I have a case class RDD in Scala and need to find the earliest date by each group (patientID).
Here is the input:
patientID date
000000047-01 2008-03-21T21:00:00Z
000000047-01 2007-10-24T19:45:00Z
000000485-01 2011-06-17T21:00:00Z
000000485-01 2006-02-22T18:45:00Z
The expected should be:
patientID date
000000047-01 2007-10-24T19:45:00Z
000000485-01 2006-02-22T18:45:00Z
I tried something like following but didn't work
val out = medication.groupBy(x => x.patientID).sortBy(x => x.date).take(1)
Okay!
So I understand your question correctly you want the top from every record, if that's the case then here I have created the solution.
val dataDF = Seq(
("000000047-01", "2008-03-21T21:00:00Z"),
("000000047-01" , "2007-10-24T19:45:00Z"),
("000000485-01", "2011-06-17T21:00:00Z"),
("000000485-01", "2006-02-22T18:45:00Z"))
import spark.implicits._
val dfWithSchema = dataDF.toDF("patientId", "date")
val winSpec = Window.partitionBy("patientId").orderBy("date")
val rank_df = dfWithSchema.withColumn("rank", rank().over(winSpec)).orderBy(col("patientId"))
val result = rank_df.select(col("patientId"),col("date")).where(col("rank") === 1)
result.show()
Please ignore the steps for creating the DF with the schema if you have already schema defined with your data.
Related
I have a Spark dataframe where few columns having a different type of date format.
To handle this I have written below code to keep a consistent type of format for all the date columns.
As the date column date format may get change every time hence I have defined a set of date formats in dt_formats.
def to_timestamp_multiple(s: Column, formats: Seq[String]): Column = {
coalesce(formats.map(fmt => to_timestamp(s, fmt)):_*)
}
val dt_formats= Seq("dd-MMM-yyyy", "MMM-dd-yyyy", "yyyy-MM-dd","MM/dd/yy","dd-MM-yy","dd-MM-yyyy","yyyy/MM/dd","dd/MM/yyyy")
val newDF = df.withColumn("ETD1", date_format(to_timestamp_multiple($"ETD",Seq("dd-MMM-yyyy", dt_formats)).cast("date"), "yyyy-MM-dd")).drop("ETD").withColumnRenamed("ETD1","ETD")
But here I have to create a new column then I have to drop older column then rename the new column.
that make the code unnecessary very clumsy hence I want to get override from this code.
I am trying to implement similar functionality by writing a Scala below function but it is throwing the exception org.apache.spark.sql.catalyst.parser.ParseException:, but I am unable to identify the what change I should made to make it work..
val CleansedData= rawDF.selectExpr(rawDF.columns.map(
x => { x match {
case "ETA" => s"""date_format(to_timestamp_multiple($x, dt_formats).cast("date"), "yyyy-MM-dd") as ETA"""
case _ => x
} } ) : _*)
Hence seeking help.
Thanks in advance.
Create a UDF in order to use with select. The select method takes columns and produces another DataFrame.
Also, instead of using coalesce, it might be more straightforward simply to build a parser that handles all of the formats. You can use DateTimeFormatterBuilder for this.
import java.time.format.DateTimeFormatter
import java.time.format.DateTimeFormatterBuilder
import org.apache.spark.sql.functions.udf
import java.time.LocalDate
import scala.util.Try
import java.sql.Date
val dtFormatStrings:Seq[String] = Seq("dd-MMM-yyyy", "MMM-dd-yyyy", "yyyy-MM-dd","MM/dd/yy","dd-MM-yy","dd-MM-yyyy","yyyy/MM/dd","dd/MM/yyyy")
// use foldLeft with appendOptional method, which for each format,
// returns a new builder with that additional possible format
val initBuilder = new DateTimeFormatterBuilder()
val builder: DateTimeFormatterBuilder = dtFormatStrings.foldLeft(initBuilder)(
(b: DateTimeFormatterBuilder, s:String) => b.appendOptional(DateTimeFormatter.ofPattern(s)))
val formatter = builder.toFormatter()
// Create the UDF, which just takes
// any function returning a sql-compatible type (java.sql.Date, here)
def toTimeStamp2(dateString:String): Date = {
val dateTry: Try[Date] = Try(java.sql.Date.valueOf(LocalDate.parse(dateString, formatter)))
dateTry.toOption.getOrElse(null)
}
val timeConversionUdf = udf(toTimeStamp2 _)
// example DF and new DF
val df = Seq(("05/08/20"), ("2020-04-03"), ("unparseable")).toDF("ETD")
df.select(timeConversionUdf(col("ETD"))).toDF("ETD2").show
Output:
+----------+
| ETD2|
+----------+
|2020-05-08|
|2020-04-03|
| null|
+----------+
Note that unparseable values end up null, as shown.
