I used to use df.repartition(1200).write.parquet(...) which created 1200 number of files as specified in the repartion argument. I am now using paritionBy, i.e. df.repartition(1200).write.partitionBy("mykey").parquet(...). This works fine, except that it is now creating 1200 files per bucket of mykey. I would like to have 1200 files over all.
Other posts suggest to repartition across certain keys. The relevant documentation for my spark version (2.4.0) seems to suggest that this feature was added later. Is there any other way to achieve it? I guess I could repartition to 1200/len(unique("mykey"). But that's a bit hacky. Is there a better way to do it? I am also worrying that reducing the number of partitions results in out of memory erros.
Calling partitionBy on your writer does not change the partitioning scheme of your dataframe. Instead it is used to specify the partitioning scheme of your data once it is written to disk
Say you have a dataframe with 200 parititons and you call df.write.partitionBy("mykey").parquet(...)
Each of your partitions will bucket it's data by unique values of "mykey"
Each bucket in each partition will correspond to one file being written to a disk partition
For example lets say the dataframe has 200 values of the field mykey=KEY1
And lets say that each of these 200 values are evenly spread accross the 200 partitions with 1 per partition
then when we call df.write.partitionBy("mykey").parquet(...)
We will get 200 files in the disk partition mykey=KEY1. One from each partition
To answer your question, there are a few ways of ensuring that exactly 1200 files are written to disk. All methods require precise control of the number of unique values in your parititons
method 1
# requires mykey to have exactly 1200 unique values
df = df.repartition("mykey")
df.write.partitionBy("mykey").parquet(...)
repartitions the data so that dataframe partitions match disk partitions
repartition is an expensive operation so should be used sparingly
method 2
# requires mykey to have exactly 1200 unique values
df = df.coalesce(1)
df.write.partitionBy("mykey").parquet(...)
This will only work if the final dataset you want to write is small enough to fit into a single partition.
method 3
# requires mykey to have exactly 1 unique value
df = df.repartition(1200)
df.write.partitionBy("mykey").parquet(...)
I'm not quite sure why you want to do both repartition and partitionBy, but you could do
df = df.repartition(1200)
df = your_processing(df)
df.coalesce(1).write.partitionBy("mykey").parquet(...)
coalesce(1) merges the partition into a single one that is then split up again by the partitionBy.
To me it seems the best way to handle it was to order by mykey. This way, the right data is already in the respective partitions such that partitionBy('mykey') does not create a too many partitions (but broadly as many files as num partitions).
Im using pyspark and I have a large data source that I want to repartition specifying the files size per partition explicitly.
I know using the repartition(500) function will split my parquet into 500 files with almost equal sizes.
The problem is that new data gets added to this data source every day. On some days there might be a large input, and on some days there might be smaller inputs. So when looking at the partition file size distribution over a period of time, it varies between 200KB to 700KB per file.
I was thinking of specifying the max size per partition so that I get more or less the same file size per file per day irrespective of the number of files.
This will help me when running my job on this large dataset later on to avoid skewed executor times and shuffle times etc.
Is there a way to specify it using the repartition() function or while writing the dataframe to parquet?
You could consider writing your result with the parameter maxRecordsPerFile.
storage_location = //...
estimated_records_with_desired_size = 2000
result_df.write.option(
"maxRecordsPerFile",
estimated_records_with_desired_size) \
.parquet(storage_location, compression="snappy")
Problem outline: Say I have 300+ GB of data being processed with spark on an EMR cluster in AWS. This data has three attributes used to partition on the filesystem for use in Hive: date, hour, and (let's say) anotherAttr. I want to write this data to a fs in such a way that minimizes the number of files written.
What I'm doing right now is getting the distinct combinations of date, hour, anotherAttr, and a count of how many rows make up combination. I collect them into a List on the driver, and iterate over the list, building a new DataFrame for each combination, repartitioning that DataFrame using the number of rows to guestimate file size, and writing the files to disk with DataFrameWriter, .orc finishing it off.
We aren't using Parquet for organizational reasons.
This method works reasonably well, and solves the problem that downstream teams using Hive instead of Spark don't see performance issues resulting from a high number of files. For example, if I take the whole 300 GB DataFrame, do a repartition with 1000 partitions (in spark) and the relevant columns, and dumped it to disk, it all dumps in parallel, and finishes in ~9 min with the whole thing. But that gets up to 1000 files for the larger partitions, and that destroys Hive performance. Or it destroys some kind of performance, honestly not 100% sure what. I've just been asked to keep the file count as low as possible. With the method I'm using, I can keep the files to whatever size I want (relatively close anyway), but there is no parallelism and it takes ~45 min to run, mostly waiting on file writes.
It seems to me that since there's a 1-to-1 relationship between some source row and some destination row, and that since I can organize the data into non-overlapping "folders" (partitions for Hive), I should be able to organize my code/DataFrames in such a way that I can ask spark to write all the destination files in parallel. Does anyone have suggestions for how to attack this?
Things I've tested that did not work:
Using a scala parallel collection to kick off the writes. Whatever spark was doing with the DataFrames, it didn't separate out the tasks very well and some machines were getting massive garbage collection problems.
DataFrame.map - I tried to map across a DataFrame of the unique combinations, and kickoff writes from inside there, but there's no access to the DataFrame of the data that I actually need from within that map - the DataFrame reference is null on the executor.
