I have a parquet file of around 1 GB. Each data record is a reading from an IOT device which captures the energy consumed by the device in the last one minute.
Schema: houseId, deviceId, energy
The parquet file is partitioned on houseId and deviceId. A file contains the data for the last 24 hours only.
I want to execute some queries on the data residing in this parquet file using Spark SQL An example query finds out the average energy consumed per device for a given house in the last 24 hours.
Dataset<Row> df4 = ss.read().parquet("/readings.parquet");
df4.as(encoder).registerTempTable("deviceReadings");
ss.sql("Select avg(energy) from deviceReadings where houseId=3123).show();
The above code works well. I want to understand that how spark executes this query.
Does Spark read the whole Parquet file in memory from HDFS without looking at the query? (I don't believe this to be the case)
Does Spark load only the required partitions from HDFS as per the query?
What if there are multiple queries which need to be executed? Will Spark look at multiple queries while preparing an execution plan? One query may be working with just one partition whereas the second query may need all the partitions, so a consolidated plan shall load the whole file from disk in memory (if memory limits allow this).
Will it make a difference in execution time if I cache df4 dataframe above?
Does Spark read the whole Parquet file in memory from HDFS without looking at the query?
It shouldn't scan all data files, but it might in general, access metadata of all files.
Does Spark load only the required partitions from HDFS as per the query?
Yes, it does.
Does Spark load only the required partitions from HDFS as per the query?
It does not. Each query has its own execution plan.
Will it make a difference in execution time if I cache df4 dataframe above?
Yes, at least for now, it will make a difference - Caching dataframes while keeping partitions
Related
I recently had a issue with with one of my spark jobs, where I was reading a hive table having several billion records, that resulted in job failure due to high disk utilization, But after adding AWS EBS volume, the job ran without any issues. Although it resolved the issue, I have few doubts, I tried doing some research but couldn't find any clear answers. So my question is?
when a spark SQL reads a hive table, where the data is stored for processing initially and what is the entire life cycle of data in terms of its storage , if I didn't explicitly specify anything? And How adding EBS volumes solves the issue?
Spark will read the data, if it does not fit in memory, it will spill it out on disk.
A few things to note:
Data in memory is compressed, from what I read, you gain about 20% (e.g. a 100MB file will take only 80MB of memory).
Ingestion will start as soon as you read(), it is not part of the DAG, you can limit how much you ingest in the SQL query itself. The read operation is done by the executors. This example should give you a hint: https://github.com/jgperrin/net.jgp.books.spark.ch08/blob/master/src/main/java/net/jgp/books/spark/ch08/lab300_advanced_queries/MySQLWithWhereClauseToDatasetApp.java
In latest versions of Spark, you can push down the filter (for example if you filter right after the ingestion, Spark will know and optimize the ingestion), I think this works only for CSV, Avro, and Parquet. For databases (including Hive), the previous example is what I'd recommend.
Storage MUST be seen/accessible from the executors, so if you have EBS volumes, make sure they are seen/accessible from the cluster where the executors/workers are running, vs. the node where the driver is running.
Initially the data is in table location in HDFS/S3/etc. Spark spills data on local storage if it does not fit in memory.
Read Apache Spark FAQ
Does my data need to fit in memory to use Spark?
No. Spark's operators spill data to disk if it does not fit in memory,
allowing it to run well on any sized data. Likewise, cached datasets
that do not fit in memory are either spilled to disk or recomputed on
the fly when needed, as determined by the RDD's storage level.
Whenever spark reads data from hive tables, it stores it in RDD. One point i want to make clear here is hive is just a warehouse so it is like a layer which is above HDFS, when spark interacts with hive , hive provides the spark the location where the hdfs loaction exists.
Thus, Spark reads a file from HDFS, it creates a single partition for a single input split. Input split is set by the Hadoop (whatever the InputFormat used to read this file. ex: if you use textFile() it would be TextInputFormat in Hadoop, which would return you a single partition for a single block of HDFS (note:the split between partitions would be done on line split, not the exact block split), unless you have a compressed file format like Avro/parquet.
If you manually add rdd.repartition(x) it would perform a shuffle of the data from N partititons you have in rdd to x partitions you want to have, partitioning would be done on round robin basis.
If you have a 10GB uncompressed text file stored on HDFS, then with the default HDFS block size setting (256MB) it would be stored in 40blocks, which means that the RDD you read from this file would have 40partitions. When you call repartition(1000) your RDD would be marked as to be repartitioned, but in fact it would be shuffled to 1000 partitions only when you will execute an action on top of this RDD (lazy execution concept)
Now its all up to spark that how it will process the data as Spark is doing lazy evaluation , before doing the processing, spark prepare a DAG for optimal processing. One more point spark need configuration for driver memory, no of cores , no of executors etc and if the configuration is inappropriate the job will fail.
