I have the following table
order_id
created_at
customer_id
1
2020-01-02
11
2
2020-02-03
12
3
2020-02-03
11
I would like to add a column "customer_orders_count" that will assign the number of orders that a customer passed to each transaction, ie obtain this table :
order_id
created_at
customer_id
customer_orders_count
1
2020-01-02
11
1
2
2020-02-03
12
1
2
2020-02-03
11
2
My problem it's I can't find how to calculated a local "customer_orders_count" dependind on each order, I only managed to add a column with the global "customer_orders_count" and for example for the first row order_id=1 I'll get customer_orders_count=2 whereas I'll like to be 1.
Does anyone has and idea ?
Use cumulative count:
with mytable as (
select 1 as order_id, date '2020-01-02' as created_at, 11 as customer_id union all
select 2, '2020-02-03', 12 union all
select 3 , '2020-02-03', 11
)
select *, count(*) over (partition by customer_id order by created_at) as customer_orders_count
from mytable
order by order_id
Use row_number():
select t.*,
row_number() over (partition by customer_id order by created_at) as customer_order_count
from t;
This is subtly different from using a cumulative count(). This version guarantees that the numbers for a given customer are never duplicated, even when the dates are the same. A cumulative count has no such guarantee.
Related
In PostgreSQL I have an orders table that represents orders made by customers of a store:
SELECT * FROM orders
order_id
customer_id
value
created_at
1
1
188.01
2020-11-24
2
2
25.74
2022-10-13
3
1
159.64
2022-09-23
4
1
201.41
2022-04-01
5
3
357.80
2022-09-05
6
2
386.72
2022-02-16
7
1
200.00
2022-01-16
8
1
19.99
2020-02-20
For a specified time range (e.g. 2022-01-01 to 2022-12-31), I need to find the following:
Average 1st order value
Average 2nd order value
Average 3rd order value
Average 4th order value
E.g. the 1st purchases for each customer are:
for customer_id 1, order_id 8 is their first purchase
customer 2, order 6
customer 3, order 5
So, the 1st-purchase average order value is (19.99 + 386.72 + 357.80) / 3 = $254.84
This needs to be found for the 2nd, 3rd and 4th purchases also.
I also need to find the average time between purchases:
order 1 to order 2
order 2 to order 3
order 3 to order 4
The final result would ideally look something like this:
order_number
AOV
av_days_since_last_order
1
254.84
0
2
300.00
28
3
322.22
21
4
350.00
20
Note that average days since last order for order 1 would always be 0 as it's the 1st purchase.
Thanks.
select order_number
,round(avg(value),2) as AOV
,coalesce(round(avg(days_between_orders),0),0) as av_days_since_last_order
from
(
select *
,row_number() over(partition by customer_id order by created_at) as order_number
,created_at - lag(created_at) over(partition by customer_id order by created_at) as days_between_orders
from t
) t
where created_at between '2022-01-01' and '2022-12-31'
group by order_number
order by order_number
order_number
aov
av_days_since_last_order
1
372.26
0
2
25.74
239
3
200.00
418
4
201.41
75
5
159.64
175
Fiddle
Im suppose it should be something like this
WITH prep_data AS (
SELECT order_id,
cuntomer_id,
ROW_NUMBER() OVER(PARTITION BY order_id, cuntomer_id ORDER BY created_at) AS pushcase_num,
created_at,
value
FROM pushcases
WHERE created_at BETWEEN :date_from AND :date_to
), prep_data2 AS (
SELECT pd1.order_id,
pd1.cuntomer_id,
pd1.pushcase_num
pd2.created_at - pd1.created_at AS date_diff,
pd1.value
FROM prep_data pd1
LEFT JOIN prep_data pd2 ON (pd1.order_id = pd2.order_id AND pd1.cuntomer_id = pd2.cuntomer_id AND pd1.pushcase_num = pd2.pushcase_num+1)
)
SELECT order_id,
cuntomer_id,
pushcase_num,
avg(value) AS avg_val,
avg(date_diff) AS avg_date_diff
FROM prep_data2
GROUP BY pushcase_num
I have the following table, where ID is the unique identifier. An can move from category to category, both up and down. My table records each day an ID stays in a given category. I am trying to identify the start date and the end date of an ID in a given category. The problem is that an ID can move up a category, and move back down to its original category after a certain number of days. Here is my table as an example with only 1 ID:
ID Category Date
1 1 2021-01-01
1 1 2021-01-02
...
1 1 2021-01-24
1 2 2021-01-25
...
1 2 2021-02-15
1 1 2021-02-16
...
