Say I have an dataframe which contains a column (called colA) which is a seq of row. I want to to append a new field to each record of colA. (And the new filed is associated with the former record, so I have to write an udf.)
How should I write this udf?
I have tried to write a udf, which takes colA as input, and output Seq[Row] where each record contains the new filed. But the problem is the udf cannot return Seq[Row]/ The exception is 'Schema for type org.apache.spark.sql.Row is not supported'.
What should I do?
The udf that I wrote:
val convert = udf[Seq[Row], Seq[Row]](blablabla...)
And the exception is java.lang.UnsupportedOperationException: Schema for type org.apache.spark.sql.Row is not supported
since spark 2.0 you can create UDFs which return Row / Seq[Row], but you must provide the schema for the return type, e.g. if you work with an Array of Doubles :
val schema = ArrayType(DoubleType)
val myUDF = udf((s: Seq[Row]) => {
s // just pass data without modification
}, schema)
But I cant really imagine where this is useful, I would rather return tuples or case classes (or Seq thereof) from the UDFs.
EDIT : It could be useful if your row contains more than 22 fields (limit of fields for tuples/case classes)
This is an old question, I just wanted to update it according to the new version of Spark.
Since Spark 3.0.0, the method that #Raphael Roth has mentioned is deprecated. Hence, you might get an AnalysisException. The reason is that the input closure using this method doesn't have type checking and the behavior might be different from what we expect in SQL when it comes to null values.
If you really know what you're doing, you need to set spark.sql.legacy.allowUntypedScalaUDF configuration to true.
Another solution is to use case class instead of schema. For example,
case class Foo(field1: String, field2: String)
val convertFunction: Seq[Row] => Seq[Foo] = input => {
input.map {
x => // do something with x and convert to Foo
}
}
val myUdf = udf(convertFunction)
Related
I need to iterate over data frame in specific order and apply some complex logic to calculate new column.
Also my strong preference is to do it in generic way so I do not have to list all columns of a row and do df.as[my_record] or case Row(...) => as shown here. Instead, I want to access row columns by their names and just add result column(s) to source row.
Below approach works just fine but I'd like to avoid specifying schema twice: first time so that I can access columns by name while iterating and second time to process output.
import org.apache.spark.sql.Row
import org.apache.spark.sql.types._
import org.apache.spark.sql.catalyst.encoders.RowEncoder
import org.apache.spark.sql.catalyst.expressions.GenericRowWithSchema
val q = """
select 2 part, 1 id
union all select 2 part, 4 id
union all select 2 part, 3 id
union all select 2 part, 2 id
"""
val df = spark.sql(q)
def f_row(iter: Iterator[Row]) : Iterator[Row] = {
if (iter.hasNext) {
def complex_logic(p: Int): Integer = if (p == 3) null else p * 10;
val head = iter.next
val schema = StructType(head.schema.fields :+ StructField("result", IntegerType))
val r =
new GenericRowWithSchema((head.toSeq :+ complex_logic(head.getAs("id"))).toArray, schema)
iter.scanLeft(r)((r1, r2) =>
new GenericRowWithSchema((r2.toSeq :+ complex_logic(r2.getAs("id"))).toArray, schema)
)
} else iter
}
val schema = StructType(df.schema.fields :+ StructField("result", IntegerType))
val encoder = RowEncoder(schema)
df.repartition($"part").sortWithinPartitions($"id").mapPartitions(f_row)(encoder).show
What information is lost after applying mapPartitions so output cannot be processed without explicit encoder? How to avoid specifying it?
What information is lost after applying mapPartitions so output cannot be processed without
The information is hardly lost - it wasn't there from the begining - subclasses of Row or InternalRow are basically untyped, variable shape containers, which don't provide any useful type information, that could be used to derive an Encoder.
schema in GenericRowWithSchema is inconsequential as it describes content in terms of metadata not types.
How to avoid specifying it?
Sorry, you're out of luck. If you want to use dynamically typed constructs (a bag of Any) in a statically typed language you have to pay the price, which here is providing an Encoder.
OK - I have checked some of my spark code and using .mapPartitions with the Dataset API does not require me to explicitly build/pass an encoder.
