问题
For a Big Data project, I'm planning to use spark, which has some nice features like in-memory-computations for repeated workloads. It can run on local files or on top of HDFS.
However, in the official documentation, I can't find any hint as to how to process gzipped files. In practice, it can be quite efficient to process .gz files instead of unzipped files.
Is there a way to manually implement reading of gzipped files or is unzipping already automatically done when reading a .gz file?
回答1:
From the Spark Scala Programming guide's section on "Hadoop Datasets":
Spark can create distributed datasets from any file stored in the Hadoop distributed file system (HDFS) or other storage systems supported by Hadoop (including your local file system, Amazon S3, Hypertable, HBase, etc). Spark supports text files, SequenceFiles, and any other Hadoop InputFormat.
Support for gzip input files should work the same as it does in Hadoop. For example, sc.textFile("myFile.gz") should automatically decompress and read gzip-compressed files (textFile() is actually implemented using Hadoop's TextInputFormat, which supports gzip-compressed files).
As mentioned by @nick-chammas in the comments:
note that if you call
sc.textFile()on a gzipped file, Spark will give you an RDD with only 1 partition (as of 0.9.0). This is because gzipped files are not splittable. If you don't repartition the RDD somehow, any operations on that RDD will be limited to a single core
来源：https://stackoverflow.com/questions/16302385/is-gzip-format-supported-in-spark