Parquet file is not partitioned Spark - scala

I am trying to save a parquet Spark dataframe with partitioning to the temporary directory for unit tests, however, for some reason partitions are not created. The data itself is saved into the directory and can be used for tests.
Here is the method I have created for that:
def saveParquet(df: DataFrame, partitions: String*): String = {
val path = createTempDir()
df.repartition(1).parquet(path)(partitions: _*)
path
}
val feedPath: String = saveParquet(feedDF.select(feed.schema), "processing_time")
This method works for various dataframe with various schemas but for some reason does not generate partitions for this one. I have logged out the resulting path and it looks like this:
/var/folders/xg/fur_diuhg83b2ba15ih2rt822000dhst/T/testutils-samples8512758291/jf81n7bsj-95hs-573n-b73h-7531ug04515
But it should look like this:
/var/folders/xg/fur_diuhg83b2ba15ih2rt822000dhst/T/testutils-samples8512758291/jf81n7bsj-95hs-573n-b73h-7531ug04515/processing_time=1591714800000/part-some-random-numbersnappy.parquet
I have checked that the data and all the columns are read just fine before partitioning, as soon as partition call is created this problem occurs. Also, I ran a regex on directories which failed with match error on test samples - s".*processing_time=([0-9]+)/.*parquet".r
So what could be the reason of this problem? How else can I partition the dataframe?
Dataframe schema looks like this:
val schema: StructType = StructType(
Seq(
StructField("field1", StringType),
StructField("field2", LongType),
StructField("field3", StringType),
StructField("field4Id", IntegerType, nullable = true),
StructField("field4", FloatType, nullable = true),
StructField("field5Id", IntegerType, nullable = true),
StructField("field5", FloatType, nullable = true),
StructField("field6Id", IntegerType, nullable = true),
StructField("field6", FloatType, nullable = true),
//partition keys
StructField("processing_time", LongType)
)
)

Related

Cannot resolve overloaded method 'createDataFrame'

The following code:
val data1 = Seq(("Android", 1, "2021-07-24 12:01:19.000", "play"), ("Android", 1, "2021-07-24 12:02:19.000", "stop"),
("Apple", 1, "2021-07-24 12:03:19.000", "play"), ("Apple", 1, "2021-07-24 12:04:19.000", "stop"))
val schema1 = StructType(Array(StructField("device_id", StringType, true),
StructField("video_id", IntegerType, true),
StructField("event_timestamp", StringType, true),
StructField("event_type", StringType, true)
))
val spark = SparkSession.builder()
.enableHiveSupport()
.appName("PlayStop")
.getOrCreate()
var transaction=spark.createDataFrame(data1, schema1)
produces the error:
Cannot resolve overloaded method 'createDataFrame'
Why?
And how to fix it?
If your schema consists of default StructField settings, the easiest way to create a DataFrame would be to simply apply toDF():
val transaction = data1.toDF("device_id", "video_id", "event_timestamp", "event_type")
To specify custom schema definition, note that createDataFrame() takes a RDD[Row] and schema as its parameters. In your case, you could transform data1 into a RDD[Row] like below:
val transaction = spark.createDataFrame(sc.parallelize(data1.map(Row(_))), schema1)
An alternative is to use toDF, followed by rdd which represents a DataFrame (i.e. Dataset[Row]) as RDD[Row]:
val transaction = spark.createDataFrame(data1.toDF.rdd, schema1)

Define StructType as input datatype of a Function Spark-Scala 2.11 [duplicate]

This question already has an answer here:
Defining a UDF that accepts an Array of objects in a Spark DataFrame?
(1 answer)
Closed 3 years ago.
I'm trying to write a Spark UDF in scala, I need to define a Function's input datatype
I have a schema variable with the StructType, mentioned the same below.
import org.apache.spark.sql.types._
val relationsSchema = StructType(
Seq(
StructField("relation", ArrayType(
StructType(Seq(
StructField("attribute", StringType, true),
StructField("email", StringType, true),
StructField("fname", StringType, true),
StructField("lname", StringType, true)
)
), true
), true)
)
)
I'm trying to write a Function like below
val relationsFunc: Array[Map[String,String]] => Array[String] = _.map(do something)
val relationUDF = udf(relationsFunc)
input.withColumn("relation",relationUDF(col("relation")))
above code throws below exception
org.apache.spark.sql.AnalysisException: cannot resolve 'UDF(relation)' due to data type mismatch: argument 1 requires array<map<string,string>> type, however, '`relation`' is of array<struct<attribute:string,email:string,fname:string,lname:string>> type.;;
'Project [relation#89, UDF(relation#89) AS proc#273]
if I give the input type as
val relationsFunc: StructType => Array[String] =
I'm not able to implement the logic, as _.map gives me metadata, filed names, etc.
Please advice how to define relationsSchema as input datatype in the below function.
val relationsFunc: ? => Array[String] = _.map(somelogic)
Your structure under relation is a Row, so your function should have the following signature :
val relationsFunc: Array[Row] => Array[String]
then you can access your data either by position or by name, ie :
{r:Row => r.getAs[String]("email")}
Check the mapping table in the documentation to determine the data type representations between Spark SQL and Scala: https://spark.apache.org/docs/2.4.4/sql-reference.html#data-types
Your relation field is a Spark SQL complex type of type StructType, which is represented by Scala type org.apache.spark.sql.Row so this is the input type you should be using.
I used your code to create this complete working example that extracts email values:
import org.apache.spark.sql.types._
import org.apache.spark.sql.Row
val relationsSchema = StructType(
Seq(
StructField("relation", ArrayType(
StructType(
Seq(
StructField("attribute", StringType, true),
StructField("email", StringType, true),
StructField("fname", StringType, true),
StructField("lname", StringType, true)
)
), true
), true)
)
)
val data = Seq(
Row("{'relation':[{'attribute':'1','email':'johnny#example.com','fname': 'Johnny','lname': 'Appleseed'}]}")
)
val df = spark.createDataFrame(
spark.sparkContext.parallelize(data),
relationsSchema
)
val relationsFunc = (relation: Array[Row]) => relation.map(_.getAs[String]("email"))
val relationUdf = udf(relationsFunc)
df.withColumn("relation", relationUdf(col("relation")))

