I am having issues reading an ORC file directly from the Spark shell. Note: running Hadoop 1.2, and Spark 1.2, using pyspark shell, can use spark-shell (runs scala).
I have used this resource http://docs.hortonworks.com/HDPDocuments/HDP2/HDP-2.2.4/Apache_Spark_Quickstart_v224/content/ch_orc-spark-quickstart.html .
from pyspark.sql import HiveContext
hiveCtx = HiveContext(sc)
inputRead = sc.hadoopFile("hdfs://user#server:/file_path",
classOf[inputFormat:org.apache.hadoop.hive.ql.io.orc.OrcInputFormat],
classOf[outputFormat:org.apache.hadoop.hive.ql.io.orc.OrcOutputFormat])
I get an error generally saying wrong syntax. One time, the code seemed to work, I used just the 1st of three arguments passed to hadoopFile, but when I tried to use
inputRead.first()
the output was RDD[nothing, nothing]. I don't know if this is because the inputRead variable did not get created as an RDD or if it was not created at all.
I appreciate any help!
In Spark 1.5, I'm able to load my ORC file as:
val orcfile = "hdfs:///ORC_FILE_PATH"
val df = sqlContext.read.format("orc").load(orcfile)
df.show
You can try this code, it's working for me.
val LoadOrc = spark.read.option("inferSchema", true).orc("filepath")
LoadOrc.show()
you can also add the multiple path to read from
val df = sqlContext.read.format("orc").load("hdfs://localhost:8020/user/aks/input1/*","hdfs://localhost:8020/aks/input2/*/part-r-*.orc")
Related
I am just learning Spark and started with RDDs and now moving on to DataFrames. In my current pyspark project, I am reading an S3 file into an RDD and running some simple transformations on them. Here is the code.
segmentsRDD = sc.textFile(fileLocation). \
filter(lambda line: line.split(",")[6] in INCLUDE_SITES). \
filter(lambda line: line.split(",")[2] not in EXCLUDE_MARKETS). \
filter(lambda line: "null" not in line). \
map(splitComma). \
filter(lambda line: line.split(",")[5] == '1')
SplitComma is a function that does some date calculations on the row data and return 10 comma-delimited fields back. Once I get that I run the last filter as shown to only pickup rows where value in field [5] = 1. So far everything is fine.
Next, I would like to convert the segmentsRDD to DF with schema as shown below.
interim_segmentsDF = segmentsRDD.map(lambda x: x.split(",")).toDF("itemid","market","itemkey","start_offset","end_offset","time_shifted","day_shifted","tmsmarketid","caption","itemstarttime")
But I get an error about unable to convert a "pyspark.rdd.PipelinedRDD" to DataFrame. Can you please explain the difference between "pyspark.rdd.PipelinedRDD" and "row RDD"? I am attempting to convert to DF with a schema as shown. What am I missing here?
Thanks
You have to add the following lines in your code:
from pyspark.sql import SparkSession
spark = SparkSession(sc)
The method .toDF() is not an original method of the rdd.
If you take a look in Spark source code you will see that the method .toDF() is a monkey patch.
So, with SparkSession initialization you call this monkey pached method; in other words when you run rdd.toDF() you run directly the method .toDF() from Dataframe API.
I can successfully load a text file into a DataFrame with the following Apache Spark Scala code:
val df = spark.read.text("first.txt")
.withColumn("fileName", input_file_name())
.withColumn("unique_id", monotonically_increasing_id())
Is there any way to provide the multiple files in the single run? Something like this:
val df = spark.read.text("first.txt,second.txt,someother.txt")
.withColumn("fileName", input_file_name())
.withColumn("unique_id", monotonically_increasing_id())
Right now the following code doesn't work with the following error:
Exception in thread "main" org.apache.spark.sql.AnalysisException: Path does not exist: file:first.txt,second.txt,someother.txt;
at org.apache.spark.sql.execution.datasources.DataSource$$anonfun$org$apache$spark$sql$execution$datasources$DataSource$$checkAndGlobPathIfNecessary$1.apply(DataSource.scala:558)
at org.apache.spark.sql.execution.datasources.DataSource$$anonfun$org$apache$spark$sql$execution$datasources$DataSource$$checkAndGlobPathIfNecessary$1.apply(DataSource.scala:545)
How to properly load multiple text files?
The function spark.read.text() have a varargs parameter, from the docs:
def text(paths: String*): DataFrame
This means that to read multiple files you only need to supply them to the function separated by commas, i.e.
val df = spark.read.text("first.txt", "second.txt", "someother.txt")
My requirement is to write only Header CSV record using Spark Scala DataFrame. Can any one help me on this.
val OHead1 = "/xxxxx/xxxx/xxxx/xxx/OHead1/"
val sc = sparkFile.sparkContext
val outDF = csvDF.select("col_01", "col_02", "col_03").schema
sc.parallelize(Seq(outDF.fieldNames.mkString("\t"))).coalesce(1).saveAsTextFile(s"$OHead1")
The above one is working and able to create header in the CSV with tab delimiter. Since I am using spark session I am creating sparkContext in the second line. outDF is my dataframe created before these statements.
