I have read about Spark's support for gzip-kind input files here, and I wonder if the same support exists for different kind of compressed files, such as .zip files. So far I have tried computing a file compressed under a zip file, but Spark seems unable to read its contents successfully.
I have taken a look to Hadoop's newAPIHadoopFile
and newAPIHadoopRDD
, but so far I have not been able to get anything working.
In addition, Spark supports creating a partition for every file under a specified folder, like in the example below:
SparkConf SpkCnf = new SparkConf().setAppName("SparkApp")
.setMaster("local[4]");
JavaSparkContext Ctx = new JavaSparkContext(SpkCnf);
JavaRDD<String> FirstRDD = Ctx.textFile("C:\input\).cache();
Where C:\input\
points to a directory with multiple files.
In the case computing zipped files would be possible, would it also be possible to pack every file under a single compressed file and follow the same pattern of one partition per file?
According to Spark Programming Guide
All of Spark’s file-based input methods, including textFile, support running on directories, compressed files, and wildcards as well. For example, you can use textFile("/my/directory"), textFile("/my/directory/.txt"), and textFile("/my/directory/.gz").
This could be expanded by providing information about what compression formats are supported by Hadoop, which basically can be checked by finding all classes extending CompressionCodec
(docs)
name | ext | codec class
-------------------------------------------------------------
bzip2 | .bz2 | org.apache.hadoop.io.compress.BZip2Codec
default | .deflate | org.apache.hadoop.io.compress.DefaultCodec
deflate | .deflate | org.apache.hadoop.io.compress.DeflateCodec
gzip | .gz | org.apache.hadoop.io.compress.GzipCodec
lz4 | .lz4 | org.apache.hadoop.io.compress.Lz4Codec
snappy | .snappy | org.apache.hadoop.io.compress.SnappyCodec
Source : List the available hadoop codecs
So the above formats and much more possibilities could be achieved simply by calling:
sc.readFile(path)
Unfortunately, zip
is not on the supported list by default.
I have found a great article: Hadoop: Processing ZIP files in Map/Reduce and some answers (example) explaining how to use imported ZipFileInputFormat
together with sc.newAPIHadoopFile
API. But this did not work for me.
Without any external dependencies, you can load your file with sc.binaryFiles
and later on decompress the PortableDataStream
reading the content. This is the approach I have chosen.
import java.io.{BufferedReader, InputStreamReader}
import java.util.zip.ZipInputStream
import org.apache.spark.SparkContext
import org.apache.spark.input.PortableDataStream
import org.apache.spark.rdd.RDD
implicit class ZipSparkContext(val sc: SparkContext) extends AnyVal {
def readFile(path: String,
minPartitions: Int = sc.defaultMinPartitions): RDD[String] = {
if (path.endsWith(".zip")) {
sc.binaryFiles(path, minPartitions)
.flatMap { case (name: String, content: PortableDataStream) =>
val zis = new ZipInputStream(content.open)
// this solution works only for single file in the zip
val entry = zis.getNextEntry
val br = new BufferedReader(new InputStreamReader(zis))
Stream.continually(br.readLine()).takeWhile(_ != null)
}
} else {
sc.textFile(path, minPartitions)
}
}
}
using this implicit class, you need to import it and call the readFile
method on SparkContext
:
import com.github.atais.spark.Implicits.ZipSparkContext
sc.readFile(path)
And the implicit class will load your zip
file properly and return RDD[String]
like it used to.
Note: This only works for single file in the zip archive!
For multiple files in your zip support, check this answer: https://stackoverflow.com/a/45958458/1549135
You can use sc.binaryFiles to open the zip file in binary format, then unzip it into the text format. Unfortunately, the zip file is not split-able.. So you need to wait for the decompression, then maybe call shuffle to balance the data in each partition.
Here is an example in Python. More info is in http://gregwiki.duckdns.org/index.php/2016/04/11/read-zip-file-in-spark/
file_RDD = sc.binaryFiles( HDFS_path + data_path )
def Zip_open( binary_stream_string ) : # New version, treat a stream as zipped file
try :
pseudo_file = io.BytesIO( binary_stream_string )
zf = zipfile.ZipFile( pseudo_file )
return zf
except :
return None
def read_zip_lines(zipfile_object) :
file_iter = zipfile_object.open('diff.txt')
data = file_iter.readlines()
return data
My_RDD = file_RDD.map(lambda kv: (kv[0], Zip_open(kv[1])))
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