Determine which year-month has the highest and lowest value [duplicate] - sql

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Here's my first query to shows the number of customers added per year-month
select count(name) AS CUSTOMER,
extract(year from create_date) as yr,
extract(month from create_date) as mon
from x
group by extract(year from create_date),
extract(month from create_date)
order by yr desc, mon desc;
CUSTOMER
YR
MON
3
2019
07
4
2015
02
100
2014
09
3
2014
04
I tried the query
SELECT MAX(count(*))
FROM x
GROUP BY create_date;
in the results I have;
MAX(COUNT(*))
100
need to see the year and month in the result.
How to do this?


The way I understood the question, you'd use rank analytic function in a subquery (or a CTE) and fetch rows whose count is either minimum or maximum. Something like this:
with temp as
(select to_char(create_date, 'yyyymm') yyyy_mm,
count(*) cnt,
--
rank() over (order by count(*) asc) rnk_min,
rank() over (order by count(*) desc) rnk_max
from x
group by to_char(create_date, 'yyyymm')
)
select yyyy_mm,
cnt
from temp
where rnk_min = 1
or rnk_max = 1;


You can use two levels of aggregation and put the results all in one row using keep (which implements a "first" aggregation function):
select max(num_customers) as max_num_customers,
max(yyyymm) keep (dense_rank first order by num_customers desc) as max_yyyymm,
min(num_customers) as max_num_customers,
max(yyyymm) keep (dense_rank first order by num_customers asc) as in_yyyymm,
from (select to_char(create_date, 'YYYY-MM') as yyyymm,
count(*) AS num_customers
from x
group by to_char(create_date, 'YYYY-MM'
) ym


From Oracle 12, you can use FETCH FIRST ROW ONLY to get the row with the highest number of customers (and, in the case of ties, the latest date):
SELECT count(name) AS CUSTOMER,
extract(year from create_date) as yr,
extract(month from create_date) as mon
FROM x
GROUP BY
extract(year from create_date),
extract(month from create_date)
ORDER BY
customer DESC,
yr DESC,
mon DESC
FETCH FIRST ROW ONLY;
If you want to include ties for the highest number of customers then:
SELECT count(name) AS CUSTOMER,
extract(year from create_date) as yr,
extract(month from create_date) as mon
FROM x
GROUP BY
extract(year from create_date),
extract(month from create_date)
ORDER BY
customer DESC
FETCH FIRST ROW WITH TIES;

Related

RANK() over (PARTITION BY) To show only TOP 3 rows for each month

I have a question about ranking . (My using Pgadmin for my SQL codes)
Mange to get my sum of sales in DESC order and rank 1 to 3 for the month of APR
But how can I achieve my result by showing only rank 1 to 3 for the month of Apr , May and June.
I need to reflect only 9 rows in my table .
SELECT restaurant_id,
EXTRACT(year FROM submitted_on) AS year,
EXTRACT(month FROM submitted_on) AS month,
SUM(total_amount),
RANK() OVER (PARTITION BY(extract(month from submitted_on))
ORDER BY SUM(total_amount) DESC) rank
FROM orders
WHERE submitted_on::date BETWEEN '2021-04-01' AND '2021-06-30'
GROUP BY restaurant_id, year, month

If you just want 3 records you should use row_number instead of rank. for your requirement you can do it in this way:
select t.* from (
SELECT restaurant_id,
EXTRACT(year FROM submitted_on) AS year,
EXTRACT(month FROM submitted_on) AS month,
SUM(total_amount),
RANK() OVER (PARTITION BY(extract(month from submitted_on))
ORDER BY SUM(total_amount) DESC) rank
FROM orders
WHERE submitted_on::date BETWEEN '2021-04-01' AND '2021-06-30'
GROUP BY restaurant_id, year, month
) t
where rank <=3;

Selecting month in each year with maximum number of projects

I have the following scenario
img
For each year I would like to display the month with the highest number of projects that have ended
I have tried the following so far:
SELECT COUNT(proj.projno) nr_proj, extract(month from proj.end_date) month
, extract(year from proj.end_date) year
FROM PROJ
GROUP BY extract(month from proj.end_date)
,extract(year from proj.end_date)
I am getting the information about the number of projects per month, per year.
Could any one give me hints how for each of the years I would select only the records with the highest count of projects?

