SQL min / max with all fields - sql

Question

I am facing a simple problem with an SQL query that I do not know how to tackle.
I have a table with the following structure
CITY COUNTRY DATES TEMPERATURE
Note that for a given country, I can have several cities. And, for a given city, I have several rows giving me the TEMPERATURE at each available DATE. This is just a time serie.
I would like to write a query which gives me for every cities the DATE where the TEMPERATURE is the MIN and the DATE where the TEMPERATURE is the MAX. The query should return something like that:
CITY COUNTRY DATE_MIN_TEMPERATURE MIN_TEMPERATURE DATE_MAX_TEMPERATURE MAX_TEMPERATURE
Any idea on how to achieve this?
Best regards,
Deny

Answer 1

Oracle provides keep/dense_rank first for this purpose:
select city,
min(temperature) as min_temperature,
max(date) keep (dense_rank first order by temperature asc) as min_temperature_date,
max(temperature) as max_temperature,
max(date) keep (dense_rank first order by temperature desc) as max_temperature_date
from t
group by city;
Note that this returns only one date if there are ties. If you want to handle that, more logic is needed:
select city, min(temperature) as min_temperature,
listagg(case when seqnum_min = 1 then date end, ',') within group (order by date) as mindates,
max(temperature) as max_temperature,
listagg(case when seqnum_max = 1 then date end, ',') within group (order by date) as maxdates,
from (select t.*,
rank() over (partition by city order by temperature) as seqnum_min,
rank() over (partition by city order by temperature desc) as seqnum_max
from t
) t
where seqnum_min = 1 or seqnum_max = 1
group by city;

Answer 2

In Oracle 11 and above, you can use PIVOT. In the solution below I use LISTAGG to show all the dates in case of ties. Another option is, in the case of ties, to show the most recent date when the extreme temperature was reached; if that is preferred, simply replace LISTAGG(dt, ....) (including the WITHIN GROUP clause) with MAX(dt). However, in that case the first solution offered by Gordon (using the first function) is more efficient anyway - no need for pivoting.
Note that I changed "date" to "dt" - DATE is a reserved word in Oracle. I also show the rows by country first, then city (the more logical ordering). I created test data in a WITH clause, but the solution is everything below the comment line.
with
inputs ( city, country, dt, temperature ) as (
select 'Palermo', 'Italy' , date '2014-02-13', 3 from dual union all
select 'Palermo', 'Italy' , date '2002-01-23', 3 from dual union all
select 'Palermo', 'Italy' , date '1998-07-22', 42 from dual union all
select 'Palermo', 'Italy' , date '1993-08-24', 30 from dual union all
select 'Maseru' , 'Lesotho', date '1994-01-11', 34 from dual union all
select 'Maseru' , 'Lesotho', date '2004-08-13', 12 from dual
)
-- >> end test data; solution (SQL query) begins with the next line
select country, city,
"'min'_DT" as date_min_temp, "'min'_TEMP" as min_temp,
"'max'_DT" as date_max_temp, "'max'_TEMP" as max_temp
from (
select city, country, dt, temperature,
case when temperature = min(temperature)
over (partition by city, country) then 'min'
when temperature = max(temperature)
over (partition by city, country) then 'max'
end as flag
from inputs
)
pivot ( listagg(to_char(dt, 'dd-MON-yyyy'), ', ')
within group (order by dt) as dt, min(temperature) as temp
for flag in ('min', 'max'))
order by country, city -- ORDER BY is optional
;
COUNTRY CITY DATE_MIN_TEMP MIN_TEMP DATE_MAX_TEMP MAX_TEMP
------- ------- ------------------------ ---------- -------------- ----------
Italy Palermo 23-JAN-2002, 13-FEB-2014 3 22-JUL-1998 42
Lesotho Maseru 13-AUG-2004 12 11-JAN-1994 34
2 rows selected.

Answer 3

Instead of keep/dense_rank first function you can also use FIRST_VALUE and LAST_VALUE:
select distinct city,
MIN(temperature) OVER (PARTITION BY city) as min_temperature,
FIRST_VALUE(date) OVER (PARTITION BY city ORDER BY temperature) AS min_temperature_date,
MAX(temperature) OVER (PARTITION BY city) as max_temperature,
LAST_VALUE(date) OVER (PARTITION BY city ORDER BY temperature) AS max_temperature_date
FROM t;