DataFrame.mapPartitions - a non-starter, couldn't come up with any ideas for doing what I want from inside mapPartitions
The word 'partition' is also not especially helpful here because it refers both to the concept of spark splitting up the data by some criteria, and to the way that the data will be organized on disk for Hive. I think I was pretty clear in the usages above. So if I'm imagining a perfect solution to this problem, it's that I can create one DataFrame that has 1000 partitions based on the three attributes for fast querying, then from that create another collection of DataFrames, each one having exactly one unique combination of those attributes, repartitioned (in spark, but for Hive) with the number of partitions appropriate to the size of the data it contains. Most of the DataFrames will have 1 partition, a few will have up to 10. The files should be ~3 GB, and our EMR cluster has more RAM than that for each executor, so we shouldn't see a performance hit from these "large" partitions.
Once that list of DataFrames is created and each one is repartitioned, I could ask spark to write them all to disk in parallel.
Is something like this possible in spark?
One thing I'm conceptually unclear on: say I have
val x = spark.sql("select * from source")
and
val y = x.where(s"date=$date and hour=$hour and anotherAttr=$anotherAttr")
and
val z = x.where(s"date=$date and hour=$hour and anotherAttr=$anotherAttr2")
To what extent is y is a different DataFrame than z? If I repartition y, what effect does the shuffle have on z, and on x for that matter?
We had the same problem (almost) and we ended up by working directly with RDD (instead of DataFrames) and implementing our own partitioning mechanism (by extending org.apache.spark.Partitioner)
Details: we are reading JSON messages from Kafka. The JSON should be grouped by customerid/date/more fields and written in Hadoop using Parquet format, without creating too many small files.
The steps are (simplified version):
a)Read the messages from Kafka and transform them to a structure of RDD[(GroupBy, Message)]. GroupBy is a case class containing all the fields that are used for grouping.
b)Use a reduceByKeyLocally transformation and obtain a map of metrics (no of messages/messages size/etc) for each group - eg Map[GroupBy, GroupByMetrics]
c)Create a GroupPartitioner that's using the previously collected metrics (and some input parameters like the desired Parquet size etc) to compute how many partitions should be created for each GroupBy object. Basically we are extending org.apache.spark.Partitioner and overriding numPartitions and getPartition(key: Any)
d)we partition the RDD from a) using the previously defined partitioner: newPartitionedRdd = rdd.partitionBy(ourCustomGroupByPartitioner)
e)Invoke spark.sparkContext.runJob with two parameters: the first one is the RDD partitioned at d), the second one is a custom function (func: (TaskContext, Iterator[T]) that will write the messages taken from Iterator[T] into Hadoop/Parquet
Let's say that we have 100 mil messages, grouped like that
Group1 - 2 mil
Group2 - 80 mil
Group3 - 18 mil
and we decided that we have to use 1.5 mil messages per partition to obtain Parquet files greater than 500MB. We'll end up with 2 partitions for Group1, 54 for Group2, 12 for Group3.
This statement:
I collect them into a List on the driver, and iterate over the list,
building a new DataFrame for each combination, repartitioning that
DataFrame using the number of rows to guestimate file size, and
writing the files to disk with DataFrameWriter, .orc finishing it off.
is completely off-beam where Spark is concerned. Collecting to driver is never a good approach, volumes and OOM issues and latency in your approach is high.
Use so the below so as to simplify and get parallelism of Spark benefits saving time and money for your boss:
df.repartition(cols...)...write.partitionBy(cols...)...
shuffle occurs via repartition, no shuffling ever with partitionBy.
That simple, with Spark's default parallelism utilized.
Can we write data to say 100 files, with 10 partitions in each file?
I know we can use repartition or coalesce to reduce number of partition. But I have seen some hadoop generated avro data with much more partitions than number of files.
The number of files that get written out is controlled by the parallelization of your DataFrame or RDD. So if your data is split across 10 Spark partitions you cannot write fewer than 10 files without reducing partitioning (e.g. coalesce or repartition).
Now, having said that when data is read back in it could be split into smaller chunks based on your configured split size but depending on format and/or compression.
If instead you want to increase the number of files written per Spark partition (e.g. to prevent files that are too large), Spark 2.2 introduces a maxRecordsPerFile option when you write data out. With this you can limit the number of records that get written per file in each partition. The other option of course would be to repartition.
The following will result in 2 files being written out even though it's only got 1 partition:
val df = spark.range(100).coalesce(1)
df.write.option("maxRecordsPerFile", 50).save("/tmp/foo")
I am writing my dataframe like below
df.write().format("com.databricks.spark.avro").save("path");
However I am getting around 200 files where around 30-40 files are empty.I can understand that it might be due to empty partitions. I then updated my code like
df.coalesce(50).write().format("com.databricks.spark.avro").save("path");
But I feel it might impact performance. Is there any other better approach to limit number of output files and remove empty files
You can remove the empty partitions in your RDD before writing by using repartition method.
The default partition is 200.
The suggested number of partition is number of partitions = number of cores * 4
repartition your dataframe using this method. To eliminate skew and ensure even distribution of data choose column(s) in your dataframe with high cardinality (having unique number of values in the columns) for the partitionExprs argument to ensure even distribution.
As default no. of RDD partitions is 200; you have to do shuffle to remove skewed partitions.
You can either use repartition method on the RDD; or make use of DISTRIBUTE BY clause on dataframe - which will repartition along with distributing data among partitions evenly.
def repartition(numPartitions: Int, partitionExprs: Column*): Dataset[T]
Returns dataset instance with proper partitions.
You may use repartitionAndSortWithinPartitions - which can improve compression ratio.