How to Select the First Row Per Group in an SQL Query

Select first row in each GROUP BY group?

On databases that support CTE and windowing functions:

WITH summary AS (
    SELECT p.id, 
           p.customer, 
           p.total, 
           ROW_NUMBER() OVER(PARTITION BY p.customer 
                                 ORDER BY p.total DESC) AS rank
      FROM PURCHASES p)
 SELECT *
   FROM summary
 WHERE rank = 1

Supported by any database:

But you need to add logic to break ties:

  SELECT MIN(x.id),  -- change to MAX if you want the highest
         x.customer, 
         x.total
    FROM PURCHASES x
    JOIN (SELECT p.customer,
                 MAX(total) AS max_total
            FROM PURCHASES p
        GROUP BY p.customer) y ON y.customer = x.customer
                              AND y.max_total = x.total
GROUP BY x.customer, x.total

Get top 1 row of each group

;WITH cte AS
(
   SELECT *,
         ROW_NUMBER() OVER (PARTITION BY DocumentID ORDER BY DateCreated DESC) AS rn
   FROM DocumentStatusLogs
)
SELECT *
FROM cte
WHERE rn = 1

If you expect 2 entries per day, then this will arbitrarily pick one. To get both entries for a day, use DENSE_RANK instead

As for normalised or not, it depends if you want to:

maintain status in 2 places
preserve status history
...

As it stands, you preserve status history. If you want latest status in the parent table too (which is denormalisation) you'd need a trigger to maintain "status" in the parent. or drop this status history table.

Selecting first row per group

SELECT  a, b, c
FROM    (
        SELECT  *, ROW_NUMBER() OVER (PARTITION BY a ORDER BY b, c) rn
        FROM    mytable
        ) q
WHERE   rn = 1
ORDER BY
        a

SELECT  mi.*
FROM    (
        SELECT  DISTINCT  a
        FROM    mytable
        ) md
CROSS APPLY
        (
        SELECT  TOP 1 *
        FROM    mytable mi
        WHERE   mi.a = md.a
        ORDER BY
                b, c
        ) mi
ORDER BY
        a

Create a composite index on (a, b, c) for the queries to work faster.

Which one is more efficient depends on your data distribution.

If you have few distinct values of a but lots of records within each a, the second query would be better.

You could improve it even more by creating an indexed view:

CREATE VIEW v_mytable_da
WITH   SCHEMABINDING
AS
       SELECT  a, COUNT_BIG(*) cnt
       FROM    dbo.mytable
       GROUP BY
               a

GO

CREATE UNIQUE CLUSTERED INDEX
       pk_vmytableda_a
ON     v_mytable_da (a)

GO

SELECT  mi.*
FROM    v_mytable_da md
CROSS APPLY
        (
        SELECT  TOP 1 *
        FROM    mytable mi
        WHERE   mi.a = md.a
        ORDER BY
                b, c
        ) mi
ORDER BY
        a

How do I select the first row per group in an SQL Query?

declare @sometable table ( foo int, bar int, value int )

insert into @sometable values (47, 1, 100)
insert into @sometable values (47, 0, 10)
insert into @sometable values (47, 2, 10)
insert into @sometable values (46, 0, 100)
insert into @sometable values (46, 1, 10)
insert into @sometable values (46, 2, 10)
insert into @sometable values (44, 0, 2)

WITH cte AS 
(
    SELECT   Foo, Bar, SUM(value) AS SumValue, ROW_NUMBER() OVER(PARTITION BY Foo ORDER BY FOO DESC, SUM(value) DESC) AS RowNumber
    FROM     @SomeTable
    GROUP BY Foo, Bar
)
SELECT * 
FROM cte
WHERE RowNumber = 1

sql server select first row from a group

If as you indicated, order doesn't matter, any aggregate function on b would be sufficient.

Example Using MIN

SELECT a, b = MIN(b)
FROM   YourTable
GROUP BY
       a

How to get the first row per group?

if your MySQL version support ROW_NUMBER + window function, you can try to use ROW_NUMBER to get the biggest num by category_id

Query #1

SELECT num,business_id,category_id
FROM (
    SELECT *,ROW_NUMBER() OVER(PARTITION BY category_id ORDER BY num desc) rn
    FROM (
        select count(1) num, business_id, category_id
        from mytable
        group by business_id, category_id
    ) t1
) t1
WHERE rn = 1

num	business_id	category_id
22	5543	8
13	3242	11

SQL selecting first record per group

GROUP BY u.d (without also listing u1, u2, u3) would only work if u.d was the PRIMARY KEY (which it is not, and also wouldn't make sense in your scenario). See:

Is it possible to have an SQL query that uses AGG functions in this way?

I suggest DISTINCT ON in a subquery on UTable instead:

SELECT o.d, u.u1, u.u2, u.u3, o.n
FROM  (
   SELECT DISTINCT ON (u.d)
          u.d, u.u1, u.u2, u.u3
   FROM   UTable u
   WHERE  u.gid = 3
   AND    u.gt = 'dog night'
   ORDER  BY u.d, u.timestamp
   ) u
JOIN   OTable o USING (gid, gt, d);

See:

Select first row in each GROUP BY group?

If UTable is big, at least a multicolumn index on (gid, gt) is advisable. Same for OTable.