1 1 2021-04-20
1 2 2021-04-21
When I try to get the MIN(DATE) and MAX(DATE) and group by the category and ID, it shows me that the account was in Category 1 from 2021-01-01 to 2021-04-20, and in Category 2 from 02-25 to 04-21. I am trying to track the movements of the file in each bucket step by step, meaning in my ideal result, the movements of the account will be tracked as:
ID Category StartDate EndDate
1 1 2021-01-01 2021-01-24
1 2 2021-01-25 2021-02-15
1 1 2021-02-16 2021-04-20
1 2 2021-04-21 NULL (or GETDATE())
How can I achieve this result? Any help would be appreciated. I tried using the RANK() function but because the table records every single day, it seems useless.
This is a type of gaps-and-islands problem that is most easily solved using the difference of row numbers:
select id, category, min(date), max(date)
from (select t.*,
row_number() over (partition by id order by date) as seqnum,
row_number() over (partition by id, category order by date) as seqnum_2
from t
) t
group by id, category, (seqnum - seqnum_2);
Actually, the difference of row numbers is only simplest because you have not specified the database. You can just subtract a sequence of numbers from the date to get a constant that defines each group. That looks like:
select id, category, min(date), max(date)
from (select t.*,
row_number() over (partition by id, category order by date) as seqnum
from t
) t
group by id, category, date - seqnum * interval '1 day';
However, the date arithmetic varies by database.
In postgres, I want to output the persons who have the highest no. of "discussed" requests for each month, irrespective of the year i.e. there should be 12 outputs.
ID PERSON REQUEST DATE
4 datanoise opened 2010-09-02
5 marsuboss opened 2010-09-02
6 m3talsmith opened 2010-09-06
7 sferik opened 2010-09-08
8 sferik opened 2010-09-09
8 dtrasbo discussed 2010-09-09
8 brianmario discussed 2010-09-09
8 sferik discussed 2010-09-09
9 rsim opened 2011-09-09
.....more tuples to follow
*This is just a small part of the databse. also assume that the dataset is big enough that all months are represented in the date column.
Test data:
CREATE TEMPORARY TABLE foo( id SERIAL PRIMARY KEY, name INTEGER NOT NULL,
dt DATE NULL, request BOOL NOT NULL );
INSERT INTO foo (name,dt,request) SELECT random()*1000,
'2010-01-01'::DATE+('1 DAY'::INTERVAL)*(random()*3650), random()>0.5
FROM generate_series(1,100000) n;
SELECT * FROM foo LIMIT 10;
id | name | dt | request
----+------+------------+---------
1 | 110 | 2014-11-05 | f
2 | 747 | 2015-03-12 | t
3 | 604 | 2014-09-26 | f
4 | 211 | 2011-12-14 | t
5 | 588 | 2016-12-15 | f
6 | 96 | 2012-02-19 | f
7 | 17 | 2018-09-18 | t
8 | 591 | 2018-02-15 | t
9 | 370 | 2015-07-28 | t
10 | 844 | 2019-05-16 | f
Now you have to get the count per name and month, then get the max count, but that won't give you the name that has the maximum, which requires joining back with the previous result. In order to do the group by only once, it is done in a CTE:
WITH totals AS (
SELECT EXTRACT(month FROM dt) mon, name, count(*) cnt FROM foo
WHERE request=true GROUP BY name,mon
)
SELECT * FROM
(SELECT mon, max(cnt) cnt FROM totals GROUP BY mon) x
JOIN totals USING (mon,cnt);
If several names have the same maximum count, they will be returned both. To keep only one, you can use DISTICT ON:
WITH (same as above)
SELECT DISTINCT ON (mon) * FROM
(SELECT mon, max(cnt) cnt FROM totals GROUP BY mon) x
JOIN totals USING (mon,cnt) ORDER BY mon,name;
You can also use DISTINCT ON to keep only one row per month, specified by the ORDER clause, in this cas by count desc, so it keeps the highest count.
SELECT DISTINCT ON (mon) * FROM (
SELECT EXTRACT(month FROM dt) mon, name, count(*) cnt FROM foo
WHERE request=true GROUP BY name,mon
)x ORDER BY mon, cnt DESC;
...or you could hack an argmax() function by sticking the primary key into an array passed to max(), which means it will return the id of the row which has the maximum value:
SELECT mon, cntid[1] cnt, name FROM
(SELECT mon, max(ARRAY[cnt,id]) cntid FROM (
SELECT EXTRACT(month FROM dt) mon, name, count(*) cnt, min(id) id FROM foo
WHERE request=true GROUP BY name,mon
) x GROUP BY mon)y
JOIN foo ON (foo.id=cntid[2]);
Which one will be faster?...
given your table is named t01 and the colum date is date1 (and in string format):
create temp table t02 as
select extract(month from CAST(date1 as date)) as month, person, count(*) nb from t01 where request = 'discussed' group by 1, 2 ;
create temp table t03 as
select month, max(nb) max_nb from t02 group by 1 ;
the result is :
select month , person from t02 a natural join t03 b where a.nb = b.max_nb;
https://rextester.com/BYMM84335[ : run here]1
I would recommend distinct on. If you want to combine all the months into a single "uber-month":
select distinct on (extract(month from date)) person, extract(month from date), count(*) as num_discussed
from t
where request = 'discussed'
group by person, extract(month from date)
order by extract(month from date), num_discussed desc;
Distinct on is a very handy Postgres extension. It returns on row per "group", which is defined by the expressions in parentheses. The row is the "first" one determined by the order by clause.
