2018-05-02 | 0
2018-05-03 | 0
2018-05-04 | 1
2018-05-05 | 1
2018-05-06 | 1
2018-05-07 | 1
2018-05-08 | 2
2018-05-09 | 2
2018-05-10 | 3
2018-05-11 | 3
2018-05-12 | 2
Actually i've managed to get the good amount of actives for one day. But my problem is when i want the results for each days.
What I've tried:
I'm stuck with two solutions that each return a different error.
First solution :
WITH
dates AS(
SELECT GENERATE_DATE_ARRAY(
DATE_SUB(CURRENT_DATE(), INTERVAL 6 MONTH), CURRENT_DATE(), INTERVAL 1 DAY)
arr_dates )
SELECT
i_date date,
(
SELECT COUNT(uuid)
FROM (
SELECT
uuid, status, date,
RANK() OVER(PARTITION BY uuid ORDER BY date DESC) rank
FROM users
WHERE
PARSE_DATE("%Y-%m-%d", FORMAT_DATETIME("%Y-%m-%d",date)) <= i_date
)
WHERE
status = 'active'
and rank = 1
## rank is the condition which causes the error
) users
FROM
dates, UNNEST(arr_dates) i_date
ORDER BY i_date;
The SELECT with the RANK() OVER correctly returns the users with a rank column that allow me to know which entry is the last for each uuid.
But when I try this, I got a :
Correlated subqueries that reference other tables are not supported unless they can be de-correlated, such as by transforming them into an efficient JOIN. because of the rank = 1 condition.
Second solution :
WITH
dates AS(
SELECT GENERATE_DATE_ARRAY(
DATE_SUB(CURRENT_DATE(), INTERVAL 6 MONTH), CURRENT_DATE(), INTERVAL 1 DAY)
arr_dates )
SELECT
i_date date,
(
SELECT
COUNT(t1.uuid)
FROM
users t1
WHERE
t1.date = (
SELECT MAX(t2.date)
FROM users t2
WHERE
t2.uuid = t1.uuid
## Here that's the i_date condition which causes problem
AND PARSE_DATE("%Y-%m-%d", FORMAT_DATETIME("%Y-%m-%d", t2.date)) <= i_date
)
AND status='active' ) users
FROM
dates,
UNNEST(arr_dates) i_date
ORDER BY i_date;
Here, the second select is working too and correctly returning the number of active user for a current day.
But the problem is when i try to use i_date to retrieve datas among the multiple days.
And Here i got a LEFT OUTER JOIN cannot be used without a condition that is an equality of fields from both sides of the join. error...
Which solution is more able to succeed ? What should i change ?
And, if my way of storing the data isn't good, how should i proceed in order to keep a precise history ?
Below is for BigQuery Standard SQL
#standardSQL
SELECT date, COUNT(DISTINCT uuid) total_active
FROM `project.dataset.table`
WHERE status = 'active'
GROUP BY date
-- ORDER BY date
Update to address your "rephrased" question :o)
Below example is using dummy data from your question
#standardSQL
WITH `project.dataset.users` AS (
SELECT 3 uuid, 'inactive' status, DATE '2018-05-12' date UNION ALL
SELECT 1, 'active', '2018-05-10' UNION ALL
SELECT 1, 'inactive', '2018-05-08' UNION ALL
SELECT 2, 'active', '2018-05-08' UNION ALL
SELECT 3, 'active', '2018-05-04' UNION ALL
SELECT 2, 'inactive', '2018-04-22' UNION ALL
SELECT 3, 'inactive', '2018-04-18'
), dates AS (
SELECT day FROM UNNEST((
SELECT GENERATE_DATE_ARRAY(MIN(date), MAX(date))
FROM `project.dataset.users`
)) day
), active_users AS (
SELECT uuid, status, date first, DATE_SUB(next_status.date, INTERVAL 1 DAY) last FROM (
SELECT uuid, date, status, LEAD(STRUCT(status, date)) OVER(PARTITION BY uuid ORDER BY date ) next_status
FROM `project.dataset.users` u
)
WHERE status = 'active'
)
SELECT day, COUNT(DISTINCT uuid) actives
FROM dates d JOIN active_users u
ON day BETWEEN first AND IFNULL(last, day)
GROUP BY day
-- ORDER BY day
with result
Row day actives
1 2018-05-04 1
2 2018-05-05 1
3 2018-05-06 1
4 2018-05-07 1
5 2018-05-08 2
6 2018-05-09 2
7 2018-05-10 3
8 2018-05-11 3
9 2018-05-12 2
I think this -- or something similar -- will do what you want:
SELECT day,
coalesce(running_actives, 0) - coalesce(running_inactives, 0)
FROM UNNEST(GENERATE_DATE_ARRAY(DATE('2015-05-11'), DATE('2018-06-29'), INTERVAL 1 DAY)
) AS day left join
(select date, sum(countif(status = 'active')) over (order by date) as running_actives,
sum(countif(status = 'active')) over (order by date) as running_inactives
from t
group by date
) a
on a.date = day
order by day;
The exact solution depends on whether the "inactive" is inclusive of the day (as above) or takes effect the next day. Either is handled the same way, by using cumulative sums of actives and inactives and then taking the difference.
