I am seeing warning method toLowerCase , is there any a better approach to lowercase column header. Also df.toDF(df.columns map(_.toLowerCase): _*) did not work.
I tried:
df.select(df.columns.map(x=> col(x).as(x.toLowerCase)):_*).show(false)
warning: Passing an explicit array value to a Scala varargs method is deprecated (since 2.13.0) and will result in a defensive copy; Use the more efficient non-copying ArraySeq.unsafeWrapArray or an explicit toIndexedSeq call```
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I've been using Spark 2.4 for a while and just started switching to Spark 3.0 in these last few days. I got this error after switching to Spark 3.0 for running udf((x: Int) => x, IntegerType):
Caused by: org.apache.spark.sql.AnalysisException: You're using untyped Scala UDF, which does not have the input type information. Spark may blindly pass null to the Scala closure with primitive-type argument, and the closure will see the default value of the Java type for the null argument, e.g. `udf((x: Int) => x, IntegerType)`, the result is 0 for null input. To get rid of this error, you could:
1. use typed Scala UDF APIs(without return type parameter), e.g. `udf((x: Int) => x)`
2. use Java UDF APIs, e.g. `udf(new UDF1[String, Integer] { override def call(s: String): Integer = s.length() }, IntegerType)`, if input types are all non primitive
3. set spark.sql.legacy.allowUntypedScalaUDF to true and use this API with caution;
The solutions are proposed by Spark itself and after googling for a while I got to Spark Migration guide page:
In Spark 3.0, using org.apache.spark.sql.functions.udf(AnyRef, DataType) is not allowed by default. Remove the return type parameter to automatically switch to typed Scala udf is recommended, or set spark.sql.legacy.allowUntypedScalaUDF to true to keep using it. In Spark version 2.4 and below, if org.apache.spark.sql.functions.udf(AnyRef, DataType) gets a Scala closure with primitive-type argument, the returned UDF returns null if the input values is null. However, in Spark 3.0, the UDF returns the default value of the Java type if the input value is null. For example, val f = udf((x: Int) => x, IntegerType), f($"x") returns null in Spark 2.4 and below if column x is null, and return 0 in Spark 3.0. This behavior change is introduced because Spark 3.0 is built with Scala 2.12 by default.
source: Spark Migration Guide
I notice that my usual way of using function.udf API, which is udf(AnyRef, DataType), is called UnTyped Scala UDF and the proposed solution, which is udf(AnyRef), is called Typed Scala UDF.
To my understanding, the first one looks more strictly typed than the second one where the first one has its output type explicitly defined and the second one does not, hence my confusion on why it's called UnTyped.
Also the function got passed to udf, which is (x:Int) => x, clearly has its input type defined but Spark claiming You're using untyped Scala UDF, which does not have the input type information?
Is my understanding correct? Even after more intensive searching I still can't find any material explaining what is UnTyped Scala UDF and what is Typed Scala UDF.
So my questions are: What are they? What are their differences?
In typed scala UDF, UDF knows the types of the columns passed as argument, whereas in untyped scala UDF, UDF doesn't know the types of the columns passed as argument
When creating typed scala UDF, the types of columns passed as argument and output of the UDF are inferred from the function arguments and output types whereas when creating untyped scala UDF, there is not type inference at all, either for arguments or output.
What can be confusing is that when creating typed UDF the types are inferred from function and not explicitly passed as argument. To be more explicit, you can write typed UDF creation as follow:
val my_typed_udf = udf[Int, Int]((x: Int) => Int)
Now, let's look at the two points you raised.
To my understanding, the first one (eg udf(AnyRef, DataType)) looks more strictly typed than the second one (eg udf(AnyRef)) where the first one has its output type explicitly defined and the second one does not, hence my confusion on why it's called UnTyped.
According to spark functions scaladoc, signatures of the udf functions that transform a function to an UDF are actually, for the first one:
def udf(f: AnyRef, dataType: DataType): UserDefinedFunction
And for the second one:
def udf[RT: TypeTag, A1: TypeTag](f: Function1[A1, RT]): UserDefinedFunction
So the second one is actually more typed than the first one, as the second one takes into account the type of the function passed as argument, whereas the first one erases the type of the function.
That's why on the first one you need to define return type, because spark needs this information but can't infer it from function passed as argument as its return type is erased, whereas in the second one the return type is inferred from function passed as argument.
Also the function got passed to udf, which is (x:Int) => x, clearly has its input type defined but Spark claiming You're using untyped Scala UDF, which does not have the input type information?
What is important here is not the function, but how Spark creates an UDF from this function.
In both cases, the function to be transformed to UDF has its input and return types defined, but those types are erased and not taken into account when creating UDF using udf(AnyRef, DataType).
This doesn't answer your original question about what the different UDFs are, but if you want to get rid of the error, in Python you can include this line in your script: spark.sql("set spark.sql.legacy.allowUntypedScalaUDF=true").
Let's say I have
public enum MyEnum {
..