Once it prepare the DAG , then it start processing the data. So it divide your job into stages and stages into tasks. Each task will further use specific executors, shuffle , partitioning. So in your case when you do processing of bilions of records may be your configuration is not adequate for the processing. One more point when we say spark load the data in RDD/Dataframe , its managed by spark, there are option to keep the data in memory/disk/memory only etc ref -storage_spark.
Briefly,
Hive-->HDFS--->SPARK>>RDD(Storage depends as its a lazy evaluation).
you may refer the following link : Spark RDD - is partition(s) always in RAM?
What is the difference between append and overwrite to parquet in spark.
I'm processing huge amount of data for say 10 days. At present I'm processing daily logs into parquet files using "append" method and partitioning the data based on date. But the problem I'm facing is daily data is also very huge and taking a lot of time, contributing to high CPU usage as well while processing data using EMR cluster. This is making my job very slow and expensive. So I'm looking for a way where I can further split the data and can merge the data to day cluster.
Please see spark SaveMode docs
https://spark.apache.org/docs/latest/api/java/index.html
I use Spark 2.
Actually I am not the one executing the queries so I cannot include query plans. I have been asked this question by the data science team.
We are having hive table partitioned into 2000 partitions and stored in parquet format. When this respective table is used in spark, there are exactly 2000 tasks that are executed among the executors. But we have a block size of 256 MB and we are expecting the (total size/256) number of partitions which will be much lesser than 2000 for sure. Is there any internal logic that spark uses physical structure of data to create partitions. Any reference/help would be greatly appreciated.
UPDATE: It is the other way around. Actually our table is very huge like 3 TB having 2000 partitions. 3TB/256MB would actually come to 11720 but we are having exactly same number of partitions as the table is partitioned physically. I just want to understand how the tasks are generated on data volume.
In general Hive partitions are not mapped 1:1 to Spark partitions. 1 Hive partition can be split into multiple Spark partitions, and one Spark partition can hold multiple hive-partitions.
The number of Spark partitions when you load a hive-table depends on the parameters:
spark.files.maxPartitionBytes (default 128MB)
spark.files.openCostInBytes (default 4MB)
You can check the partitions e.g. using
spark.table(yourtable).rdd.partitions
This will give you an Array of FilePartitions which contain the physical path of your files.
Why you got exactly 2000 Spark partitions from your 2000 hive partitions seems a coincidence to me, in my experience this is very unlikely to happen. Note that the situation in spark 1.6 was different, there the number of spark partitions resembled the number of files on the filesystem (1 spark partition for 1 file, unless the file was very large)
I just want to understand how the tasks are generated on data volume.
Tasks are a runtime artifact and their number is exactly the number of partitions.
The number of tasks does not correlate to data volume in any way. It's a Spark developer's responsibility to have enough partitions to hold the data.
I have a spark streaming job with a batch interval of 2 mins(configurable).
This job reads from a Kafka topic and creates a Dataset and applies a schema on top of it and inserts these records into the Hive table.
The Spark Job creates one file per batch interval in the Hive partition like below:
dataset.coalesce(1).write().mode(SaveMode.Append).insertInto(targetEntityName);
Now the data that comes in is not that big, and if I increase the batch duration to maybe 10mins or so, then even I might end up getting only 2-3mb of data, which is way less than the block size.
This is the expected behaviour in Spark Streaming.
I am looking for efficient ways to do a post processing to merge all these small files and create one big file.
If anyone's done it before, please share your ideas.
I would encourage you to not use Spark to stream data from Kafka to HDFS.
Kafka Connect HDFS Plugin by Confluent (or Apache Gobblin by LinkedIn) exist for this very purpose. Both offer Hive integration.
Find my comments about compaction of small files in this Github issue
If you need to write Spark code to process Kafka data into a schema, then you can still do that, and write into another topic in (preferably) Avro format, which Hive can easily read without a predefined table schema
I personally have written a "compaction" process that actually grabs a bunch of hourly Avro data partitions from a Hive table, then converts into daily Parquet partitioned table for analytics. It's been working great so far.
