I have one of our Python systems generating Parquet files using Pandas and fastparquet. These are to be read by a Scala system that runs atop Akka streams.
Akka does provide a source for reading Avro Parquet files. However, when I try to read the file, I end up with
java.lang.IllegalArgumentException: INT96 not yet implemented.
This is one of the columns that does not need to be read for the Scala application to work. My question is whether I can specify a schema and get just that one column out considering that the generated file is from fastparquet.
The relevant snippet which generates a source for reading Parquet files is:
.map(result => {
val path = s"s3a://${result.bucketName}/${result.key}"
val file = HadoopInputFile.fromPath(new Path(path), hadoopConfig)
val reader: ParquetReader[GenericRecord] =
AvroParquetReader
.builder[GenericRecord](file)
.withConf(hadoopConfig)
.build()
AvroParquetSource(reader)
})
Related
I have a framework written in scala which reads a json and processes it based on the input in the json.
Val fileContent=scala.io.Source.fromFile("filename")
val jsonText=fileContent.getLines.mkstring("\n")
val parseJson=JSON.parseFull(jsonText).get.asInstanceOf[Map[String,Any]]
The input json has a section to extract a file and place it in the path with file delimiter as CEDILLA.
The json is parsed correctly in mapR evrytime. However, it fails in hortonworks intermittently.
Can anyone help me know where this could be failing.
So, I made an Scala Application to run in Spark, and created the Uber Jar using sbt> assembly.
The file I load is a lookup needed by the application, thus the idea is to package it together. It works fine from within InteliJ using the path "src/main/resources/lookup01.csv"
I am developing in Windows, testing locally, to after deploy it to a remote test server.
But when I call spark-submit on the Windows machine, I get the error :
"org.apache.spark.sql.AnalysisException: Path does not exist: file:/H:/dev/Spark/spark-2.4.3-bin-hadoop2.7/bin/src/main/resources/"
Seems it tries to find the file in the sparkhome location instead of from inside the JAr file.
How could I express the Path so it works looking the file from within the JAR package?
Example code of the way I load the Dataframe. After loading it I transform it into other structures like Maps.
val v_lookup = sparkSession.read.option( "header", true ).csv( "src/main/resources/lookup01.csv")
What I would like to achieve is getting as way to express the path so it works in every environment I try to run the JAR, ideally working also from within InteliJ while developing.
Edit: scala version is 2.11.12
Update:
Seems that to get a hand in the file inside the JAR, I have to read it as a stream, the bellow code worked, but I cant figure out a secure way to extract the headers of the file such as SparkSession.read.option has.
val fileStream = scala.io.Source.getClass.getResourceAsStream("/lookup01.csv")
val inputDF = sparkSession.sparkContext.makeRDD(scala.io.Source.fromInputStream(fileStream).getLines().toList).toDF
When the makeRDD is applied, I get the RDD and then can convert it to a dataframe, but it seems I lost the ability tu use the option from "read" that parsed out the headers as the schema.
Any way around it when using makeRDD ?
Other problem with this is that seems that I will have to manually parse the lines into columns.
You have to get the correct path from classPath
Considering that your file is under src/main/resources:
val path = getClass.getResource("/lookup01.csv")
val v_lookup = sparkSession.read.option( "header", true ).csv(path)
So, it all points to that after the file is inside JAR, it can only be accessed as a inputstream to read the chunk of data from within the compressed file.
I arrived at a solution, even though its not pretty it does what I need, that is to read a csv file, take the 2 first columns and make it into a dataframe and after load it inside a key-value structure (in this case i created a case class to hold these pairs).
I am considering migrating these lookups to a HOCON file, that may make the process less convoluted to load these lookups
import sparkSession.implicits._
val fileStream = scala.io.Source.getClass.getResourceAsStream("/lookup01.csv")
val input = sparkSession.sparkContext.makeRDD(scala.io.Source.fromInputStream(fileStream).getLines().toList).toDF()
val myRdd = input.map {
line =>
val col = utils.Utils.splitCSVString(line.getString(0))
KeyValue(col(0), col(1))
}
val myDF = myRdd.rdd.map(x => (x.key, x.value)).collectAsMap()
fileStream.close()
I am trying to load google's Pre-trained vectors 'GoogleNews-vectors-negative300.bin.gz' Google-word2vec into spark.
