Scala 使用嵌套结构作为输入参数的Spark UDF
我正在尝试使用以下数据对Scala 使用嵌套结构作为输入参数的Spark UDF,scala,apache-spark,Scala,Apache Spark,我正在尝试使用以下数据对df进行操作: +---+----------------------------------------------------+ |ka |readingsWFreq | +---+----------------------------------------------------+ |列 |[[[列,つ],220], [[列,れっ],353], [[列,れつ],47074]] |
df
进行操作:
+---+----------------------------------------------------+
|ka |readingsWFreq |
+---+----------------------------------------------------+
|列 |[[[列,つ],220], [[列,れっ],353], [[列,れつ],47074]] |
|制 |[[[制,せい],235579]] |
以及以下结构:
root
|-- ka: string (nullable = true)
|-- readingsWFreq: array (nullable = true)
| |-- element: struct (containsNull = true)
| | |-- furigana: struct (nullable = true)
| | | |-- _1: string (nullable = true)
| | | |-- _2: string (nullable = true)
| | |-- Occ: long (nullable = true)
我的目标是将readingsWFreq
的值分成三列。为此,我尝试使用udf
s,如下所示:
val uExtractK = udf((kWFreq:Seq[((String, String), Long)]) => kWFreq.map(_._1._1))
val uExtractR = udf((kWFreq:Seq[((String, String), Long)]) => kWFreq.map(_._1._2))
val uExtractN = udf((kWFreq:Seq[((String, String), Long)]) => kWFreq.map(_._2)
val df2 = df.withColumn("K", uExtractK('readingsWFreq))
.withColumn("R", uExtractR('readingsWFreq))
.withColumn("N", uExtractN('readingsWFreq))
.drop('readingsWFreq)
但是,我得到了一个与udf
s的输入参数相关的异常:
[error] (run-main-0) org.apache.spark.sql.AnalysisException: cannot resolve
'UDF(readingsWFreq)' due to data type mismatch: argument 1 requires
array<struct<_1:struct<_1:string,_2:string>,_2:bigint>> type, however,
'`readingsWFreq`' is of
array<struct<furigana:struct<_1:string,_2:string>,Occ:bigint>> type.;;
您可以先分解外部的
数组
,获取每个值,然后再分组,并使用收集列表
收集
val df1 = df.withColumn("readingsWFreq", explode($"readingsWFreq"))
df1.select("ka", "readingsWFreq.furigana.*", "readingsWFreq.Occ")
.groupBy("ka").agg(collect_list("_1").as("K"),
collect_list("_2").as("R"),
collect_list("Occ").as("N")
)
希望这有帮助 数据帧API方法: 您不需要自定义项,只需执行以下操作:
df.select(
$"readingsWFreq.furigana._1".as("K"),
$"readingsWFreq.furigana._2".as("R"),
$"i.Occ".as("N")
)
这里的技巧是数组类型的列上的
也充当映射/投影操作符。在struct
类型的列上,此运算符用于选择元素
UDF方法
您不能将元组传递到UDF中,而是需要将它们作为行
s传递,请参见例如
在本例中,您有嵌套元组,因此需要将行分解两次:
import org.apache.spark.sql.Row
val uExtractK = udf((kWFreq:Seq[Row]) => kWFreq.map(r => r.getAs[Row](0).getAs[String](0)))
val uExtractR = udf((kWFreq:Seq[Row]) => kWFreq.map(r => r.getAs[Row](0).getAs[String](1)))
val uExtractN = udf((kWFreq:Seq[Row]) => kWFreq.map(r => r.getAs[Long](1)))
或在行上进行模式匹配
:
val uExtractK = udf((kWFreq:Seq[Row]) => kWFreq.map{case Row(kr:Row,n:Long) => kr match {case Row(k:String,r:String) => k}})
val uExtractR = udf((kWFreq:Seq[Row]) => kWFreq.map{case Row(kr:Row,n:Long) => kr match {case Row(k:String,r:String) => r}})
val uExtractN = udf((kWFreq:Seq[Row]) => kWFreq.map{case Row(kr:Row,n:Long) => n})
另请参见您是否确定维持了订购?好了,你不能在spark中依赖这个。一般来说,是的,顺序是不保证的,但是没有提到有问题的顺序。你可以用它来代替有一个明确的顺序。
val uExtractK = udf((kWFreq:Seq[Row]) => kWFreq.map{case Row(kr:Row,n:Long) => kr match {case Row(k:String,r:String) => k}})
val uExtractR = udf((kWFreq:Seq[Row]) => kWFreq.map{case Row(kr:Row,n:Long) => kr match {case Row(k:String,r:String) => r}})
val uExtractN = udf((kWFreq:Seq[Row]) => kWFreq.map{case Row(kr:Row,n:Long) => n})