Two things are outstanding, can you one of you help me.
1. The above working code is not overriding the files, so every time I need to delete the files manually. I could not find override option, can you help me.
2. Since I am doing a select statement and schema, will it be consider as action and start another lineage for this statement. If it is true then this would degrade the performance.
If you need to output only header you can use this code:
df.schema.fieldNames.reduce(_ + "," + _)
It will create line of CSV with names of columns
I tested and the solution below did not affect any performance.
val OHead1 = "/xxxxx/xxxx/xxxx/xxx/OHead1/"
val sc = sparkFile.sparkContext
val outDF = csvDF.select("col_01", "col_02", "col_03").schema
sc.parallelize(Seq(outDF.fieldNames.mkString("\t"))).coalesce(1).saveAsTextFile(s"$OHead1")
I got a solution to handle this situation. Define the columns in the configuration file and write those columns in an file. Here is the snipet.
val Header = prop.getProperty("OUT_HEADER_COLUMNS").replaceAll("\"","").replaceAll(",","\t")
scala.tools.nsc.io.File(s"$HeadOPath").writeAll(s"$Header")
I'm new to Spark/Scala/Dataframes. I'm using Scala 2.10.5, Spark 1.6.0. I am trying to load in a csv file and then create a dataframe from it. Using the scala shell I execute the following in the order below. Once I execute line 6, I get an error that says:
error: value show is not a member of org.apache.spark.sql.DataFrameReader
Could someone advise what I might be missing? I understand I don't need to import sparkcontext if I'm using the REPL (shell) so sc will be automatically created, but any ideas what I'm doing wrong?
1.import org.apache.spark.sql.SQLContext
import sqlContext.implicits._
val sqlContext = new SQLContext(sc)
val csvfile = "path_to_filename in hdfs...."
val df = sqlContext.read.format(csvfile).option("header", "true").option("inferSchema", "true")
df.show()
Try this:
val df = sqlContext.read.option("header", "true").option("inferSchema", "true").csv(csvfile)
sqlContext.read gives you a DataFrameReader, and option and format both set some options and give you back a DataFrameReader. You need to call one of the methods that gives you a DataFrame (like csv) before you can do things like show with it.
See https://spark.apache.org/docs/latest/api/scala/index.html#org.apache.spark.sql.DataFrameReader for more info.
i am trying to upload a csv file into a tempTable such that I can query on it and I am having two issues.
First: I tried uploading the csv to a DataFrame, and this csv has some empty fields.... and I didn't find a way to do it. I found someone posting in another post to use :
val df = sqlContext.read.format("com.databricks.spark.csv").option("header", "true").load("cars.csv")
but it gives me an error saying "Failed to load class for data source: com.databricks.spark.csv"
Then I uploaded the file and read it as a text file, without the headings as:
val sqlContext = new org.apache.spark.sql.SQLContext(sc);
import sqlContext.implicits._;
case class cars(id: Int, name: String, licence: String);
val carsDF = sc.textFile("../myTests/cars.csv").map(_.split(",")).map(p => cars( p(0).trim.toInt, p(1).trim, p(2).trim) ).toDF();
carsDF.registerTempTable("cars");
val dgp = sqlContext.sql("SELECT * FROM cars");
dgp.show()
gives an error because one of the licence field is empty... I tried to control this issue when I build the data frame but did not work.
I can obviously go into the csv file but and fix by adding a null to it but U do not want to do this because of there are a lot of fields it could be problematic. I want to fix it programmatically either when i create the dataframe or the class...
any other thoughts please let me know as well
To be able to use spark-csv you have to make sure it is available. In an interactive mode the simplest solution is to use packages argument when you start shell:
bin/spark-shell --packages com.databricks:spark-csv_2.10:1.1.0
Regarding manual parsing working with csv files, especially malformed like cars.csv, requires much more work than simply splitting on commas. Some things to consider:
how to detect csv dialect, including method of string quoting
how to handle quotes and new line characters inside strings
how handle malformed lines
In case of example file you have to at least:
filter empty lines
read header
map lines to fields providing default value if field is missing
Here you go. Remember to check the delimiter for your CSV.
// create spark session
val spark = org.apache.spark.sql.SparkSession.builder
.master("local")
.appName("Spark CSV Reader")
.getOrCreate;
// read csv
val df = spark.read
.format("csv")
.option("header", "true") //reading the headers
.option("mode", "DROPMALFORMED")
.option("delimiter", ",")
.load("/your/csv/dir/simplecsv.csv")
// create a table from dataframe
df.createOrReplaceTempView("tableName")
// run your sql query
val sqlResults = spark.sql("SELECT * FROM tableName")
// display sql results
display(sqlResults)