Avi*_*jit 12 scala apache-spark spark-dataframe apache-spark-dataset
我正在尝试创建一个DataFrame使用RDD.
首先,我创建一个RDD使用下面的代码 -
val account = sc.parallelize(Seq(
(1, null, 2,"F"),
(2, 2, 4, "F"),
(3, 3, 6, "N"),
(4,null,8,"F")))
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它工作正常 -
account:org.apache.spark.rdd.RDD [(Int,Any,Int,String)] = ParallelCollectionRDD [0]并行化:27
但是,当尝试创建DataFrame从RDD使用下面的代码
account.toDF("ACCT_ID", "M_CD", "C_CD","IND")
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我收到了以下错误
java.lang.UnsupportedOperationException:不支持类型为Any的架构
我分析说,每当我把null值放进去的时候,Seq只有我得到了错误.
有没有办法添加空值?
Mar*_*ace 17
不使用RDD的替代方法:
import spark.implicits._
val df = spark.createDataFrame(Seq(
(1, None, 2, "F"),
(2, Some(2), 4, "F"),
(3, Some(3), 6, "N"),
(4, None, 8, "F")
)).toDF("ACCT_ID", "M_CD", "C_CD","IND")
df.show
+-------+----+----+---+
|ACCT_ID|M_CD|C_CD|IND|
+-------+----+----+---+
| 1|null| 2| F|
| 2| 2| 4| F|
| 3| 3| 6| N|
| 4|null| 8| F|
+-------+----+----+---+
df.printSchema
root
|-- ACCT_ID: integer (nullable = false)
|-- M_CD: integer (nullable = true)
|-- C_CD: integer (nullable = false)
|-- IND: string (nullable = true)
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Zyo*_*oma 13
问题是Any是太普通的类型而Spark根本不知道如何序列化它.在您的情况下,您应该明确提供一些特定类型Integer.由于无法将空值分配给Scala中的基本类型,因此可以使用java.lang.Integer.试试这个:
val account = sc.parallelize(Seq(
(1, null.asInstanceOf[Integer], 2,"F"),
(2, new Integer(2), 4, "F"),
(3, new Integer(3), 6, "N"),
(4, null.asInstanceOf[Integer],8,"F")))
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这是一个输出:
rdd: org.apache.spark.rdd.RDD[(Int, Integer, Int, String)] = ParallelCollectionRDD[0] at parallelize at <console>:24
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和相应的DataFrame:
scala> val df = rdd.toDF("ACCT_ID", "M_CD", "C_CD","IND")
df: org.apache.spark.sql.DataFrame = [ACCT_ID: int, M_CD: int ... 2 more fields]
scala> df.show
+-------+----+----+---+
|ACCT_ID|M_CD|C_CD|IND|
+-------+----+----+---+
| 1|null| 2| F|
| 2| 2| 4| F|
| 3| 3| 6| N|
| 4|null| 8| F|
+-------+----+----+---+
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你也可以考虑一些更简洁的方法来声明空整数值,如:
object Constants {
val NullInteger: java.lang.Integer = null
}
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