Maybe even on (gid, gt, d). Depends on data types, cardinalities, ...

How to select the first row of each group?

Window functions:

Something like this should do the trick:

import org.apache.spark.sql.functions.{row_number, max, broadcast}
import org.apache.spark.sql.expressions.Window

val df = sc.parallelize(Seq(
  (0,"cat26",30.9), (0,"cat13",22.1), (0,"cat95",19.6), (0,"cat105",1.3),
  (1,"cat67",28.5), (1,"cat4",26.8), (1,"cat13",12.6), (1,"cat23",5.3),
  (2,"cat56",39.6), (2,"cat40",29.7), (2,"cat187",27.9), (2,"cat68",9.8),
  (3,"cat8",35.6))).toDF("Hour", "Category", "TotalValue")

val w = Window.partitionBy($"hour").orderBy($"TotalValue".desc)

val dfTop = df.withColumn("rn", row_number.over(w)).where($"rn" === 1).drop("rn")

dfTop.show
// +----+--------+----------+
// |Hour|Category|TotalValue|
// +----+--------+----------+
// |   0|   cat26|      30.9|
// |   1|   cat67|      28.5|
// |   2|   cat56|      39.6|
// |   3|    cat8|      35.6|
// +----+--------+----------+

This method will be inefficient in case of significant data skew. This problem is tracked by SPARK-34775 and might be resolved in the future (SPARK-37099).

Plain SQL aggregation followed by join:

Alternatively you can join with aggregated data frame:

val dfMax = df.groupBy($"hour".as("max_hour")).agg(max($"TotalValue").as("max_value"))

val dfTopByJoin = df.join(broadcast(dfMax),
    ($"hour" === $"max_hour") && ($"TotalValue" === $"max_value"))
  .drop("max_hour")
  .drop("max_value")

dfTopByJoin.show

// +----+--------+----------+
// |Hour|Category|TotalValue|
// +----+--------+----------+
// |   0|   cat26|      30.9|
// |   1|   cat67|      28.5|
// |   2|   cat56|      39.6|
// |   3|    cat8|      35.6|
// +----+--------+----------+

It will keep duplicate values (if there is more than one category per hour with the same total value). You can remove these as follows:

dfTopByJoin
  .groupBy($"hour")
  .agg(
    first("category").alias("category"),
    first("TotalValue").alias("TotalValue"))

Using ordering over structs:

Neat, although not very well tested, trick which doesn't require joins or window functions:

val dfTop = df.select($"Hour", struct($"TotalValue", $"Category").alias("vs"))
  .groupBy($"hour")
  .agg(max("vs").alias("vs"))
  .select($"Hour", $"vs.Category", $"vs.TotalValue")

dfTop.show
// +----+--------+----------+
// |Hour|Category|TotalValue|
// +----+--------+----------+
// |   0|   cat26|      30.9|
// |   1|   cat67|      28.5|
// |   2|   cat56|      39.6|
// |   3|    cat8|      35.6|
// +----+--------+----------+

With DataSet API (Spark 1.6+, 2.0+):

Spark 1.6:

case class Record(Hour: Integer, Category: String, TotalValue: Double)

df.as[Record]
  .groupBy($"hour")
  .reduce((x, y) => if (x.TotalValue > y.TotalValue) x else y)
  .show

// +---+--------------+
// | _1|            _2|
// +---+--------------+
// |[0]|[0,cat26,30.9]|
// |[1]|[1,cat67,28.5]|
// |[2]|[2,cat56,39.6]|
// |[3]| [3,cat8,35.6]|
// +---+--------------+

Spark 2.0 or later:

df.as[Record]
  .groupByKey(_.Hour)
  .reduceGroups((x, y) => if (x.TotalValue > y.TotalValue) x else y)

The last two methods can leverage map side combine and don't require full shuffle so most of the time should exhibit a better performance compared to window functions and joins. These cane be also used with Structured Streaming in completed output mode.

Don't use:

df.orderBy(...).groupBy(...).agg(first(...), ...)

It may seem to work (especially in the local mode) but it is unreliable (see SPARK-16207, credits to Tzach Zohar for linking relevant JIRA issue, and SPARK-30335).

The same note applies to

df.orderBy(...).dropDuplicates(...)

which internally uses equivalent execution plan.

SQL Server Group By Query Select first row each group

You can GROUP BY StudyID, Year and then in an outer query select the first row from each StudyID, Year group:

SELECT StudyID, Year, minAccess1, minAccess2, minAccess3
FROM (
   SELECT StudyID, Year, min(Access1) minAccess1, min(Access2) minAccess2,
          min(Access3) minAccess3, 
          ROW_NUMBER() OVER (PARTITION BY StudyID ORDER BY Year DESC) AS rn 
   FROM mytable
   GROUP BY StudyID, Year ) t
WHERE t.rn = 1

ROW_NUMBER is used to assign an ordering number to each StudyID group according to Year values. The row with the maximum Year value is assigned a rn = 1.

Select the first row for each group in MySQL?

You can GROUP BY and pick the MAX position.

SELECT ri.*
FROM (
    SELECT ri.release_id, MAX(ri.position) AS position
    FROM release_image ri
    GROUP BY ri.release_id
) ri_max
INNER JOIN release_image ri ON ri_max.release_id = ri.release_id
    AND ri_max.position = ri.position