Get most common value in SQL

Question

I have a table like this:

 Column  | Type | Modifiers 
---------+------+-----------
 country | text | 
 food_id | int  | 
 eaten   | date |

And for each country, I want to get the food that is eaten most often. The best I can think of (I'm using postgres) is:

CREATE TEMP TABLE counts AS 
   SELECT country, food_id, count(*) as count FROM munch GROUP BY country, food_id;

CREATE TEMP TABLE max_counts AS 
   SELECT country, max(count) as max_count FROM counts GROUP BY country;

SELECT country, max(food_id) FROM counts 
   WHERE (country, count) IN (SELECT * from max_counts) GROUP BY country;

In that last statement, the GROUP BY and max() are needed to break ties, where two different foods have the same count.

This seems like a lot of work for something conceptually simple. Is there a more straight forward way to do it?

Answer 1

Try something like this

select country, food_id, count(*) cnt 
into #tempTbl 
from mytable 
group by country, food_id

select country, food_id
from  #tempTbl as x
where cnt = 
  (select max(cnt) 
  from mytable 
  where country=x.country 
  and food_id=x.food_id)

This could be put all into a single select, but I don't have time to muck around with it right now.

Good luck.

Answer 2

Here's how to do it without any temp tables:

Edit: simplified

select nf.country, nf.food_id as most_frequent_food_id
from national_foods nf
group by country, food_id 
having
  (country,count(*)) in (  
                        select country, max(cnt)
                        from
                          (
                          select country, food_id, count(*) as cnt
                          from national_foods nf1
                          group by country, food_id
                          )
                        group by country
                        having country = nf.country
                        )

Answer 3

SELECT DISTINCT
"F1"."food",
"F1"."country"
FROM "foo" "F1"
WHERE
"F1"."food" =
    (SELECT "food" FROM
        (
            SELECT "food", COUNT(*) AS "count"
            FROM "foo" "F2" 
            WHERE "F2"."country" = "F1"."country" 
            GROUP BY "F2"."food" 
            ORDER BY "count" DESC
        ) AS "F5"
        LIMIT 1
    )

Well, I wrote this in a hurry and didn't check it really well. The sub-select might be pretty slow, but this is shortest and most simple SQL statement that I could think of. I'll probably tell more when I'm less drunk.

PS: Oh well, "foo" is the name of my table, "food" contains the name of the food and "country" the name of the country. Sample output:

   food    |  country   
-----------+------------
 Bratwurst | Germany
 Fisch     | Frankreich

Answer 4

SELECT country, MAX( food_id )
  FROM( SELECT m1.country, m1.food_id
          FROM munch m1
         INNER JOIN ( SELECT country
                           , food_id
                           , COUNT(*) as food_counts
                        FROM munch m2
                    GROUP BY country, food_id ) as m3
                 ON m1.country = m3.country
         GROUP BY m1.country, m1.food_id 
        HAVING COUNT(*) / COUNT(DISTINCT m3.food_id) = MAX(food_counts) ) AS max_foods
  GROUP BY country

I don't like the MAX(.) GROUP BY to break ties... There's gotta be a way to incorporate eaten date into the JOIN in some way to arbitrarily select the most recent one...

I'm interested on the query plan for this thing if you run it on your live data!

Answer 5

select country,food_id, count(*) ne  
from   food f1  
group by country,food_id    
having count(*) = (select max(count(*))  
                   from   food f2  
                   where  country = f1.country  
                   group by food_id)

Answer 6

try this:

Select Country, Food_id
From Munch T1
Where Food_id= 
    (Select Food_id
     from Munch T2
     where T1.Country= T2.Country
     group by Food_id
     order by count(Food_id) desc
      limit 1)
group by Country, Food_id