合并spark scala Dataframe中的行

Posted 2021-03-27

合并spark Dataframe中的行

我有以下数据

ID  Name    Passport    Country  License    UpdatedtimeStamp
1   Ostrich 12345       -       ABC         11-02-2018
1   -       -           -       BCD         10-02-2018
1   Shah    12345       -       -           12-02-2018
2   PJ      -           ANB     a           10-02-2018

需要的输出是

ID  Name    Passport    Country  License    UpdatedtimeStamp
1   Shah    12345       -       ABC         12-02-2018
2   PJ      -           ANB     a           10-02-2018

基本上，相同的ID中的数据应该合并，并且最新更新而不是null记录应该在输出中，如果所有值都是null，则应保留null ..

请建议...另外，建议不使用SparkSQL Window函数，因为我需要它非常快

答案

如果你想完全留在sparkSQL中

val df= Seq((1,Some("ostrich"), Some(12345), None, Some("ABC")," 11-02-2018" ),
(1,None, None, None, Some("BCD"), "10-02-2018"),(1,Some("Shah"), Some(12345), None,None, "12-02-2018"),
(2,Some("PJ"), None, Some("ANB"), Some("a"), "10-02-2018")).toDF("ID","Name","Passport","Country","License","UpdatedtimeStamp")


val df1= df.withColumn("date", to_date($"UpdatedtimeStamp","MM-dd-yyyy" )).drop($"UpdatedtimeStamp")

val win = Window.partitionBy("ID").orderBy($"date".desc)

val df2=df1.select($"*", row_number.over(win).as("r")).orderBy($"ID", $"r").drop("r")
val exprs= df2.columns.drop(1).map(x=>collect_list(x).as(x+"_grp"))

val df3=df2.groupBy("ID").agg(exprs.head,exprs.tail: _*)

val exprs2= df3.columns.drop(1).map(x=> col(x)(0).as(x))

df3.select((Array(col(df2.columns(0)))++exprs2): _*).show


+---+----+--------+-------+-------+----------+
| ID|Name|Passport|Country|License|      date|
+---+----+--------+-------+-------+----------+
|  1|Shah|   12345|   null|    ABC|2018-12-02|
|  2|  PJ|    null|    ANB|      a|2018-10-02|
+---+----+--------+-------+-------+----------+

另一答案

您可以通过定义udf函数并将收集的struct列传递给udf函数来实现结果，以便对非零值的空值进行排序和填充。 （注释代码中提供了注释）

import org.apache.spark.sql.functions._
//udf function definition
def sortAndAggUdf = udf((structs: Seq[Row])=>{
  //sorting the collected list by timestamp in descending order
  val sortedStruct = structs.sortBy(str => str.getAs[Long]("UpdatedtimeStamp"))(Ordering[Long].reverse)
  //selecting the first struct and casting to out case class
  val first = out(sortedStruct(0).getAs[String]("Name"), sortedStruct(0).getAs[String]("Passport"), sortedStruct(0).getAs[String]("Country"), sortedStruct(0).getAs[String]("License"), sortedStruct(0).getAs[Long]("UpdatedtimeStamp"))
  //aggregation for checking nulls and populating first not null value
  sortedStruct
    .foldLeft(first)((x, y) => {
      out(
        if(x.Name == null || x.Name.isEmpty) y.getAs[String]("Name") else x.Name,
        if(x.Passport == null || x.Passport.isEmpty) y.getAs[String]("Passport") else x.Passport,
        if(x.Country == null || x.Country.isEmpty) y.getAs[String]("Country") else x.Country,
        if(x.License == null || x.License.isEmpty) y.getAs[String]("License") else x.License,
        x.UpdatedtimeStamp)
    })
})
//making the rest of the columns as one column and changing the UpdatedtimeStamp column to long for sorting in udf
df.select(col("ID"), struct(col("Name"), col("Passport"), col("Country"), col("License"), unix_timestamp(col("UpdatedtimeStamp"), "MM-dd-yyyy").as("UpdatedtimeStamp")).as("struct"))
    //grouping and collecting the structs and passing to udf function for manipulation
    .groupBy("ID").agg(sortAndAggUdf(collect_list("struct")).as("struct"))
    //separating the aggregated columns to separate columns
    .select(col("ID"), col("struct.*"))
    //getting the date in correct format
    .withColumn("UpdatedtimeStamp", date_format(col("UpdatedtimeStamp").cast("timestamp"), "MM-dd-yyyy"))
 .show(false)

哪个应该给你

+---+----+--------+-------+-------+----------------+
|ID |Name|Passport|Country|License|UpdatedtimeStamp|
+---+----+--------+-------+-------+----------------+
|1  |Shah|12345   |null   |ABC    |12-02-2018      |
|2  |PJ  |null    |ANB    |a      |10-02-2018      |
+---+----+--------+-------+-------+----------------+

当然还需要一个案例类

case class out(Name: String, Passport: String, Country: String, License: String, UpdatedtimeStamp: Long)