I converted the bin file to txt and created a smaller chunk for testing that I called 'vectors.txt'. I tried loading it as the following:
val sparkSession = SparkSession.builder
.master("local[*]")
.appName("Word2VecExample")
.getOrCreate()
val model2= Word2VecModel.load(sparkSession.sparkContext, "src/main/resources/vectors.txt")
val synonyms = model2.findSynonyms("the", 5)
for((synonym, cosineSimilarity) <- synonyms) {
println(s"$synonym $cosineSimilarity")
}
and to my surprise I am faced with the following error:
Exception in thread "main" org.apache.hadoop.mapred.InvalidInputException: Input path does not exist: file:/home/elievex/Repository/ARCANA/src/main/resources/vectors.txt/metadata
I'm not sure where did the 'metadata' after 'vectors.txt' came from.
I am using Spark, Scala and Scala IDE for Eclipse.
What am I doing wrong? is there a different way to load a pre-trained model in spark? Would appreciate any tips.
How exactly did you get vector.txt? If you read JavaDoc for Word2VecModel.save you may see that:
This saves: - human-readable (JSON) model metadata to path/metadata/ - Parquet formatted data to path/data/
The model may be loaded using Loader.load.
So what you need is model in Parquet format which is standard for Spark ML models.
Unfortunately load from Google's native format has not been implemented yet (see SPARK-9484).
I need to save DataFrame in CSV or parquet format (as a single file) and then open it again. The amount of data will not exceed 60Mb, so a single file is reasonable solution. This simple task provides me a lot of headache... This is what I tried:
To read the file if it exists:
df = sqlContext
.read.parquet("s3n://bucket/myTest.parquet")
.toDF("key", "value", "date", "qty")
To write the file:
df.write.parquet("s3n://bucket/myTest.parquet")
This does not work because:
1) write creates the folder myTest.parquet with hadoopish files that later I cannot read with .read.parquet("s3n://bucket/myTest.parquet"). In fact I don't care about multiple hadoopish files, unless I can later read them easily into DataFrame. Is it possible?
2) I am always working with the same file myTest.parquet that I am updating and overwriting in S3. It tells me that the file cannot be saved because it already exists.
So, can someone indicate me a right way to do the read/write loop? The file format doesn't matter for me (csv,parquet,csv,hadoopish files) unleass I can make the read and write loop.
You can save your DataFrame with saveAsTable("TableName") and read it with table("TableName"). And the location can be set by spark.sql.warehouse.dir. And you can overwrite a file with mode(SaveMode.Ignore). You can read here more from the official documentation.
In Java it would look like this:
SparkSession spark = ...
spark.conf().set("spark.sql.warehouse.dir", "hdfs://localhost:9000/tables");
Dataset<Row> data = ...
data.write().mode(SaveMode.Overwrite).saveAsTable("TableName");
Now you can read from the Data with:
spark.read().table("TableName");
I am new to Spark and Scala. We have ad event log files formatted as CSV's and then compressed using pkzip. I have seen many examples on how to decompress zipped files using Java, but how would I do this using Scala for Spark? We, ultimately, want to get, extract, and load the data from each incoming file into an Hbase destination table. Maybe this can this be done with the HadoopRDD? After this, we are going to introduce Spark streaming to watch for these files.
Thanks,
Ben
Default compression support
#samthebest answer is correct, if you are using compression format that is by default available in Spark (Hadoop). Which are:
bzip2
gzip
lz4
snappy
I have explained this topic deeper in my other answer: https://stackoverflow.com/a/45958182/1549135
Reading zip
However, if you are trying to read a zip file you need to create a custom solution. One is mentioned in the answer I have already provided.
If you need to read multiple files from your archive, you might be interested in the answer I have provided: https://stackoverflow.com/a/45958458/1549135
Basically, all the time, using sc.binaryFiles and later on decompressing the PortableDataStream, like in the sample:
sc.binaryFiles(path, minPartitions)
.flatMap { case (name: String, content: PortableDataStream) =>
val zis = new ZipInputStream(content.open)
Stream.continually(zis.getNextEntry)
.takeWhile(_ != null)
.flatMap { _ =>
val br = new BufferedReader(new InputStreamReader(zis))
Stream.continually(br.readLine()).takeWhile(_ != null)
}
In Spark, provided your files have the correct filename suffix (e.g. .gz for gzipped), and it's supported by org.apache.hadoop.io.compress.CompressionCodecFactory, then you can just use
sc.textFile(path)
UPDATE: At time of writing their is a bug in Hadoop bzip2 library which means trying to read bzip2 files using spark results in weird exceptions - usually ArrayIndexOutOfBounds.