I have a file which is file1snappy.parquet. It is having a complex data structure like a map, array inside that.After processing that I got final result.while writing that results to csv I am getting some error saying
"Exception in thread "main" java.lang.UnsupportedOperationException: CSV data source does not support map<string,bigint> data type."
Code which I have used:
val conf=new SparkConf().setAppName("student-example").setMaster("local")
val sc = new SparkContext(conf)
val sqlcontext = new org.apache.spark.sql.SQLContext(sc)
val datadf = sqlcontext.read.parquet("C:\\file1.snappy.parquet")
def sumaggr=udf((aggr: Map[String, collection.mutable.WrappedArray[Long]]) => if (aggr.keySet.contains("aggr")) aggr("aggr").sum else 0)
datadf.select(col("neid"),sumaggr(col("marks")).as("sum")).filter(col("sum") =!= 0).show(false)
datadf.write.format("com.databricks.spark.csv").option("header", "true").save("C:\\myfile.csv")
I tried converting datadf.toString() but still I am facing same issue.
How can write that result to CSV.
spark version 2.1.1
Spark CSV source supports only atomic types. You cannot store any columns that are non-atomic
I think best is to create a JSON for the column that has map<string,bigint> as a datatype and save it in csv as below.
import spark.implicits._
import org.apache.spark.sql.functions._
datadf.withColumn("column_name_with_map_type", to_json(struct($"column_name_with_map_type"))).write.csv("outputpath")
Hope this helps!
You are trying to save the output of
val datadf = sqlcontext.read.parquet("C:\\file1.snappy.parquet")
which I guess is a mistake as the udf function and all the aggregation done would go in vain if you do so
So I think you want to save the output of
datadf.select(col("neid"),sumaggr(col("marks")).as("sum")).filter(col("sum") =!= 0).show(false)
So you need to save it in a new dataframe variable and use that variable to save.
val finalDF = datadf.select(col("neid"),sumaggr(col("marks")).as("sum")).filter(col("sum") =!= 0)
finalDF.write.format("com.databricks.spark.csv").option("header", "true").save("C:\\myfile.csv")
And you should be fine.
Related
I'm trying to read the CSV files stored on HDFS using sparkSession and count the number of lines and print the value on the console. However, I'm constantly getting NullPointerException while calculating the count. Below is the code snippet,
val validEmployeeIds = Set("12345", "6789")
val count = sparkSession
.read
.option("escape", "\"")
.option("quote", "\"")
.csv(inputPath)
.filter(row => validEmployeeIds.contains(row.getString(0)))
.distinct()
.count()
println(count)
I'm getting an NPE exactly at .filter condition. If I remove .filter in the code, it runs fine and prints the count. How can I handle this NPE?
The inputPath is a folder that contains contains multiple CSV files. Each CSV file has two columns, one represents Id and other represents name of the employee. A sample CSV extract is below:
12345,Employee1
AA888,Employee2
I'm using Spark version 2.3.1.
Try using isin function.
import spark.implicits._
val validEmployeeIds = List("12345", "6789")
val df = // Read CSV
df.filter('_c0.isin(validEmployeeIds:_*)).distinct().count()
I am trying to create an empty dataframe and simply writing it to csv file.I was expecting schema will get written to file as I have specified header=true while writing but its creating empty .csv file.
I have tried setting different properties but nothing is working.
object HeaderTest extends App {
val spark = SparkSession.builder.master("local").appName("learning
spark").getOrCreate
val sc = spark.sparkContext
import spark.implicits._
val df: DataFrame = Seq.empty[(String, Int)].toDF("k", "v")
val f = "E:\\data.csv"
df.write.mode("overwrite").option("header", "true").csv(f)
}
I am new to spark, I am trying to save my text file to orc using spark-shell is their any way to do that?
vall data =sc.textFile("/yyy/yyy/yyy")
data.saveAsOrcFile("/yyy/yyy/yyy")
You can convert the RDD to DataFrame and then save it.
data.toDF().write.format("orc").save("/path/to/save/file")
To read it back, use sqlContext
import org.apache.spark.sql._
val sqlContext = new org.apache.spark.sql.hive.HiveContext(sc)
val data = sqlContext.read.format("orc").load("/path/to/file/*")
*Hi all,
I have an easy question for you all.
I have an RDD, created from kafka streaming using createStream method.
Now i want to add a timestamp as a value to this rdd before converting in to dataframe.
