I have a streaming csv data set that comes in this format
2,C4653,C5030
2,C5782,C16712
6,C1191,C419
15,C3380,C22841
18,C2436,C5030
I am trying to take the Dstream and convert it into a DataFrame where i should get each field as a column. something like this.
col1 col2 col3
2 C4653 C5030
2 C5782 C16712
and so on.
I am using the following code but cannot get it to work. This is the code that I am using.
val messages = KafkaUtils.createDirectStream[String, String, StringDecoder, StringDecoder](
ssc, kafkaParams, topicsSet)
val lines = messages.map(_._2)
val seperator = lines.map(_.split(","))
lines.foreachRDD { rdd =>
// Get the singleton instance of SparkSession
val spark = SparkSession.builder.config(rdd.sparkContext.getConf).getOrCreate()
import spark.implicits._
// Convert RDD[String] to DataFrame
val wordsDataFrame = rdd.map(_.split(",")).toDF().show();
}
I am getting the following as output for the code I am using.
+-----------------+
| value|
+-----------------+
|[2, C4653, C5030]|
+-----------------+
However, I am trying to make it into three columns. Please help.
You can try something like this.
val wordsDataFrame = rdd.map { record => {
val recordArr = record.split(",")
(recordArr(0),recordArr(1),recordArr(2))
} }.toDF("col1","col2","col3")
Please provide a schema with toDF . Something like this val wordsDataFrame = rdd.map(_.split(",")).toDF("col1","col2","col3").show() should work then
Related
Link to (data.csv) and (output.csv)
import org.apache.spark.sql._
object Test {
def main(args: Array[String]) {
val spark = SparkSession.builder()
.appName("Test")
.master("local[*]")
.getOrCreate()
val sc = spark.sparkContext
val tempDF=spark.read.csv("data.csv")
tempDF.coalesce(1).write.parquet("Parquet")
val rdd = sc.textFile("Parquet")
I Convert data.csv into optimised parquet file and then loaded it and now i want to do all the transformation on parquet file just like i did on csv file given below and then save it as a parquet file.Link of (data.csv) and (output.csv)
val header = rdd.first
val rdd1 = rdd.filter(_ != header)
val resultRDD = rdd1.map { r =>
val Array(country, values) = r.split(",")
country -> values
}.reduceByKey((a, b) => a.split(";").zip(b.split(";")).map { case (i1, i2) => i1.toInt + i2.toInt }.mkString(";"))
import spark.sqlContext.implicits._
val dataSet = resultRDD.map { case (country: String, values: String) => CountryAgg(country, values) }.toDS()
dataSet.coalesce(1).write.option("header","true").csv("output")
}
case class CountryAgg(country: String, values: String)
}
I reckon, you are trying to add up corresponding elements from the array based on Country. I have done this using DataFrame APIs, which makes the job easier.
Code for your reference:
import org.apache.spark.sql.functions._
import org.apache.spark.sql.expressions.Window
val df = spark.read
.format("csv")
.option("header", "true")
.option("inferSchema", "true")
.option("path", "/path/to/input/data.csv")
.load()
val df1 = df.select(
$"Country",
(split($"Values", ";"))(0).alias("c1"),
(split($"Values", ";"))(1).alias("c2"),
(split($"Values", ";"))(2).alias("c3"),
(split($"Values", ";"))(3).alias("c4"),
(split($"Values", ";"))(4).alias("c5")
)
.groupBy($"Country")
.agg(
sum($"c1" cast "int").alias("s1"),
sum($"c2" cast "int").alias("s2"),
sum($"c3" cast "int").alias("s3"),
sum($"c4" cast "int").alias("s4"),
sum($"c5" cast "int").alias("s5")
)
.select(
$"Country",
concat(
$"s1", lit(";"),
$"s2", lit(";"),
$"s3", lit(";"),
$"s4", lit(";"),
$"s5"
).alias("Values")
)
df1.repartition(1)
.write
.format("csv")
.option("delimiter",",")
.option("header", "true")
.option("path", "/path/to/output")
.save()
Here is the output for your reference.
scala> df1.show()
+-------+-------------------+
|Country| Values|
+-------+-------------------+
|Germany| 144;166;151;172;70|
| China| 218;239;234;209;75|
| India| 246;153;148;100;90|
| Canada| 183;258;150;263;71|
|England|178;114;175;173;153|
+-------+-------------------+
P.S.:
You can change the output format to parquet/orc or anything you wish.