try withColumn(...) with same name and coalesce as below-
val dt_formats= Seq("dd-MMM-yyyy", "MMM-dd-yyyy", "yyyy-MM-dd","MM/dd/yy","dd-MM-yy","dd-MM-yyyy","yyyy/MM/dd","dd/MM/yyyy")
val newDF = df.withColumn("ETD", coalesce(dt_formats.map(fmt => to_date($"ETD", fmt)):_*))
ErrorHi I am trying to a new column to a Spark. I am trying in a data set where I want to add the percentage made by in all games.
The data set looks like this:
Name, Platform, Year, Genre, Publisher, NA_Sales, EU_Sales, JP_Sales, Other_Sales
val vgdataLines = sc.textFile("hdfs:///user/ashhall1616/bdc_data/t1/vgsales-small.csv")
val vgdata = vgdataLines.map(_.split(";"))
def toPercentage(x: Double): Double = {x * 100} val countPubl = vgdata.map(r => (r(4),1)).reduceByKey(_+_)
val addpercen = countPubl.withColumn("count", toPercentage($"count"/countPubl.count(_._2)))
I used withColumn() to add new column 'count' and expected output to be like:
(Ubisoft,3,15.0)
Can anyone tell whats wrong here?
You cannot use "withColumn" with an RDD.
You could do as follow
val addpercen = countPubl.map({case(key, value) => (key, value, toPercentage(value))})
use map to add a calculated value as new column and convert to a DataFrame if you want
import spark.implicits._
val myDf = addpercen.toDF("key","value","myNewColumn")
myDf.show()
Hope it helps.
You can not use withColumn with an RDD hence convert it to DataFrame as below and then use it
val countPubl : DataFrame = vgdata.map(r => (r(4),1)).reduceByKey(_+_).toDF()
If you still looking to use RDD then just converto it back to RDD once you add the with column as
val javaRdd : JavaRDD[Row] = countPubl.withColumn("...",col("...")).toJavaRDD
Would you be able to help in this spark prob statement
Data -
empno|ename|designation|manager|hire_date|sal|deptno
7369|SMITH|CLERK|9902|2010-12-17|800.00|20
7499|ALLEN|SALESMAN|9698|2011-02-20|1600.00|30
Code:
val rawrdd = spark.sparkContext.textFile("C:\\Users\\cmohamma\\data\\delta scenarios\\emp_20191010.txt")
val refinedRDD = rawrdd.map( lines => {
val fields = lines.split("\\|") (fields(0).toInt,fields(1),fields(2),fields(3).toInt,fields(4).toDate,fields(5).toFloat,fields(6).toInt)
})
Problem Statement - This is not working -fields(4).toDate , whats is the alternative or what is the usage ?
What i have tried ?
tried replacing it to - to_date(col(fields(4)) , "yyy-MM-dd") - Not working
2.
Step 1.
val refinedRDD = rawrdd.map( lines => {
val fields = lines.split("\\|")
(fields(0),fields(1),fields(2),fields(3),fields(4),fields(5),fields(6))
})
Now this tuples are all strings
Step 2.
mySchema = StructType(StructField(empno,IntegerType,true), StructField(ename,StringType,true), StructField(designation,StringType,true), StructField(manager,IntegerType,true), StructField(hire_date,DateType,true), StructField(sal,DoubleType,true), StructField(deptno,IntegerType,true))
Step 3. converting the string tuples to Rows
val rowRDD = refinedRDD.map(attributes => Row(attributes._1, attributes._2, attributes._3, attributes._4, attributes._5 , attributes._6, attributes._7))
Step 4.
val empDF = spark.createDataFrame(rowRDD, mySchema)
This is also not working and gives error related to types. to solve this i changed the step 1 as
(fields(0).toInt,fields(1),fields(2),fields(3).toInt,fields(4),fields(5).toFloat,fields(6).toInt)
Now this is giving error for the date type column and i am again at the main problem.