If you want to batch the records before they land on HDFS, that's where Kafka Connect or Apache Nifi (mentioned in the link) can help, given that you have enough memory to store records before they are flushed to HDFS
I have exactly the same situation as you. I solved it by:
Lets assume that your new coming data are stored in a dataset: dataset1
1- Partition the table with a good partition key, in my case I have found that I can partition using a combination of keys to have around 100MB per partition.
2- Save using spark core not using spark sql:
a- load the whole partition in you memory (inside a dataset: dataset2) when you want to save
b- Then apply dataset union function: dataset3 = dataset1.union(dataset2)
c- make sure that the resulted dataset is partitioned as you wish e.g: dataset3.repartition(1)
d - save the resulting dataset in "OverWrite" mode to replace the existing file
If you need more details about any step please reach out.
I am working with a large dataset, that is partitioned by two columns - plant_name and tag_id. The second partition - tag_id has 200000 unique values, and I mostly access the data by specific tag_id values. If I use the following Spark commands:
sqlContext.setConf("spark.sql.hive.metastorePartitionPruning", "true")
sqlContext.setConf("spark.sql.parquet.filterPushdown", "true")
val df = sqlContext.sql("select * from tag_data where plant_name='PLANT01' and tag_id='1000'")
I would expect a fast response as this resolves to a single partition. In Hive and Presto this takes seconds, however in Spark it runs for hours.
The actual data is held in a S3 bucket, and when I submit the sql query, Spark goes off and first gets all the partitions from the Hive metastore (200000 of them), and then calls refresh() to force a full status list of all these files in the S3 object store (actually calling listLeafFilesInParallel).
It is these two operations that are so expensive, are there any settings that can get Spark to prune the partitions earlier - either during the call to the metadata store, or immediately afterwards?
Yes, spark supports partition pruning.
Spark does a listing of partitions directories (sequential or parallel listLeafFilesInParallel) to build a cache of all partitions first time around. The queries in the same application, that scan data takes advantage of this cache. So the slowness that you see could be because of this cache building. The subsequent queries that scan data make use of the cache to prune partitions.
These are the logs which shows partitions being listed to populate the cache.
App > 16/11/14 10:45:24 main INFO ParquetRelation: Listing s3://test-bucket/test_parquet_pruning/month=2015-01 on driver
App > 16/11/14 10:45:24 main INFO ParquetRelation: Listing s3://test-bucket/test_parquet_pruning/month=2015-02 on driver
App > 16/11/14 10:45:24 main INFO ParquetRelation: Listing s3://test-bucket/test_parquet_pruning/month=2015-03 on driver
These are the logs showing pruning is happening.
App > 16/11/10 12:29:16 main INFO DataSourceStrategy: Selected 1 partitions out of 20, pruned 95.0% partitions.
Refer convertToParquetRelation and getHiveQlPartitions in HiveMetastoreCatalog.scala.
Just a thought:
Spark API documentation for HadoopFsRelation says,
( https://spark.apache.org/docs/1.6.2/api/java/org/apache/spark/sql/sources/HadoopFsRelation.html )
"...when reading from Hive style partitioned tables stored in file
systems, it's able to discover partitioning information from the paths
of input directories, and perform partition pruning before start
reading the data..."
So, i guess "listLeafFilesInParallel" could not be a problem.
A similar issue is already in spark jira: https://issues.apache.org/jira/browse/SPARK-10673
In spite of "spark.sql.hive.verifyPartitionPath" set to false and, there is no effect in performance, I suspect that the
issue might have been caused by unregistered partitions. Please list out the partitions of the table and verify if all
the partitions are registered. Else, recover your partitions as shown in this link:
Hive doesn't read partitioned parquet files generated by Spark
Update:
I guess appropriate parquet block size and page size were set while writing the data.
Create a fresh hive table with partitions mentioned, and file-format as parquet, load it from non-partitioned table using dynamic partition approach.
( https://cwiki.apache.org/confluence/display/Hive/DynamicPartitions )
Run a plain hive query and then compare by running a spark program.
Disclaimer: I am not a spark/parquet expert. The problem sounded interesting, and hence responded.
similar question popped up here recently:
http://apache-spark-user-list.1001560.n3.nabble.com/Spark-SQL-reads-all-leaf-directories-on-a-partitioned-Hive-table-td35997.html#a36007
This question is old but I thought I'd post the solution here as well.
spark.sql.hive.convertMetastoreParquet=false
will use the Hive parquet serde instead of the spark inbuilt parquet serde. Hive's Parquet serde will not do a listLeafFiles on all partitions, but only and directly read from the selected partitions. On tables with many partitions and files, this is much faster (and cheaper, too). Feel free to try it ou! :)