scala-如何计算单个groupby中的sum和count?


0

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|
+---+-----+----+

我想按类别获取身份证号码和总金额:

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

是否可以在不进行 join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")

可能是这样的:

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

3 答案

0

我举的例子和你的不同

像这样可以有多个组函数。试一试吧

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

0

您可以对给定的表执行如下聚合:

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

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


0

有多种方法可以在spark中执行聚合函数,

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

1。

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 |
+-----+-----+---+

2。

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 |
+-----+-----+---+

三。

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。如果您是SQL专家,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|
+-----+---------+---------+


我来回答