Use Case - use textFile Api, convert this to a dataframe using custom schema (StructType) on top of it.
This can be done using the case class but in case class also i would be stuck where i would need to do a fields(4).toDate (i know i can cast string to date later in code but if the above problem solutionis possible)
You can use the following code snippet
import org.apache.spark.sql.functions.to_timestamp
scala> val df = spark.read.format("csv").option("header", "true").option("delimiter", "|").load("gs://otif-etl-input/test.csv")
df: org.apache.spark.sql.DataFrame = [empno: string, ename: string ... 5 more fields]
scala> val ts = to_timestamp($"hire_date", "yyyy-MM-dd")
ts: org.apache.spark.sql.Column = to_timestamp(`hire_date`, 'yyyy-MM-dd')
scala> val enriched_df = df.withColumn("ts", ts).show(2, false)
+-----+-----+-----------+-------+----------+-------+----------+-------------------+
|empno|ename|designation|manager|hire_date |sal |deptno |ts |
+-----+-----+-----------+-------+----------+-------+----------+-------------------+
|7369 |SMITH|CLERK |9902 |2010-12-17|800.00 |20 |2010-12-17 00:00:00|
|7499 |ALLEN|SALESMAN |9698 |2011-02-20|1600.00|30 |2011-02-20 00:00:00|
+-----+-----+-----------+-------+----------+-------+----------+-------------------+
enriched_df: Unit = ()
There are multiple ways to cast your data to proper data types.
First : use InferSchema
val df = spark.read .option("delimiter", "\\|").option("header", true) .option("inferSchema", "true").csv(path)
df.printSchema
Some time it doesn't work as expected. see details here
Second : provide your own Datatype conversion template
val rawDF = Seq(("7369", "SMITH" , "2010-12-17", "800.00"), ("7499", "ALLEN","2011-02-20", "1600.00")).toDF("empno", "ename","hire_date", "sal")
//define schema in DF , hire_date as Date
val schemaDF = Seq(("empno", "INT"), ("ename", "STRING"), (**"hire_date", "date"**) , ("sal", "double")).toDF("columnName", "columnType")
rawDF.printSchema
//fetch schema details
val dataTypes = schemaDF.select("columnName", "columnType")
val listOfElements = dataTypes.collect.map(_.toSeq.toList)
//creating a map friendly template
val validationTemplate = (c: Any, t: Any) => {
val column = c.asInstanceOf[String]
val typ = t.asInstanceOf[String]
col(column).cast(typ)
}
//Apply datatype conversion template on rawDF
val convertedDF = rawDF.select(listOfElements.map(element => validationTemplate(element(0), element(1))): _*)
println("Conversion done!")
convertedDF.show()
convertedDF.printSchema
Third : Case Class
Create schema from caseclass with ScalaReflection and provide this customized schema while loading DF.
import org.apache.spark.sql.catalyst.ScalaReflection
import org.apache.spark.sql.types._
case class MySchema(empno: int, ename: String, hire_date: Date, sal: Double)
val schema = ScalaReflection.schemaFor[MySchema].dataType.asInstanceOf[StructType]
val rawDF = spark.read.schema(schema).option("header", "true").load(path)
rawDF.printSchema
Hope this will help.