You need something like:
case class Before(part: Int, id: Int)
case class After(part: Int, id: Int, newCol: String)
import spark.implicits._
// Note column names/types must match case class constructor parameters.
val beforeDS = <however you obtain your input DF>.as[Before]
def f_row(it: Iterator[Before]): Iterator[After] = ???
beforeDS.reparition($"part").sortWithinPartitions($"id").mapPartitions(f_row).show
I found below explanation sufficient, maybe it will be useful for others.
mapPartitions requires Encoder because otherwise it cannot construct Dataset from iterator or Rows. Even though each row has a schema, that shema cannot be derived (used) by constructor of Dataset[U].
def mapPartitions[U : Encoder](func: Iterator[T] => Iterator[U]): Dataset[U] = {
new Dataset[U](
sparkSession,
MapPartitions[T, U](func, logicalPlan),
implicitly[Encoder[U]])
}
On the other hand, without calling mapPartitions Spark can use the schema derived from initial query because structure (metadata) of the original columns is not changed.
I described alternatives in this answer: https://stackoverflow.com/a/53177628/7869491.
I'd like to create a Row with a schema from a case class to test one of my map functions. The most straightforward way I can think of doing this is:
import org.apache.spark.sql.Row
case class MyCaseClass(foo: String, bar: Option[String])
def buildRowWithSchema(record: MyCaseClass): Row = {
sparkSession.createDataFrame(Seq(record)).collect.head
}
However, this seemed like a lot of overhead to just get a single Row, so I looked into how I could directly create a Row with a schema. This led me to:
import org.apache.spark.sql.catalyst.expressions.GenericRowWithSchema
import org.apache.spark.sql.{Encoders, Row}
def buildRowWithSchemaV2(record: MyCaseClass): Row = {
val recordValues: Array[Any] = record.getClass.getDeclaredFields.map((field) => {
field.setAccessible(true)
field.get(record)
})
new GenericRowWithSchema(recordValues, Encoders.product[MyCaseClass].schema)
}
Unfortunately, the Row that the second version returns is different from the first Row. Option fields in the first version are reduced to their primitive values, while they are still Options in the second version. Also, the second version is quite unwieldy.
Is there a better way to do this?
The second version is returning Option itself for the bar case class field, thus you are not getting primitive value as the first version. you can use the following code for primitive values
def buildRowWithSchemaV2(record: MyCaseClass): Row = {
val recordValues: Array[Any] = record.getClass.getDeclaredFields.map((field) => {
field.setAccessible(true)
val returnValue = field.get(record)
if(returnValue.isInstanceOf[Option[String]]){
returnValue.asInstanceOf[Option[String]].get
}
else
returnValue
})
new GenericRowWithSchema(recordValues, Encoders.product[MyCaseClass].schema)
}
But meanwhile I would suggest you to use DataFrame or DataSet.
DataFrame and DataSet are themselves collections of Row with schema.
So when you have a case class defined, you just need to encode your input data into case class
For example:
lets say you have input data as
val data = Seq(("test1", "value1"),("test2", "value2"),("test3", "value3"),("test4", null))
If you have a text file you can read it with sparkContext.textFile and split according to your need.
Now when you have converted your data to RDD, converting it to dataframe or dataset is two lines code
import sqlContext.implicits._
val dataFrame = data.map(d => MyCaseClass(d._1, Option(d._2))).toDF
.toDS would generate dataset
Thus you have collection of Rows with schema
for validation you can do the followings
println(dataFrame.schema) //for checking if there is schema
println(dataFrame.take(1).getClass.getName) //for checking if it is a collection of Rows
Hope you have the right answer.
I have a dataframe which have a complex column datatype of Arraytype>. For transforming this dataframe I have created udf which can consume this column using Array [case class] as parameter. The main bottle neck here is when I create case class according to stucttype, the structfield name contains special characters for example "##field". So I provide same name to case class like this way case class (##field) and attach this to udf parameter. After interpreted in spark udf definition change name of case class field to this "$hash$hashfield". When performing transform using this dataframe it is failing because of this miss match. Please help ...
Due JVM limitations Scala stores identifiers in encoded form and currently Spark can't map ##field to $hash$hashfield.
