I want to do a join operation between two very big key-value pair RDDs. The keys of these two RDD comes from the same set. To reduce data shuffle, I wish I could add a pre-distribute phase so that partitions with the same key will be distributed on the same machine. Hopefully this could reduce some shuffle time.
I want to know is spark smart enough to do that for me or I have to implement this logic myself?
I know when I join two RDD, one preprocess with partitionBy. Spark is smart enough to use this information and only shuffle the other RDD. But I don't know what will happen if I use partitionBy on two RDD at the same time and then do the join.
If you use the same partitioner for both RDDs you achieve co-partitioning of your data sets. That does not necessarily mean that your RDDs are co-located - that is, that the partitioned data is located on the same node.
Nevertheless, the performance should be better as if both RDDs would have different partitioner.
I have seen this, Speeding Up Joins by Assigning a Known Partitioner that would be helpful to understand the effect of using the same partitioner for both RDDs;
Speeding Up Joins by Assigning a Known Partitioner
If you have to do an operation before the join that requires a
shuffle, such as aggregateByKey or reduceByKey, you can prevent the
shuffle by adding a hash partitioner with the same number of
partitions as an explicit argument to the first operation and
persisting the RDD before the join.
Related
Looking into the Spark UI and physical plan, I found that orderBy is accomplished by Exchange rangepartitioning(col#0000 ACS NULLS FIRST, 200) and then Sort [col#0000 ACS NULLS FIRST], true, 0.
From what I understand, rangepartitioning would define minimum and maximum values for each partition and order the data with column value within the min and max into that partition so as to achieve global ordering.
But now I have 200 partitions and I want to output to a single csv file. If I do a repartition(1), spark will trigger a shuffle and the ordering will be gone. However, I tried coalesce(1) and it retained the global ordering. Yet I don't know if it was merely pure luck since coalesce does not necessarily decrease number of partitions and keep the ordering of partitions. Does anyone know how to repartition to keep the ordering after rangepartitioning? Thanks a lot.
As you state yourself maintaining order is not part of the coalesce API contract. You you have to choose:
collect the ordered dataframe as a list of Row instances and write to csv outside spark
write the partitions to individual CSV files with spark and concatenate the partitions with some other tool, e.g. "hadoop dfs getmerge" on the command line.
I am having a lot trouble finding the answer to this question. Let's say I write a dataframe to parquet and I use repartition combined with partitionBy to get a nicely partitioned parquet file. See Below:
df.repartition(col("DATE")).write.partitionBy("DATE").parquet("/path/to/parquet/file")
Now later on I would like to read the parquet file so I do something like this:
val df = spark.read.parquet("/path/to/parquet/file")
Is the dataframe partitioned by "DATE"? In other words if a parquet file is partitioned does spark maintain that partitioning when reading it into a spark dataframe. Or is it randomly partitioned?
Also the why and why not to this answer would be helpful as well.
The number of partitions acquired when reading data stored as parquet follows many of the same rules as reading partitioned text:
If SparkContext.minPartitions >= partitions count in data, SparkContext.minPartitions will be returned.
If partitions count in data >= SparkContext.parallelism, SparkContext.parallelism will be returned, though in some very small partition cases, #3 may be true instead.
Finally, if the partitions count in data is somewhere between SparkContext.minPartitions and SparkContext.parallelism, generally you'll see the partitions reflected in the dataset partitioning.
Note that it's rare for a partitioned parquet file to have full data locality for a partition, meaning that, even when the partitions count in data matches the read partition count, there is a strong likelihood that the dataset should be repartitioned in memory if you're trying to achieve partition data locality for performance.
Given your use case above, I'd recommend immediately repartitioning on the "DATE" column if you're planning to leverage partition-local operations on that basis. The above caveats regarding minPartitions and parallelism settings apply here as well.
val df = spark.read.parquet("/path/to/parquet/file")
df.repartition(col("DATE"))
You would get the number of partitions based on the spark config spark.sql.files.maxPartitionBytes which defaults to 128MB. And the data would not be partitioned as per the partition column which was used while writing.
