在没有公共列的情况下连接两个数据流

Question

val df1 has
Customer_name 
Customer_phone
Customer_age

val df2 has
Order_name
Order_ID

这两个dataframe没有任何公共列。两个数据流中的行数和列数也不同。我尝试插入一个新的虚拟列，以增加row_index值，如下val dfr=df1.withcolumn（“row_index”，monotonically_increasing_id())所示。

但由于我使用的是Spark2，因此不支持monotonically_increasing_id方法。有没有办法把两个dataframe连接起来，这样我就可以在一张单张的excel文件中创建两个dataframe的值。

例如：

val df1:
Customer_name  Customer_phone  Customer_age
karti           9685684551     24      
raja            8595456552     22

val df2:
Order_name Order_ID
watch       1
cattoy     2

Customer_name  Customer_phone  Customer_age   Order_name  Order_ID

karti          9685684551      24             watch        1
   
raja           8595456552      22             cattoy      2

Answer 1

monotonically_increasing_id()是递增且唯一的，但不是连续的。

您可以使用zipwithindex，方法是转换为rdd并为这两个Dataframe用相同的架构重构Dataframe。

import spark.implicits._


val df1 = Seq(
  ("karti", "9685684551", 24),
  ("raja", "8595456552", 22)
).toDF("Customer_name", "Customer_phone", "Customer_age")


val df2 = Seq(
  ("watch", 1),
  ("cattoy", 2)
).toDF("Order_name", "Order_ID")

val df11 = spark.sqlContext.createDataFrame(
  df1.rdd.zipWithIndex.map {
    case (row, index) => Row.fromSeq(row.toSeq :+ index)
  },
  // Create schema for index column
  StructType(df1.schema.fields :+ StructField("index", LongType, false))
)


val df22 = spark.sqlContext.createDataFrame(
  df2.rdd.zipWithIndex.map {
    case (row, index) => Row.fromSeq(row.toSeq :+ index)
  },
  // Create schema for index column
  StructType(df2.schema.fields :+ StructField("index", LongType, false))
)

现在加入最后的数据流

df11.join(df22, Seq("index")).drop("index")

+-------------+--------------+------------+----------+--------+
|Customer_name|Customer_phone|Customer_age|Order_name|Order_ID|
+-------------+--------------+------------+----------+--------+
|karti        |9685684551    |24          |watch     |1       |
|raja         |8595456552    |22          |cattoy    |2       |
+-------------+--------------+------------+----------+--------+