You can use this solution using max analytic function to get max nr_proj value per year (partition by clause), then keep only rows where nr_proj = mx.
select t.nr_proj, t.month, t.year
from (
SELECT COUNT(proj.projno) nr_proj
, extract(month from proj.end_date) month
, extract(year from proj.end_date) year
, max( COUNT(proj.projno) ) over(partition by extract(year from proj.end_date)) mx
FROM PROJ
GROUP BY extract(month from proj.end_date), extract(year from proj.end_date)
) t
where nr_proj = mx
;
demo

I think the following will give you what you are after (if I understood the requirements). It fist counts the projects for each month then ranks the months by year, finally it selects the first rank.
select dt "Most Projects Month", cnt "Monthly Projects"
from ( -- Rank month Count by Year
select to_char( dt, 'yyyy-mm') dt
, cnt
, rank() over (partition by extract(year from dt)
order by cnt desc) rnk
from (-- count number of in month projects for each year
select trunc(end_date,'mon') dt, count(*) cnt
from projects
group by trunc(end_date,'mon')
)
)
where rnk = 1
order by dt;
NOTE: Not tested, no data supplied. In future do not post images, see Why No Images.

Top 2 per month in SQL

I have this dataset, which has dates and products for cities:
CREATE TABLE my_table (
the_id varchar(5) NOT NULL,
the_date timestamp NOT NULL,
the_city varchar(5) NOT NULL,
the_product varchar(1) NOT NULL
);
INSERT INTO my_table
VALUES ('VIS01', '2019-05-02 09:00:00','LISBO','A'),
('VIS02', '2019-05-04 12:00:00','EVORA','A'),
('VIS03', '2019-05-05 18:00:00','LISBO','B'),
('VIS04', '2019-05-06 18:30:00','PORTO','B'),
('VIS05', '2019-05-15 12:05:00','PORTO','C'),
('VIS06', '2019-06-02 18:06:00','EVORA','C'),
('VIS07', '2019-06-02 18:07:00','PORTO','A'),
('VIS08', '2019-06-04 18:08:00','EVORA','B'),
('VIS09', '2019-06-07 18:09:00','LISBO','B'),
('VIS10', '2019-06-09 18:10:00','LISBO','D'),
('VIS11', '2019-06-12 18:11:00','EVORA','D'),
('VIS12', '2019-06-15 18:12:00','LISBO','E'),
('VIS13', '2019-06-15 18:13:00','EVORA','F'),
('VIS14', '2019-06-18 18:14:00','PORTO','G'),
('VIS15', '2019-06-23 18:15:00','LISBO','A'),
('VIS16', '2019-06-25 18:16:00','LISBO','A'),
('VIS17', '2019-06-27 18:17:00','LISBO','F'),
('VIS18', '2019-06-27 18:18:00','LISBO','A'),
('VIS19', '2019-06-28 18:19:00','LISBO','A'),
('VIS20', '2019-06-30 18:20:00','EVORA','D'),
('VIS21', '2019-07-01 18:21:00','EVORA','D'),
('VIS22', '2019-07-04 18:30:00','EVORA','D'),
('VIS23', '2019-07-04 18:31:00','EVORA','B'),
('VIS24', '2019-07-06 18:40:00','EVORA','K'),
('VIS25', '2019-07-12 18:50:00','EVORA','G'),
('VIS26', '2019-07-15 18:00:00','PORTO','C'),
('VIS27', '2019-07-18 18:00:00','PORTO','C'),
('VIS28', '2019-07-25 18:00:00','PORTO','B'),
('VIS29', '2019-07-30 18:00:00','PORTO','M');
And I want the top two per month. The expected result should be:
month product count
2019-05 A 2
2019-05 B 2
2019-06 A 5
2019-06 D 3
2019-07 C 2
2019-07 D 2
But I'm not quite sure how to group by month. Please, any help will be greatly appreciated.

First, you can use to_char(the_date,'YYYY-MM') to get the year and month in the right format.
Next, you can use count(*) to group by the month and product, and row_number() to give a sequence number to each row in the groups.
SELECT to_char(the_date,'YYYY-MM') as month,
the_product as product,
count(*) as p_count,
row_number() over (partition by to_char(the_date,'YYYY-MM') order by count(*) desc) as seq
FROM my_table
group by month, product
Last, you can wrap that in an outer query to select just the columns and rows that you want.
SELECT month, product, p_count as count
FROM (
SELECT to_char(the_date,'YYYY-MM') as month,
the_product as product,
count(*) as p_count,
row_number() over (partition by to_char(the_date,'YYYY-MM') order by count(*) desc) as seq
FROM my_table
group by month, product
) as foo
where foo.seq <= 2;

You can use aggregation and window functions:
select mp.*
from (select date_trunc('month', the_date) as yyyymm,
the_product, count(*) as cnt,
row_number() over (partition by date_trunc('month', the_date) order by count(*) desc) as seqnum
from my_table
group by yyyymm, the_product
) mp
where seqnum <= 2;