In order to get data on all days, this generates the days using arrays and unnest(). If you have data on all days, that step may be unnecessary
Dear Stack Overflow community,
I am looking for the patient id where the two consecutive dates after the very first one are less than 7 days.
So differences between 2nd and 1st date <= 7 days
and differences between 3rd and 2nd date <= 7 days
Example:
ID Date
1 9/8/2014
1 9/9/2014
1 9/10/2014
2 5/31/2014
2 7/20/2014
2 9/8/2014
For patient 1, the two dates following it are less than 7 days apart.
For patient 2 however, the following date are more than 7 days apart (50 days).
I am trying to write an SQL query that just output the patient id "1".
Thanks for your help :)
You want to use lead(), but this is complicated because you want this only for the first three rows. I think I would go for:
select t.*
from (select t.*,
lead(date, 1) over (partition by id order by date) as next_date,
lead(date, 2) over (partition by id order by date) as next_date_2,
row_number() over (partition by id order by date) as seqnum
from t
) t
where seqnum = 1 and
next_date <= date + interval '7' day and
next_date2 <= next_date + interval '7' day;
You can try using window function lag()
select * from
(
select id,date,lag(date) over(order by date) as prevdate
from tablename
)A where datediff(day,date,prevdate)<=7
I am trying to get the following query on Google Merchandise Store public dataset in BigQuery:
Date
Number of distinct users
Running sum of the number of distinct users in the last 30 days
For eg (I used 3 days in the example for simplicity):
date distinct_users distinct_users_3days
15/07/2018 8 15
14/07/2018 2 12
13/07/2018 5 20
12/07/2018 5 15
11/07/2018 10 10
...
This is my current SQL code which gets the first two columns, but I can't figure out how to get the running sum:
SELECT
date,
COUNT(DISTINCT(fullVisitorId)) as daily_active_user
FROM
`bigquery-public-data.google_analytics_sample.ga_sessions_2017*`
WHERE
_table_suffix BETWEEN "0101"
AND "0715"
GROUP BY
date
Any help is appreciated! :)
I managed to figure out the answer to my question so I would like to share with the others who may encounter this problem in future.
The SQL code is:
SELECT
date,
COUNT(DISTINCT(fullVisitorId)) as daily_active_user,
SUM(count(Distinct(fullVisitorId))) OVER (ORDER BY date ROWS BETWEEN 29 PRECEDING AND CURRENT ROW) AS monthly_active_user
FROM
`bigquery-public-data.google_analytics_sample.ga_sessions_2017*`,
unnest(hits) as h
WHERE
_table_suffix BETWEEN "0101" AND "0715"
GROUP BY
date
This gives a column which sums the distinct users in a 30 day window.
Please try the following query for 3 days (SQL server 2014 )-:
SELECT date,COUNT(DISTINCT(fullVisitorId)) as daily_active_user,sum(COUNT(DISTINCT(fullVisitorId))) over (PARTITION BY null ORDER BY date desc ROWS
BETWEEN CURRENT ROW AND 2 FOLLOWING) AS distinct_users_3days FROM YOUR_TABLE_NAME WHERE _table_suffix BETWEEN '0101' AND '715' GROUP BY date
For 30 days-:
SELECT
date,COUNT(DISTINCT(fullVisitorId)) as daily_active_user,
sum(COUNT(DISTINCT(fullVisitorId))) over (PARTITION BY null ORDER BY date desc ROWS
BETWEEN CURRENT ROW AND 29 FOLLOWING) AS distinct_users_3days
FROM YOUR_TABLE_NAME
WHERE _table_suffix
BETWEEN '0101' AND '715'
GROUP BY date
Looking to compute a moving sum day by day over a date range. i.e. Looking to sum all values greater than or equal to the date but do it row by row. I know that a window function is needed, but need some help with the actual function.
** I need to compute the sum greater than each date in a row. Notice on 2017-08-02 I do not count the value from the day before
Example data:
2017-08-1, 1
2017-08-2, 5
2017-08-3, 4
2017-08-4, 3
2017-08-5, 2
Desired Result:
2017-08-1, 15
2017-08-2, 14
2017-08-3, 9
2017-08-4, 5
2017-08-5, 2
Here is what I have to produce this data.
SELECT DATE_TRUNC('day', created_at),
COUNT(*)
FROM table
GROUP BY 1
ORDER BY 1 DESC
Just use cumulative sums:
SELECT DATE_TRUNC('day', created_at),
COUNT(*),
SUM(COUNT(*)) OVER (ORDER BY DATE_TRUNC('day', created_at) DESC) as sum_greater_than
FROM table
GROUP BY 1
ORDER BY 1 DESC;