如何计算单个 groupBy 中的总和和计数？答案

【问题标题】：How to calculate sum and count in a single groupBy?如何计算单个 groupBy 中的总和和计数？
【发布时间】：2016-11-06 12:05:17
【问题描述】：

基于以下DataFrame：

val client = Seq((1,"A",10),(2,"A",5),(3,"B",56)).toDF("ID","Categ","Amnt")
+---+-----+----+
| ID|Categ|Amnt|
+---+-----+----+
|  1|    A|  10|
|  2|    A|   5|
|  3|    B|  56|
+---+-----+----+

我想按类别获取ID个数和总金额：

+-----+-----+---------+
|Categ|count|sum(Amnt)|
+-----+-----+---------+
|    B|    1|       56|
|    A|    2|       15|
+-----+-----+---------+

是否可以在不进行连接的情况下进行计数和求和？

client.groupBy("Categ").count
      .join(client.withColumnRenamed("Categ","cat")
           .groupBy("cat")
           .sum("Amnt"), 'Categ === 'cat)
      .drop("cat")

也许是这样的：

client.createOrReplaceTempView("client")
spark.sql("SELECT Categ count(Categ) sum(Amnt) FROM client GROUP BY Categ").show()

【问题讨论】：

标签： scala apache-spark apache-spark-sql

【解决方案1】：

我给出的例子与你的不同

multiple group functions are possible like this. try it accordingly

  // In 1.3.x, in order for the grouping column "department" to show up,
// it must be included explicitly as part of the agg function call.
df.groupBy("department").agg($"department", max("age"), sum("expense"))

// In 1.4+, grouping column "department" is included automatically.
df.groupBy("department").agg(max("age"), sum("expense"))

import org.apache.spark.sql.{DataFrame, SparkSession}
import org.apache.spark.sql.functions._

val spark: SparkSession = SparkSession
      .builder.master("local")
      .appName("MyGroup")
      .getOrCreate()
import spark.implicits._
    val client: DataFrame = spark.sparkContext.parallelize(
Seq((1,"A",10),(2,"A",5),(3,"B",56))
).toDF("ID","Categ","Amnt")

client.groupBy("Categ").agg(sum("Amnt"),count("ID")).show()

+-----+---------+---------+
|Categ|sum(Amnt)|count(ID)|
+-----+---------+---------+
|    B|       56|        1|
|    A|       15|        2|
+-----+---------+---------+

【讨论】：

我们可以用 reduceBy 代替 groupBy 吗？

【解决方案2】：

您可以在给定的表上进行如下聚合：

client.groupBy("Categ").agg(sum("Amnt"),count("ID")).show()

+-----+---------+---------+
|Categ|sum(Amnt)|count(ID)|
+-----+---------+---------+
|    A|       15|        2|
|    B|       56|        1|
+-----+---------+---------+

【讨论】：

【解决方案3】：

spark中有多种方法可以做聚合函数，

val client = Seq((1,"A",10),(2,"A",5),(3,"B",56)).toDF("ID","Categ","Amnt")

val aggdf = client.groupBy('Categ).agg(Map("ID"->"count","Amnt"->"sum"))

+-----+---------+---------+
|Categ|count(ID)|sum(Amnt)|
+-----+---------+---------+
|B    |1        |56       |
|A    |2        |15       |
+-----+---------+---------+

//Rename and sort as needed.
aggdf.sort('Categ).withColumnRenamed("count(ID)","Count").withColumnRenamed("sum(Amnt)","sum")
+-----+-----+---+
|Categ|Count|sum|
+-----+-----+---+
|A    |2    |15 |
|B    |1    |56 |
+-----+-----+---+

import org.apache.spark.sql.functions._
client.groupBy('Categ).agg(count("ID").as("count"),sum("Amnt").as("sum"))
+-----+-----+---+
|Categ|count|sum|
+-----+-----+---+
|B    |1    |56 |
|A    |2    |15 |
+-----+-----+---+

import com.google.common.collect.ImmutableMap;
client.groupBy('Categ).agg(ImmutableMap.of("ID", "count", "Amnt", "sum"))
+-----+---------+---------+
|Categ|count(ID)|sum(Amnt)|
+-----+---------+---------+
|B    |1        |56       |
|A    |2        |15       |
+-----+---------+---------+
//Use column rename is required.

4。如果你是 SQL 专家，你也可以这样做

client.createOrReplaceTempView("df")

 val aggdf = spark.sql("select Categ, count(ID),sum(Amnt) from df group by Categ")
 aggdf.show()

    +-----+---------+---------+
    |Categ|count(ID)|sum(Amnt)|
    +-----+---------+---------+
    |    B|        1|       56|
    |    A|        2|       15|
    +-----+---------+---------+

【讨论】：