In postgresql, I believe you can extract every parts of the timestamp using the Extract function.
e.g.:
SELECT the_date, EXTRACT(MONTH from the_date) as MONTH
the_date
MONTH
'2019-08-05'
08
that said, you can then group by Product, then Month, and Select the TOP 2
SELECT EXTRACT(MONTH from the_date) as month, the_product, count (*) FROM my_table
GROUP BY EXTRACT(MONTH from the_date), the_product
ORDER BY count(*)
LIMIT 2
There might be some optimization to do since I don't have a Database to test the query, but it might give you a good start

Find max value for each year

I have a question that is asking:
-List the max sales for each year?
I think I have the starter query but I can't figure out how to get all the years in my answer:
SELECT TO_CHAR(stockdate,'YYYY') AS year, sales
FROM sample_newbooks
WHERE sales = (SELECT MAX(sales) FROM sample_newbooks);
This query gives me the year with the max sales. I need max sales for EACH year. Thanks for your help!

Use group by and max if all you need is year and max sales of the year.
select
to_char(stockdate, 'yyyy') year,
max(sales) sales
from sample_newbooks
group by to_char(stockdate, 'yyyy')
If you need rows with all the columns with max sales for the year, you can use window function row_number:
select
*
from (
select
t.*,
row_number() over (partition by to_char(stockdate, 'yyyy') order by sales desc) rn
from sample_newbooks t
) t where rn = 1;
If you want to get the rows with ties on sales, use rank:
select
*
from (
select
t.*,
rank() over (partition by to_char(stockdate, 'yyyy') order by sales desc) rn
from sample_newbooks t
) t where rn = 1;

To find the last updated record of each month for each policy(another field)

I have a table named a, and other fields as eff_date,policy no.
Now for each policy, consider all the records, and take out the last updated one (eff_date) from each month.
So I need the last updated record for each month for each policy. How would I write a query for this?

I'm not 100 percent on Teradata syntax, but I believe you're after this:
SELECT policy_no,eff_date
FROM (SELECT policy_no,eff_date, ROW_NUMBER() OVER (PARTITION BY policy no, EXTRACT(YEAR FROM eff_date),EXTRACT(MONTH FROM eff_date) ORDER BY eff_date DESC) as RowRank
FROM a) as sub
WHERE RowRank = 1
I'm assuming when you say by month you also want to differentiate by year, but if not, just remove the EXTRACT(YEAR FROM eff_date) from the PARTITION BY section.
Edit: Update for Teradata syntax.

SELECT * from a
qualify ROW_NUMBER() OVER (PARTITION BY policy no, EXTRACT(YEAR FROM eff_date),
EXTRACT(MONTH FROM eff_date) ORDER BY eff_date DESC) = 1

The main difficulty, is that the group by needs to be made both the conbination of policy_no, but also the month (extracted from the date). For example:
In Mysql
SELECT policy_no,
month(eff_date),
year(eff_date),
max(eff_date)
FROM myTable
GROUP BY policy_no,
month(eff_date),
year(eff_date);
Update
I saw derived tables are allowed in teradata. Using a join to a derived table, here is how to access the full rows:
select * from a,
(SELECT policy_no,
month(eff_date),
year(eff_date),
max(eff_date) as MaxMonthDate
FROM a
GROUP BY policy_no,
month(eff_date),
year(eff_date)
) as b
where a.policy_no = b.policy_no and
a.eff_date = b.MaxMonthDate;
http://www.sqlfiddle.com/#!2/1f728/5
Update (Using Extract)
select * from a,
(SELECT a2.policy_no,
EXTRACT(MONTH FROM a2.eff_date),
EXTRACT(YEAR FROM a2.eff_date),
max(a2.eff_date) as MaxMonthDate
FROM a as a2
GROUP BY a2.policy_no,
EXTRACT(MONTH FROM a2.eff_date),
EXTRACT(YEAR FROM a2.eff_date)
) as b
where a.policy_no = b.policy_no and
a.eff_date = b.MaxMonthDate;

I'm going to suggest looking into Windows Aggregate functions and the QUALIFY statement. I believe the following SQL will work.
SELECT Policy_No
, EXTRACT(MONTH FROM Eff_Date) AS Eff_Month_
, Eff_Date
FROM TableA
QUALIFY ROW_NUMBER() OVER (PARTITION BY Policy_No, EXTRACT(MONTH FROM Eff_Date)
ORDER BY Eff_Date DESC) = 1;