If you want the highest month regardless of year:
select distinct on (extract(month from date)) person, date_trunc('month', date), count(*) as num_discussed
from t
where request = 'discussed'
group by person, date_trunc('month', date)
order by extract(month from date), num_discussed desc;
I have the following table,
id status price date
2 complete 10 2020-01-01 10:10:10
2 complete 20 2020-02-02 10:10:10
2 complete 10 2020-03-03 10:10:10
3 complete 10 2020-04-04 10:10:10
4 complete 10 2020-05-05 10:10:10
Required output,
id status_count price ratio
2 0 0 0
2 1 10 0
2 2 30 0.33
I am looking to add the price for previous row. Row 1 is 0 because it has no previous row value.
Find ratio ie 10/30=0.33
You can use analytical function ROW_NUMBER and SUM as follows:
SELECT
id,
ROW_NUMBER() OVER (PARTITION BY id ORDER BY date) - 1 AS status_count,
COALESCE(SUM(price) OVER (PARTITION BY id ORDER BY date), 0) - price as price
FROM yourTable;
DB<>Fiddle demo
I think you want something like this:
SELECT
id,
COUNT(*) OVER (PARTITION BY id ORDER BY date) - 1 AS status_count,
COALESCE(SUM(price) OVER (PARTITION BY id
ORDER BY date ROWS BETWEEN
UNBOUNDED PRECEDING AND 1 PRECEDING), 0) price
FROM yourTable;
Demo
Please also check another method:
with cte
as(*,ROW_NUMBER() OVER (PARTITION BY id ORDER BY date) - 1 AS status_count,
SUM(price) OVER (PARTITION BY id ORDER BY date) ss from yourTable)
select id,status_count,isnull(ss,0)-price price
from cte
I have the following table:
id partid orderdate qty price
1 10 01/01/2017 10 3
2 10 02/01/2017 5 9
3 11 01/01/2017 0.5 0.001
4 145 02/01/2017 5 18
5 10 12/12/2016 8 7
6 10 05/07/2010 81 7.5
Basically I want to compare the most recent purchasing of parts to the other purchasing of the same part in a period of 24 months. For that matter compare id=2 to id = 1,5.
I want to check if the price of the latest orderdate (per part) is larger than the average price of that part in the last 24 months.
So first I need to calculate the avg price:
partid avgprice
10 (3+9+7)/3=6.33 (7.5 is out of range)
11 0.001
145 18
I also need to know the latest orderdate of each part:
id partid
2 10
3 11
4 145
and then I need to check if id=2, id=3, id=6 (latest purchases) are bigger than the average. If they are I need to return their partid.
So I should have something like this:
id partid avgprice lastprice
2 10 6.33 9
3 11 0.001 0.001
4 145 18 18
Finally I need to return partid=10 since 9>6.33
Now to my questions...
I'm not sure how I can find the latest order in PostgreSQL.
I tried:
select id, distinct partid,orderdate
from table
where orderdate> current_date - interval '24 months'
order by orderdate desc
This gives :
ERROR: syntax error at or near "distinct".
I'm a bit of a lost here. I know what I want to do but I cant translate it to SQL. Any one can help?
Get the avarage per part and the last order per price and join these:
select
lastorder.id,
lastorder.partid,
lastorder.orderdate,
lastorder.price as lastprice,
avgorder.price as avgprice
from
(
select
partid,
avg(price) as price
from mytable
where orderdate >= current_date - interval '24 months'
group by partid
) avgorder
join
(
select distinct on (partid)
id,
partid,
orderdate,
price
from mytable
order by partid, orderdate desc
) lastorder on lastorder.partid = avgorder.partid
and lastorder.price > avgorder.price;
This can be solved without distinct (which is heavy on the DB anyways):
with avg_price as (
select partid, avg(price) as price
from table
where orderdate> current_date - interval '24 months'
group by partid
)
select f.id, f.partid, av.price, f.price
from (
select id, partid, orderdate, price, rank() over (partition by partid order by orderdate desc)
from table
) as f
join avg_price av on f.partid = av.partid
where f.rank = 1
and av.price < f.price