}
When I invoke
java.util.Arrays.asList(MyEnum.values)
I get back
java.util.List[Array[MyEnum]]
instead of
java.util.List[MyEnum]
Why is that?
Am I missing some JavaConverter magic?
The signature of the java method is
static <T> List<T> asList(T... a)
, i.e. it takes variable number of arguments. You are passing a single array, so you are getting a list with a single array.
If you want to pass each element of the array as a separate argument, use the _*-annotation:
java.util.Arrays.asList(java.util.concurrent.TimeUnit.values: _*)
If you want a Scala List instead, use toList directly on array:
java.util.concurrent.TimeUnit.values.toList
Thanks to Andrey Tyukin for pointing out the correct answer already.
I'm just copy-pasting the scala documentation for Ascription for completeness
Ascription
Type ascription is often confused with type annotation, as the syntax in Scala is identical. The following are examples of
ascription:
Nil: List[String]
Set(values: _*)
"Daniel": AnyRef
Ascription is basically just an up-cast performed at compile-time for
the sake of the type checker. Its use is not common, but it does
happen on occasion. The most often seen case of ascription is
invoking a varargs method with a single Seq parameter. This is done
by ascribing the _* type (as in the second example above).
Ascription follows the type annotation conventions; a space follows
the colon
I am trying to call collect() over a Dataframe in Scala 2.12. Instead of returning an Array[Row], it returns me this - [Long.apache.spark.sql.Row;#58131fc
It's annoying, but on the JVM, in both Java and Scala, that's just how the toString method on arrays works. Instead of seeing the contents, you get a cryptic thing beginning with e.g. [L:
scala 2.12.10> Array("foo").toString
res0: String = [Ljava.lang.String;#8bffb8b
So it appears to me that you do in fact have an Array[Row].
See also Why does the toString method in java not seem to work for an array
I am a bit confused about using $ to reference columns in DataFrame operators like select or filter.
The following statements work:
df.select("app", "renders").show
df.select($"app", $"renders").show
But, only the first statement in the following works:
df.filter("renders = 265").show // <-- this works
df.filter($"renders" = 265).show // <-- this does not work (!) Why?!
However, this again works:
df.filter($"renders" > 265).show
Basically, what is this $ in DataFrame's operators and when/how should I use it?
Implicits are a major feature of the Scala language that take a lot of different forms--like implicit classes as we will see shortly. They have different purposes, and they all come with varying levels of debate regarding how useful or dangerous they are. Ultimately though, implicits generally come down to simply having the compiler convert one class to another when you bring them into scope.
Why does this matter? Because in Spark there is an implicitclass called StringToColumn that endows a StringContext with additional functionality. As you can see, StringToColumn adds the $ method to the Scala class StringContext. This method produces a ColumnName, which extends Column.
The end result of all this is that the $ method allows you to treat the name of a column, represented as a String, as if it were the Column itself. Implicits, when used wisely, can produce convenient conversions like this to make development easier.
So let's use this to understand what you found:
df.select("app","renders").show -- succeeds because select takes multiple Strings
df.select($"app",$"renders").show -- succeeds because select takes multiple Columnss that result after the implicit conversions are applied
df.filter("renders = 265").show -- succeeds because Spark supports SQL-like filters
df.filter($"renders" = 265).show -- fails because $"renders" is of type Column after implicit conversion, and Columns use the custom === operator for equality (unlike the case in SQL).
df.filter($"renders" > 265).show -- succeeds because you're using a Column after implicit conversion and > is a function on Column.
$ is a way to convert a string to the column with that name.
Both options of select work originally because select can receive either a column or a string.
When you do the filter $"renders" = 265 is an attempt at assigning a number to the column. > on the other hand is a comparison method. You should be using === instead of =.
val prefix = "ABC"
val renamedColumns = df.columns.map(c=> df(c).as(s"$prefix$c"))
val dfNew = df.select(renamedColumns: _*)
Hi,
I am fairly new to scala and the code above works perfectly to add a prefix to all columns. Can someone please explain the breakdown of how it works ?
The second line above will return a map of col1 as ABCcol1, col2 as ABCcol2.... etc
I have trouble understanding what the third line is doing , especailly the ":_* at the end.
thanks for your help in advance.
The third line is an example of Scala's syntactic sugar. Essentially, Scala has ways to shorten just exactly what you are typing, and you have discovered the dreaded :_*.
There are two portions to this small bit - the : and the _* serve two different purposes. The : is typically for ascription, which tells the compiler "this is the type that I need to use for this method". The _* however, is your type - in Scala this is the type varargs. Varargs is a type that has an arbitrary number of values (good resource here). It allows you to pass a method a list that you do not know the number of elements in.
In your example, you are creating a variable called renamedColumns from the columns of your original dataframe, with the new string appendage. Although you may know just how many columns are in your df, Scala does not. When you create dfNew, you are running a select statement on that and passing in your new column names, of which there could be an arbitrary number.
Essentially, you do not know how many columns you may have, so you pass in your varargs to allow the number to be arbitrary, thus determined by the compiler.