【问题标题】:How to use separate key lists to perform a join between two DataFrames?如何使用单独的键列表在两个 DataFrame 之间执行连接?
【发布时间】:2022-01-14 13:26:21
【问题描述】:

我想加入两个不同的 DataFrame(dfAdfB),构建如下:

dfA.show()
+-----+-------+-------+
| id_A| name_A|address|
+-----+-------+-------+
|    1|   AAAA|  Paris|
|    4|   DDDD| Sydney|
+-----+-------+-------+

dfB.show()
+-----+-------+---------+
| id_B| name_B|      job|
+-----+-------+---------+
|    1|   AAAA|  Analyst|
|    2|   AERF| Engineer|
|    3|   UOPY| Gardener|
|    4|   DDDD|  Insurer|
+-----+-------+---------+

我需要使用以下列表来进行连接:

val keyListA = List("id_A", "name_A")
val keyListB = List("id_B", "name_B")

一个简单的解决方案是:

val join = dfA.join(
  dfA("id_A") === dfB("id_B") &&
  dfA("name_A") === dfB("name_B"),
"left_outer")

是否有一种语法允许您通过使用 keyListAkeyListB 列表来执行此连接?

【问题讨论】:

    标签: dataframe scala apache-spark apache-spark-sql


    【解决方案1】:

    如果您真的想从列名列表构建连接表达式:

    import org.apache.spark.sql.{Column, DataFrame}
    import org.apache.spark.sql.functions._
    
    val dfA: DataFrame = ???
    val dfB: DataFrame = ???
    
    val keyListA = List("id_A", "name_A", "property1_A", "property2_A", "property3_A")
    val keyListB = List("id_B", "name_B", "property1_B", "property2_B", "property3_B")
    
    
    def joinExprsFrom(keyListA: List[String], keyListB: List[String]): Column = 
      keyListA
        .zip(keyListB)
        .map { case (fromA, fromB) => col(fromA) === col(fromB) }
        .reduce((acc, expr) => acc && expr )
    
    dfA.join(
      dfB,
      joinExprsFrom(keyListA, keyListB),
      "left_outer")
    

    您需要确保keyListAkeyListB 大小相同且非空。

    【讨论】:

      猜你喜欢
      • 1970-01-01
      • 2021-04-07
      • 1970-01-01
      • 2019-06-12
      • 1970-01-01
      • 2019-06-29
      • 2016-10-01
      • 1970-01-01
      • 2017-12-06
      相关资源
      最近更新 更多