I have repartitioned df1 into 1 partition just so that you could get a single output file. You can choose to repartition or not based
on your usecase
Hope this helps.
You could just read the file as parquet and perform the same operations on the resulting dataframe:
val spark = SparkSession.builder()
.appName("Test")
.master("local[*]")
.getOrCreate()
// Read in the parquet file created above
// Parquet files are self-describing so the schema is preserved
// The result of loading a Parquet file is also a DataFrame
val parquetFileDF = spark.read.parquet("data.parquet")
If you need an rdd you can then just call:
val rdd = parquetFileDF.rdd
The you can proceed with the transformations as before and write as parquet like you have in your question.
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.
I am reading a CSV file from my local machine using spark and scala and storing into a dataframe (called df). I have to select only few selected columns with new aliasing names from the df and save to new dataframe newDf. I have tried to do the same but I am getting the error below.
main" org.apache.spark.sql.AnalysisException: cannot resolve '`history_temp.time`' given input columns: [history_temp.time, history_temp.poc]
Below is the code written to read the csv file from my local machine.
import org.apache.spark.sql.SparkSession
object DataLoadConversion {
def main(args: Array[String]): Unit = {
System.setProperty("spark.sql.warehouse.dir", "file:///C:/spark-warehouse")
val spark = SparkSession.builder().master("local").appName("DataConversion").getOrCreate()
val df = spark.read.format("com.databricks.spark.csv")
.option("quote", "\"")
.option("escape", "\"")
.option("delimiter", ",")
.option("header", "true")
.option("mode", "FAILFAST")
.option("inferSchema","true")
.load("file:///C:/Users/an/Desktop/ct_temp.csv")
df.show(5) // Till this code is working fine
val newDf = df.select("history_temp.time","history_temp.poc")
Below are the code which I tried but not working.
// val newDf = df.select($"history_temp.time",$"history_temp.poc")
// val newDf = df.select("history_temp.time","history_temp.poc")
// val newDf = df.select( df("history_temp.time").as("TIME"))
// val newDf = df.select(df.col("history_temp.time"))
// df.select(df.col("*")) // This is working
newDf.show(10)
}
}
from the looks of it. your column name format is the issue here. i am guessing they are just regular stringType but when you have something like history_temp.time spark thinks it as an arrayed column. which is not the case. I would rename all of the columns and replace "." to "". then you can run the same select and it should work. you can use foldleft to rplace all "." with "" like below.
val replacedDF = df.columns.foldleft(df){ (newdf, colname)=>
newdf.withColumnRenamed (colname, colname.replace(".","_"))
}
With that done you can select from replacedDF with below
val newDf= replacedDf.select("history_temp_time","history_temp_poc")
Let me know how it works out for you.
I would like to create dataframe names dynamically from a collection.
Please see below:
val set1 = Set("category1","category2","category3")
The following is a UDF which takes a string x from the set as input and generate the dataframe accordingly:
def catDfgen(x: String): DataFrame = {
spark.sql(s"select * from table where col1 = '$x'")
}
Now I need help here, to create not only DataFrame but also the DataFrame name should be dynamically generated in order to achieve
val category1DF = catDfgen($x)
val category2DF = catDfgen($x)
...etc. Would it be possible to do it using the code below?
set1.map( x => val $x+"DF" = catDfgen($x))
If not please suggest an effective method.
Suman, I believe the below might help your use-case
import org.apache.spark.sql.{DataFrame, SparkSession}
object Test extends App {
val spark: SparkSession = SparkSession.builder().master("local").getOrCreate()
val set1 = Set("category1","category2","category3")
val dfs: Map[String, DataFrame] = set1.map(x =>
(s"${x}DF", spark.sql(s"select * from table where col1 = '$x'").alias(s"${x}DF").toDF())
).toMap
dfs("category1DF").show()
spark.stop()
}
In EMR Spark, I have a HadoopRDD
org.apache.spark.rdd.RDD[(org.apache.hadoop.io.Text, org.apache.hadoop.dynamodb.DynamoDBItemWritable)] = HadoopRDD[0] at hadoopRDD
I want to convert this to DataFrame org.apache.spark.sql.DataFrame.
Does anyone know how to do this?
First convert it to simple types. Let's say your DynamoDBItemWritable has just one string column:
val simple: RDD[(String, String)] = rdd.map {
case (text, dbwritable) => (text.toString, dbwritable.getString(0))
}
Then you can use toDF to get a DataFrame:
import sqlContext.implicits._
val df: DataFrame = simple.toDF()