Reference https://spark.apache.org/docs/latest/sql-performance-tuning.html
In your question, there are two ways we could say the data are being "partitioned", which are:
via repartition, which uses a hash partitioner to distribute the data into a specific number of partitions. If, as in your question, you don't specify a number, the value in spark.sql.shuffle.partitions is used, which has default value 200. A call to .repartition will usually trigger a shuffle, which means the partitions are now spread across your pool of executors.
via partitionBy, which is a method specific to a DataFrameWriter that tells it to partition the data on disk according to a key. This means the data written are split across subdirectories named according to your partition column, e.g. /path/to/parquet/file/DATE=<individual DATE value>. In this example, only rows with a particular DATE value are stored in each DATE= subdirectory.
Given these two uses of the term "partitioning," there are subtle aspects in answering your question. Since you used partitionBy and asked if Spark "maintain's the partitioning", I suspect what you're really curious about is if Spark will do partition pruning, which is a technique used drastically improve the performance of queries that have filters on a partition column. If Spark knows values you seek cannot be in specific subdirectories, it won't waste any time reading those files and hence your query completes much quicker.
If the way you're reading the data isn't partition aware, you'll get a number of partitions something like what's in bsplosion's answer. Spark won't employ partition pruning, and hence you won't get the benefit of Spark automatically ignoring reading certain files to speed things up1.
Fortunately, reading parquet files in Spark that were written with partitionBy is a partition-aware read. Even without a metastore like Hive that tells Spark the files are partitioned on disk, Spark will discover the partitioning automatically. Please see partition discovery in Spark for how this works in parquet.
I recommend testing reading your dataset in spark-shell so that you can easily see the output of .explain, which will let you verify that Spark correctly finds the partitions and can prune out the ones that don't contain data of interest in your query. A nice writeup on this can be found here. In short, if you see PartitionFilters: [], it means that Spark isn't doing any partition pruning. But if you see something like PartitionFilters: [isnotnull(date#3), (date#3 = 2021-01-01)], Spark is only reading in a specific set of DATE partitions, and hence the query execution is usually a lot faster.
1A separate detail is that parquet stores statistics about data in its columns inside of the files themselves. If these statistics can be used to eliminate chunks of data that can't match whatever filtering you're doing, e.g. on DATE, then you'll see some speedup even if the way you read the data isn't partition-aware. This is called predicate pushdown. It works because the files on disk will still contain only specific values of DATE when using .partitionBy. More info can be found here.
I have an input A which I convert into an rdd X spread across the cluster.
I perform certain operations on it.
Then I do .repartition(1) on the output rdd.
Will my output rdd be in the same order that input A.
Does spark handle this automatically? If yes, then how?
The documentation doesn't guarantee that order will be kept, so you can assume it won't be. If you look at the implementation, you'll see it certainly won't be (unless your original RDD already has 1 partition for some reason): repartition calls coalesce(shuffle = true), which
Distributes elements evenly across output partitions, starting from a random partition.
I have a pair RDD[String,String] where key is a string and value is html. I want to split this rdd into n RDDS based on n keys and store them in HDFS.
htmlRDD = [key1,html
key2,html
key3,html
key4,html
........]
Split this RDD based on keys and store html from each RDD individually on HDFS. Why I want to do that? When, I'm trying to store the html from the main RDD to HDFS,it takes a lot of time as some tasks are denied committing by output coordinator.
I'm doing this in Scala.
htmlRDD.saveAsHadoopFile("hdfs:///Path/",classOf[String],classOf[String], classOf[Formatter])
You can also try this in place of breaking RDD:
htmlRDD.saveAsTextFile("hdfs://HOST:PORT/path/");
I tried this and it worked for me. I had RDD[JSONObject] and it wrote toString() of JSON Object very well.
Spark saves each RDD partition into 1 hdfs file partition. So to achieve good parallelism your source RDD should have many partitions(actually depends on size of whole data). So I think you want to split your RDD not into several RDDs, but rather to have RDD with many partitions.
You you can do it with repartition() or coallesce()
Assume I have RDD of Integer from 1 to 1,000,000,000 and I want to print them ordered using foreachPartition. There might be situation that the partition of 5-6-7-8 will be printed before 1-2-3-4. How can I prevent this?
Thanks,
Maya
I think the only way to do this would be to ensure there is only one partition, and then you could print your data. You can call repartition(1) or coalesce(1) on your RDD to reduce the number of partitions. For your use case I think coalesce is better as it avoids a shuffle.
https://spark.apache.org/docs/1.3.1/programming-guide.html#transformations