Converting RDD into Dataframe

I am new in spark/scala.
I have a created below RDD by loading data from multiple paths. Now i want to create dataframe from same for further operations.
below should be the schema of dataframe
schema[UserId, EntityId, WebSessionId, ProductId]
rdd.foreach(println)
545456,5615615,DIKFH6545614561456,PR5454564656445454
875643,5485254,JHDSFJD543514KJKJ4
545456,5615615,DIKFH6545614561456,PR5454564656445454
545456,5615615,DIKFH6545614561456,PR5454564656445454
545456,5615615,DIKFH6545614561456,PR54545DSKJD541054
264264,3254564,MNXZCBMNABC5645SAD,PR5142545564542515
732543,8765984,UJHSG4240323545144
564574,6276832,KJDXSGFJFS2545DSAS
Will anyone please help me....!!!
I have tried same by defining schema class and mapping same against rdd but getting error
"ArrayIndexOutOfBoundsException :3"
If you treat your columns as String you can create with the following:
import org.apache.spark.sql.Row
val rdd : RDD[Row] = ???
val df = spark.createDataFrame(rdd, StructType(Seq(
StructField("userId", StringType, false),
StructField("EntityId", StringType, false),
StructField("WebSessionId", StringType, false),
StructField("ProductId", StringType, true))))
Note that you must "map" your RDD to a RDD[Row] for the compiler to allow to use the "createDataFrame" method. For the missing fields you can declare the columns as nullable in the DataFrame Schema.
In your example you are using the RDD method spark.sparkContext.textFile(). This method returns a RDD[String] that means that each element of your RDD is a line. But, you need a RDD[Row]. So you need to split your string by commas like:
val list =
List("545456,5615615,DIKFH6545614561456,PR5454564656445454",
"875643,5485254,JHDSFJD543514KJKJ4",
"545456,5615615,DIKFH6545614561456,PR5454564656445454",
"545456,5615615,DIKFH6545614561456,PR5454564656445454",
"545456,5615615,DIKFH6545614561456,PR54545DSKJD541054",
"264264,3254564,MNXZCBMNABC5645SAD,PR5142545564542515",
"732543,8765984,UJHSG4240323545144","564574,6276832,KJDXSGFJFS2545DSAS")
val FilterReadClicks = spark.sparkContext.parallelize(list)
val rows: RDD[Row] = FilterReadClicks.map(line => line.split(",")).map { arr =>
val array = Row.fromSeq(arr.foldLeft(List[Any]())((a, b) => b :: a))
if(array.length == 4)
array
else Row.fromSeq(array.toSeq.:+(""))
}
rows.foreach(el => println(el.toSeq))
val df = spark.createDataFrame(rows, StructType(Seq(
StructField("userId", StringType, false),
StructField("EntityId", StringType, false),
StructField("WebSessionId", StringType, false),
StructField("ProductId", StringType, true))))
df.show()
+------------------+------------------+------------+---------+
| userId| EntityId|WebSessionId|ProductId|
+------------------+------------------+------------+---------+
|PR5454564656445454|DIKFH6545614561456| 5615615| 545456|
|JHDSFJD543514KJKJ4| 5485254| 875643| |
|PR5454564656445454|DIKFH6545614561456| 5615615| 545456|
|PR5454564656445454|DIKFH6545614561456| 5615615| 545456|
|PR54545DSKJD541054|DIKFH6545614561456| 5615615| 545456|
|PR5142545564542515|MNXZCBMNABC5645SAD| 3254564| 264264|
|UJHSG4240323545144| 8765984| 732543| |
|KJDXSGFJFS2545DSAS| 6276832| 564574| |
+------------------+------------------+------------+---------+
With rows rdd you will be able to create the dataframe.

Spark error when using except on a dataframe with MapType

I am seeing the error Cannot have map type columns in DataFrame which calls set operations when using Spark MapType.
Below is the sample code I wrote to reproduce it. I understand this is happening because the MapType objects are not hashable but I have an use case where I need to do the following.
val schema1 = StructType(Seq(
StructField("a", MapType(StringType, StringType, true)),
StructField("b", StringType, true)
))
val df = spark.read.schema(schema1).json("path")
val filteredDF = df.filter($"b" === "apple")
val otherDF = df.except(filteredDF)
Any suggestions for workarounds?

Spark Scala read custom file format to dataframe with schema

I have an input file which looks much like csv but with custom header:
FIELDS-START
field1
field2
field3
FIELDS-END-DATA-START
val1,2,3
val2,4,5
DATA-END
Task:
To read data to a typed dataframe, schema is obtained dynamically, example for this specific file:
val schema = StructType(
StructField("field1", StringType, true) ::
StructField("field2", IntegerType, true) ::
StructField("field3", IntegerType, true) :: Nil
)
So bacause of custom header I can't use spark csv reader. Other thing I tried:
val file = spark.sparkContext.textFile(...)
val data: RDD[List[String]] = file.filter(_.contains(",")).map(_.split(',').toList)
val df: DataFrame = spark.sqlContext.createDataFrame(data.map(Row.fromSeq(_)), schema)
It fails with runtime exception
java.lang.String is not a valid external type for schema of int which is because createDataFrame doesn't do any casting.
NOTE: Schema is obtained at runtime
Thanks in advance!