分组并选择最大对SQL

Question

I have a table which has two columns.

create table txns( 
  person varchar(255),
  fruit varchar(255)
  );

This is a log table. I have sqlfiddle here .

This is as far as I am able to get with the sql query.

In essence, For every person, which is the most frequent fruit he has eaten.

I have both Oracle and MySql at my place. In the future, it would also be deployed on hadoop (via Hive/Impala etc). Thus a non-db centric answer would be best. But pls also do provide a db centric answer if there is such only.

Answer 1

SQL Fiddle

Oracle 11g R2 Schema Setup :

create table txns(

  person varchar(255),
  fruit varchar(255)
  );

insert into txns
values ('alpha','apple');

insert into txns
values ('charlie','cherry');

insert into txns
values ('bravo','banana');

insert into txns
values ('alpha','apple');

insert into txns
values ('bravo','banana');

insert into txns
values ('alpha','apricot');

insert into txns
values ('bravo','berry');

Query 1 :

with tab as (
select person, fruit,count(1) cnt,  
       max(count(1)) over (partition by person) m_cnt
  from txns
 group by person, fruit)
select person, fruit, cnt, m_cnt
  from tab
 where cnt = m_cnt

Results :

|  PERSON |  FRUIT | CNT | M_CNT |
|---------|--------|-----|-------|
|   alpha |  apple |   2 |     2 |
|   bravo | banana |   2 |     2 |
| charlie | cherry |   1 |     1 |

Answer 2

For Oracle -

select x.person,x.fruit
from ( select person, fruit, count(*) ct,rank() over (partition by person 
                                                      order by count(*) desc) as rank
         from txns
group by person, fruit) x
where rank=1;

SQL Fiddle

The "non DB centric" idea is-

1. You first find out for each person, how many time a particular fruit appears in the table (or how many time the fruit was eaten). This is done by person, fruit,count(*) .
2. Then you need to find out which fruit was most eaten by the PERSON ie essentially the fruit that holds the highest RANK or position, in this case a DESC ending order of count(*) would place the most eaten fruit in RANK =1 for each person ( partition by person ).
3. Once you are done ranking, you just need to select the first RANK for each Person which would essentially be the most eaten fruit by the person.

This is a perfect example of Oracle's analytical function RANK() .

Why use RANK , you ask?

Your boss may change his mind & may ask you "Hey prog_guy, I changed my mind, I not only want the most eaten fruit by the fruit eater, I also want the 3rd most eaten fruit as well". What do you do? Scramble to write another query? No, you take the same query and change rank=1 to rank in (1,3) and BAM! you now have the most eaten fruit and the third most eaten fruit (if any) by the fruit eaters.

AND / OR

Your boss may change his mind again and say "Hey prog_guy, forget about most eaten fruits, now I want you to get the least eaten fruit" What do you do? Scram again for a new query? Nope! You change the desc to asc in the partition by and BAM! you now have the least eaten fruit by the fruit eaters.

---

Some detail about RANK() equivalent of Oracle in MySQL here . A little bit of info related to RANK() equivalent in Hive, here .

Answer 3

Following query would run both in Oracle and MySQL.

select k.person, k.fruit from
(
  select person,fruit,count(fruit) as cnt
  from txns
  group by person,fruit
) k
join
(
  select t.person,max(t.cnt) mxCnt
  from
  (
    select person,fruit,count(fruit) as cnt
    from txns
    group by person,fruit
  )t
group by t.person
) s
on s.person = k.person
and s.mxCnt = k.cnt 
order by k.person