Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
335 views
in Technique[技术] by (71.8m points)

Spark parquet partitioning : Large number of files

I am trying to leverage spark partitioning. I was trying to do something like

data.write.partitionBy("key").parquet("/location")

The issue here each partition creates huge number of parquet files which result slow read if I am trying to read from the root directory.

To avoid that I tried

data.coalese(numPart).write.partitionBy("key").parquet("/location")

This however creates numPart number of parquet files in each partition. Now my partition size is different. SO I would ideally like to have separate coalesce per partition. This is however doesn't look like an easy thing. I need to visit all the partition coalesce to a certain number and store at a separate location.

How should I use partitioning to avoid many files after write?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

First I would really avoid using coalesce, as this is often pushed up further in the chain of transformation and may destroy the parallelism of your job (I asked about this issue here : Coalesce reduces parallelism of entire stage (spark))

Writing 1 file per parquet-partition is realtively easy (see Spark dataframe write method writing many small files):

data.repartition($"key").write.partitionBy("key").parquet("/location")

If you want to set an arbitrary number of files (or files which have all the same size), you need to further repartition your data using another attribute which could be used (I cannot tell you what this might be in your case):

data.repartition($"key",$"another_key").write.partitionBy("key").parquet("/location")

another_key could be another attribute of your dataset, or a derived attribute using some modulo or rounding-operations on existing attributes. You could even use window-functions with row_number over key and then round this by something like

data.repartition($"key",floor($"row_number"/N)*N).write.partitionBy("key").parquet("/location")

This would put you N records into 1 parquet file

using orderBy

You can also control the number of files without repartitioning by ordering your dataframe accordingly:

data.orderBy($"key").write.partitionBy("key").parquet("/location")

This will lead to a total of (at least, but not much more than) spark.sql.shuffle.partitions files across all partitions (by default 200). It's even beneficial to add a second ordering column after $key, as parquet will remember the ordering of the dataframe and will write the statistics accordingly. For example, you can order by an ID:

data.orderBy($"key",$"id").write.partitionBy("key").parquet("/location")

This will not change the number of files, but it will improve the performance when you query your parquet file for a given key and id. See e.g. https://www.slideshare.net/RyanBlue3/parquet-performance-tuning-the-missing-guide and https://db-blog.web.cern.ch/blog/luca-canali/2017-06-diving-spark-and-parquet-workloads-example

Spark 2.2+

From Spark 2.2 on, you can also play with the new option maxRecordsPerFile to limit the number of records per file if you have too large files. You will still get at least N files if you have N partitions, but you can split the file written by 1 partition (task) into smaller chunks:

df.write
.option("maxRecordsPerFile", 10000)
...

See e.g. http://www.gatorsmile.io/anticipated-feature-in-spark-2-2-max-records-written-per-file/ and spark write to disk with N files less than N partitions


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...