I have tried doing to add a value to the dataframe using with withColumn() but returning this error*
val topicMaps = Map("topic" -> 1)
val now = java.util.Calendar.getInstance().getTime()
val messages = KafkaUtils.createStream[String, String, StringDecoder, StringDecoder](ssc, kafkaConf, topicMaps, StorageLevel.MEMORY_ONLY_SER)
messages.foreachRDD(rdd =>
{
val sqlContext = new org.apache.spark.sql.SQLContext(sc)
import sqlContext.implicits._
val dataframe = sqlContext.read.json(rdd.map(_._2))
val d =dataframe.withColumn("timeStamp_column",dataframe.col("now"))
val d =dataframe.withColumn("timeStamp_column",dataframe.col("now"))
org.apache.spark.sql.AnalysisException: Cannot resolve column name "now" among (action, device_os_ver, device_type, event_name,
item_name, lat, lon, memberid, productUpccd, tenantid);
at org.apache.spark.sql.DataFrame$$anonfun$resolve$1.apply(DataFrame.scala:15
As i came to know that DataFrames cannot be altered as they are immutable, but RDDs are immutable as well.
Then what is the best way to do it.
How to a value to the RDD(adding timestamp to an RDD dynamically).
Try current_timestamp function.
import org.apache.spark.sql.functions.current_timestamp
df.withColumn("time_stamp", current_timestamp())
For add a new column with a constant like timestamp, you can use litfunction:
import org.apache.spark.sql.functions._
val newDF = oldDF.withColumn("timeStamp_column", lit(System.currentTimeMillis))
This works for me. I usually perform a write after this.
val d = dataframe.withColumn("SparkLoadedAt", current_timestamp())
In Scala/Databricks:
import org.apache.spark.sql.functions._
val newDF = oldDF.withColumn("Timestamp",current_timestamp())
See my output
I see in comments that some folks are having trouble getting the timestamp to string. Here is a way to do that using spark 3 datetime format
import org.apache.spark.sql.functions._
val d =dataframe.
.withColumn("timeStamp_column", date_format(current_timestamp(), "y-M-d'T'H:m:sX"))
I know how to read a csv file into spark using spark-csv (https://github.com/databricks/spark-csv), but I already have the csv file represented as a string and would like to convert this string directly to dataframe. Is this possible?
Update : Starting from Spark 2.2.x
there is finally a proper way to do it using Dataset.
import org.apache.spark.sql.{Dataset, SparkSession}
val spark = SparkSession.builder().appName("CsvExample").master("local").getOrCreate()
import spark.implicits._
val csvData: Dataset[String] = spark.sparkContext.parallelize(
"""
|id, date, timedump
|1, "2014/01/01 23:00:01",1499959917383
|2, "2014/11/31 12:40:32",1198138008843
""".stripMargin.lines.toList).toDS()
val frame = spark.read.option("header", true).option("inferSchema",true).csv(csvData)
frame.show()
frame.printSchema()
Old spark versions
Actually you can, though it's using library internals and not widely advertised. Just create and use your own CsvParser instance.
Example that works for me on spark 1.6.0 and spark-csv_2.10-1.4.0 below
import com.databricks.spark.csv.CsvParser
val csvData = """
|userid,organizationid,userfirstname,usermiddlename,userlastname,usertitle
|1,1,user1,m1,l1,mr
|2,2,user2,m2,l2,mr
|3,3,user3,m3,l3,mr
|""".stripMargin
val rdd = sc.parallelize(csvData.lines.toList)
val csvParser = new CsvParser()
.withUseHeader(true)
.withInferSchema(true)
val csvDataFrame: DataFrame = csvParser.csvRdd(sqlContext, rdd)
You can parse your string into a csv using, e.g. scala-csv:
val myCSVdata : Array[List[String]] =
myCSVString.split('\n').flatMap(CSVParser.parseLine(_))
Here you can do a bit more processing, data cleaning, verifying that every line parses well and has the same number of fields, etc ...
You can then make this an RDD of records:
val myCSVRDD : RDD[List[String]] = sparkContext.parallelize(msCSVdata)
Here you can massage your lists of Strings into a case class, to reflect the fields of your csv data better. You should get some inspiration from the creations of Persons in this example:
https://spark.apache.org/docs/latest/sql-programming-guide.html#inferring-the-schema-using-reflection
I omit this step.
You can then convert to a DataFrame:
import spark.implicits._
myCSVDataframe = myCSVRDD.toDF()
The accepted answer wasn't working for me in spark 2.2.0 but lead me to what I needed with csvData.lines.toList
val fileUrl = getClass.getResource(s"/file_in_resources.csv")
val stream = fileUrl.getContent.asInstanceOf[InputStream]
val streamString = Source.fromInputStream(stream).mkString
val csvList = streamString.lines.toList
spark.read
.option("header", "true")
.option("inferSchema", "true")
.csv(csvList.toDS())
.as[SomeCaseClass]