In Spark 1.6 , Basically I would like to apply partition by and then do order by using two columns so that I can apply rank logic for each partition
val str = "insertdatetime,a_load_dt"
val orderByList = str.split(",")
val ptr = "memberidnum"
val partitionsColumnsList = ptr.split(",").toList
val landingDF = hc.sql("""select memberidnum,insertdatetime,'2019-09-26' as a_load_dt from landing_omega.omegamaster""")
val stagingDF = hc.sql("""select memberidnum,insertdatetime,a_load_dt from staging_omega.omegamaster where recordstatus ='current'""")
val unionedDF = landingDF.unionAll(stagingDF)
unionedDF.registerTempTable("temp_table")
val windowFunction = Window.partitionBy(partitionsColumnsList.map(elem => col(elem)):_*).orderBy(unionedDF(orderByList(0),orderByList(1)).desc)
But it throws the below error
scala> val windowFunction = Window.partitionBy(partitionsColumnsList.map(elem => col(elem)):_*).orderBy(unionedDF(orderByList(0),orderByList(1)).desc)
<console>:56: error: too many arguments for method apply: (colName: String)org.apache.spark.sql.Column in class DataFrame
val windowFunction = Window.partitionBy(partitionsColumnsList.map(elem => col(elem)):_*).orderBy(unionedDF(orderByList(0),orderByList(1)).desc)
How do I fix this issue . I want to apply order by on two columns desc order
Please help
^
You can simply do the below change:
val windowFunction = Window.partitionBy(partitionsColumnsList.head, partitionsColumnsList.tail:_*).orderBy(unionedDF(orderByList(0),orderByList(1)).desc)
You can use the below snippet:
import org.apache.spark.sql.functions.col
import org.apache.spark.sql.expressions.Window
Window.partitionBy(partitionsColumnsList.map(col): _*)
.orderBy(array_union(orderByList.map(col): _*).desc)
If this did not work. Please let me know.
I can load data from database, and I do some process with this data.
The problem is some table has date column as 'String', but some others trait it as 'timestamp'.
I cannot know what type of date column is until loading data.
> x.getAs[String]("date") // could be error when date column is timestamp type
> x.getAs[Timestamp]("date") // could be error when date column is string type
This is how I load data from spark.
spark.read
.format("jdbc")
.option("url", url)
.option("dbtable", table)
.option("user", user)
.option("password", password)
.load()
Is there any way to trait them together? or convert it as string always?
You can pattern-match on the type of the column (using the DataFrame's schema) to decide whether to parse the String into a Timestamp or just use the Timestamp as is - and use the unix_timestamp function to do the actual conversion:
import org.apache.spark.sql.functions._
import org.apache.spark.sql.types.StringType
// preparing some example data - df1 with String type and df2 with Timestamp type
val df1 = Seq(("a", "2016-02-01"), ("b", "2016-02-02")).toDF("key", "date")
val df2 = Seq(
("a", new Timestamp(new SimpleDateFormat("yyyy-MM-dd").parse("2016-02-01").getTime)),
("b", new Timestamp(new SimpleDateFormat("yyyy-MM-dd").parse("2016-02-02").getTime))
).toDF("key", "date")
// If column is String, converts it to Timestamp
def normalizeDate(df: DataFrame): DataFrame = {
df.schema("date").dataType match {
case StringType => df.withColumn("date", unix_timestamp($"date", "yyyy-MM-dd").cast("timestamp"))
case _ => df
}
}
// after "normalizing", you can assume date has Timestamp type -
// both would print the same thing:
normalizeDate(df1).rdd.map(r => r.getAs[Timestamp]("date")).foreach(println)
normalizeDate(df2).rdd.map(r => r.getAs[Timestamp]("date")).foreach(println)
Here are a few things you can try:
(1) Start utilizing the inferSchema function during load if you have a version that supports it. This will have spark figure the data type of columns, this doesn't work in all scenarios. Also look at the input data, if you have quotes I advise adding an extra argument to account for them during the load.
val inputDF = spark.read.format("csv").option("header","true").option("inferSchema","true").load(fileLocation)
(2) To identify the data type of a column you can use the below code, it will place all of the column name and data types into their own Arrays of Strings.
val columnNames : Array[String] = inputDF.columns
val columnDataTypes : Array[String] = inputDF.schema.fields.map(x=>x.dataType).map(x=>x.toString)
It has a easy way to address this which is get(i: Int): Any. And it will be map between Spark SQL types and return types automatically. e.g.
val fieldIndex = row.fieldIndex("date")
val date = row.get(fieldIndex)
def parseLocationColumn(df: DataFrame): DataFrame = {
df.schema("location").dataType match {
case StringType => df.withColumn("locationTemp", $"location")
.withColumn("countryTemp", lit("Unknown"))
.withColumn("regionTemp", lit("Unknown"))
.withColumn("zoneTemp", lit("Unknown"))
case _ => df.withColumn("locationTemp", $"location.location")
.withColumn("countryTemp", $"location.country")
.withColumn("regionTemp", $"location.region")
.withColumn("zoneTemp", $"location.zone")
}
}