使用dataframeapi的一个好方法是使用argmax逻辑，如下所示

val df = 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")

  df.groupBy($"Hour")
    .agg(max(struct($"TotalValue", $"Category")).as("argmax"))
    .select($"Hour", $"argmax.*").show

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

下面的解决方案只执行一次groupby并在一次快照中提取包含maxvalue的Dataframe行。不需要进一步的连接或窗口。

import org.apache.spark.sql.Row
import org.apache.spark.sql.catalyst.encoders.RowEncoder
import org.apache.spark.sql.DataFrame

//df is the dataframe with Day, Category, TotalValue

implicit val dfEnc = RowEncoder(df.schema)

val res: DataFrame = df.groupByKey{(r) => r.getInt(0)}.mapGroups[Row]{(day: Int, rows: Iterator[Row]) => i.maxBy{(r) => r.getDouble(2)}}

你可以这样做-

val data = df.groupBy("Hour").agg(first("Hour").as("_1"),first("Category").as("Category"),first("TotalValue").as("TotalValue")).drop("Hour")

data.withColumnRenamed("_1","Hour").show

模式是按键分组=>对每个组执行一些操作，例如reduce=>返回Dataframe
在这种情况下，我认为Dataframe抽象有点麻烦，所以我使用了rdd功能

val rdd: RDD[Row] = originalDf
  .rdd
  .groupBy(row => row.getAs[String]("grouping_row"))
  .map(iterableTuple => {
    iterableTuple._2.reduce(reduceFunction)
  })

val productDf = sqlContext.createDataFrame(rdd, originalDf.schema)

这与zero323的答案完全相同，但采用sql查询方式。
假设dataframe被创建并注册为

df.createOrReplaceTempView("table")
//+----+--------+----------+
//|Hour|Category|TotalValue|
//+----+--------+----------+
//|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      |
//+----+--------+----------+

窗口功能：

sqlContext.sql("select Hour, Category, TotalValue from (select *, row_number() OVER (PARTITION BY Hour ORDER BY TotalValue DESC) as rn  FROM table) tmp where rn = 1").show(false)
//+----+--------+----------+
//|Hour|Category|TotalValue|
//+----+--------+----------+
//|1   |cat67   |28.5      |
//|3   |cat8    |35.6      |
//|2   |cat56   |39.6      |
//|0   |cat26   |30.9      |
//+----+--------+----------+

纯sql聚合，后跟联接：

sqlContext.sql("select Hour, first(Category) as Category, first(TotalValue) as TotalValue from " +
  "(select Hour, Category, TotalValue from table tmp1 " +
  "join " +
  "(select Hour as max_hour, max(TotalValue) as max_value from table group by Hour) tmp2 " +
  "on " +
  "tmp1.Hour = tmp2.max_hour and tmp1.TotalValue = tmp2.max_value) tmp3 " +
  "group by tmp3.Hour")
  .show(false)
//+----+--------+----------+
//|Hour|Category|TotalValue|
//+----+--------+----------+
//|1   |cat67   |28.5      |
//|3   |cat8    |35.6      |
//|2   |cat56   |39.6      |
//|0   |cat26   |30.9      |
//+----+--------+----------+

使用结构上的排序：

sqlContext.sql("select Hour, vs.Category, vs.TotalValue from (select Hour, max(struct(TotalValue, Category)) as vs from table group by Hour)").show(false)
//+----+--------+----------+
//|Hour|Category|TotalValue|
//+----+--------+----------+
//|1   |cat67   |28.5      |
//|3   |cat8    |35.6      |
//|2   |cat56   |39.6      |
//|0   |cat26   |30.9      |
//+----+--------+----------+

数据集的方式和不dos与原始答案相同

窗口功能：
像这样的事情应该可以做到：

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|
// +----+--------+----------+

这种方法在数据严重倾斜的情况下效率很低。
纯sql聚合，后跟 join :
或者，您可以加入聚合Dataframe：

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|
// +----+--------+----------+

它将保留重复的值（如果每小时有多个类别具有相同的总值）。您可以按如下方式删除它们：

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

使用排序结束 structs :
整洁，虽然没有很好的测试，技巧不需要连接或窗口函数：

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|
// +----+--------+----------+

使用数据集api（spark 1.6+、2.0+）：
Spark1.6：

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或更高版本：

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

最后两种方法可以利用map-side-combine，并且不需要完全洗牌，因此大多数情况下，与窗口函数和连接相比，应该表现出更好的性能。这些命令也可用于结构化流式输入 completed 输出模式。
不要使用：

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

这似乎是可行的（尤其是在 local 模式），但它是不可靠的（见Spark-16207，信贷tzach-zohar链接相关的jira问题，和Spark-30335）。
同样的注解也适用于

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

内部使用等效的执行计划。

对于按多列分组的spark 2.0.2：

import org.apache.spark.sql.functions.row_number
import org.apache.spark.sql.expressions.Window

val w = Window.partitionBy($"col1", $"col2", $"col3").orderBy($"timestamp".desc)

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

我们可以使用rank（）窗口函数（您可以选择rank=1）rank只为组的每一行添加一个数字（在本例中是小时）
举个例子从https://github.com/jaceklaskowski/mastering-apache-spark-book/blob/master/spark-sql-functions.adoc#rank )

val dataset = spark.range(9).withColumn("bucket", 'id % 3)

import org.apache.spark.sql.expressions.Window
val byBucket = Window.partitionBy('bucket).orderBy('id)

scala> dataset.withColumn("rank", rank over byBucket).show
+---+------+----+
| id|bucket|rank|
+---+------+----+
|  0|     0|   1|
|  3|     0|   2|
|  6|     0|   3|
|  1|     1|   1|
|  4|     1|   2|
|  7|     1|   3|
|  2|     2|   1|
|  5|     2|   2|
|  8|     2|   3|
+---+------+----+