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)
Related
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$1.apply(DataSource.scala:545)
How to properly load multiple text files?
The function spark.read.text() have a varargs parameter, from the docs:
def text(paths: String*): DataFrame
This means that to read multiple files you only need to supply them to the function separated by commas, i.e.
val df = spark.read.text("first.txt", "second.txt", "someother.txt")
I'm reading from a path say /json//myfiles_.json
I'm then flattening the json using explode. This causes an error since I have some empty files. How do I tell it to ignore empty files of somehow filter them out?
I can detect individual files checking if the head is empty but I need to do this on the collection of files iterated in the dataframe with the use of the wildcard path.
So the answer seems to be that I need to provide a schema explicitly because it can't infer one from empty file - as you would expect!
e.g.
val schemadf = sqlContext.read.json(schemapath) //infer schema from file with data or do manually
val schema = schemadf.schema
val raw = sqlContext.read.schema(schema).json(monthfile)
val prep = raw.withColumn("MyArray", explode($"MyArray"))
.select($"ID", $"name", $"CreatedAt")
display(prep)
My requirement is to write only Header CSV record using Spark Scala DataFrame. Can any one help me on this.
val OHead1 = "/xxxxx/xxxx/xxxx/xxx/OHead1/"
val sc = sparkFile.sparkContext
val outDF = csvDF.select("col_01", "col_02", "col_03").schema
sc.parallelize(Seq(outDF.fieldNames.mkString("\t"))).coalesce(1).saveAsTextFile(s"$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")
This question already has answers here:
Spark - load CSV file as DataFrame?
(14 answers)
Closed 4 years ago.
How do I read the data from hdfs data sets using scala language? data is any "CSV" file with limited records.
You tagged the question with Spark, so I'm assuming you are trying to use that. I would recommend you start by reading through the Spark documentation here to get an idea of how to use Spark to interact with your data.
https://spark.apache.org/docs/latest/quick-start.html
https://spark.apache.org/docs/latest/sql-programming-guide.html
But, to answer your specific question, in Spark you would read in the CSV file using code like this:
val csvDf = spark.read.format("csv")
.option("sep", ",")
.option("header", "true")
.load("hdfs://some/path/to/data.csv/")
The path your provide will be to a CSV file on HDFS, or a folder containing multiple CSV files. Also, Spark will accept other types of file systems. For example you could also use "file://" to access the local file system, or "s3://" to use S3. Once you have loaded the data you will have a Spark DataFrame object with SQL like methods available to interact with it.
Note, I provided an option for separator just to show you how to do it, but it defaults to "," anyways, so it is not required. Also, if your CSV files do not include a header, you will need to specify the Schema yourself and set header to false instead.
You can read data from HDFS by following this approach :-
val hdfs = FileSystem.get(new URI("hdfs://hdfsUrl:port/"), new Configuration())
val path = new Path("/pathOfTheFileInHDFS/")
val stream = hdfs.open(path)
def readLines = Stream.cons(stream.readLine, Stream.continually( stream.readLine))
//This example checks line for null and prints every existing line consequentally
readLines.takeWhile(_ != null).foreach(line => println(line))
Also please have a look at this article https://blog.matthewrathbone.com/2013/12/28/reading-data-from-hdfs-even-if-it-is-compressed
Please let me know if this answers your question.
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")