One possible solution is to extract fields manually from raw row (but you need to know order of the fields in df, you can use df.schema for that):
val myUdf = udf { (struct: Row) =>
// Pattern match struct:
struct match {
case Row(a: String) => Foo(a)
}
// .. or extract values from Row
val `##a` = struct.getAs[String](0)
}
I am reading in a file that has many spaces and need to filter out the space. Afterwards we need to convert it to a dataframe. Example input below.
2017123 ¦ ¦10¦running¦00000¦111¦-EXAMPLE
My solution to this was the following function which parses out all spaces and trims the file.
def truncateRDD(fileName : String): RDD[String] = {
val example = sc.textFile(fileName)
example.map(lines => lines.replaceAll("""[\t\p{Zs}]+""", ""))
}
However, I am not sure how to get it into a dataframe. sc.textFile returns a RDD[String]. I tried the case class way but the issue is we have 800 field schema, case class cannot go beyond 22.
I was thinking of somehow converting RDD[String] to RDD[Row] so I can use the createDataFrame function.
val DF = spark.createDataFrame(rowRDD, schema)
Any suggestions on how to do this?
First split/parse your strings into the fields.
rdd.map( line => parse(line)) where parse is some parsing function. It could be as simple as split but you may want something more robust. This will get you an RDD[Array[String]] or similar.
You can then convert to an RDD[Row] with rdd.map(a => Row.fromSeq(a))
From there you can convert to DataFrame wising sqlContext.createDataFrame(rdd, schema) where rdd is your RDD[Row] and schema is your schema StructType.
In your case simple way :
val RowOfRDD = truncateRDD("yourfilename").map(r => Row.fromSeq(r))
How to solve productarity issue if you are using scala 2.10 ?
However, I am not sure how to get it into a dataframe. sc.textFile
returns a RDD[String]. I tried the case class way but the issue is we
have 800 field schema, case class cannot go beyond 22.
Yes, There are some limitations like productarity but we can overcome...
you can do like below example for < versions 2.11 :
prepare a case class which extends Product and overrides methods.
like...
productArity():Int: This returns the size of the attributes. In our case, it's 33. So, our implementation looks like this:
productElement(n:Int):Any: Given an index, this returns the attribute. As protection, we also have a default case, which throws an IndexOutOfBoundsException exception:
canEqual (that:Any):Boolean: This is the last of the three functions, and it serves as a boundary condition when an equality check is being done against class:
Example implementation you can refer this Student case class which has 33 fields in it
Example student dataset description here
I am trying to convert a dataframe of multiple case classes to an rdd of these multiple cases classes. I cant find any solution. This wrappedArray has drived me crazy :P
For example, assuming I am having the following:
case class randomClass(a:String,b: Double)
case class randomClass2(a:String,b: Seq[randomClass])
case class randomClass3(a:String,b:String)
val anRDD = sc.parallelize(Seq(
(randomClass2("a",Seq(randomClass("a1",1.1),randomClass("a2",1.1))),randomClass3("aa","aaa")),
(randomClass2("b",Seq(randomClass("b1",1.2),randomClass("b2",1.2))),randomClass3("bb","bbb")),
(randomClass2("c",Seq(randomClass("c1",3.2),randomClass("c2",1.2))),randomClass3("cc","Ccc"))))
val aDF = anRDD.toDF()
Assuming that I am having the aDF how can I get the anRDD???
I tried something like this just to get the second column but it was giving an error:
aDF.map { case r:Row => r.getAs[randomClass3]("_2")}
You can convert indirectly using Dataset[randomClass3]:
aDF.select($"_2.*").as[randomClass3].rdd
Spark DatataFrame / Dataset[Row] represents data as the Row objects using mapping described in Spark SQL, DataFrames and Datasets Guide Any call to getAs should use this mapping.
For the second column, which is struct<a: string, b: string>, it would be a Row as well:
aDF.rdd.map { _.getAs[Row]("_2") }
As commented by Tzach Zohar to get back a full RDD you'll need:
aDF.as[(randomClass2, randomClass3)].rdd
I don't know the scala API but have you considered the rdd value?
Maybe something like :
aDR.rdd.map { case r:Row => r.getAs[randomClass3]("_2")}