Let us assume I have two pyspark dataframes with three partitions,
df1=[[1,2,3],[3,2,1],[2,3,1]]
df2=[[3,2,1],[2,3,1],[1,2,3]]
df1.join(df2,"id").groupby("id").count()
I am performing join and group by operations which means it can have two stages.
after the first stage 200 shuffle partitions will be created in my example 3 partitions will be created and rest are empty partitions
the shuffle partitions looks like this
partition1 :[1,1,1]
partition2 :[2,2,2]
partition3 :[3,3,3]
are these shuffle partitions needs to written to executor disks? so spark in that case is not in-memory computations? why it needs to write the shuffle partitions to the disk? does it use stage1 shuffle partitions in stage 2(group by )?
My initial answer was incorrect. I misread your question, apologies.
You are correct, Spark writes Shuffle results to disk for optimization purposes. Spark computes in memory but stores intermediate results on disk.
It does that, because shuffling is very expensive and you can avoid doing it more often by reusing shuffling results that have been persisted on disk.
This is an example where you can leverage that behaviour:
df \
.join(df2, ["id"]) \
.join(df3, ["id"]) \
.join(df4, ["id2"])
is faster than
df \
.join(df2, ["id"]) \
.join(df3, ["id2"]) \
.join(df4, ["id"])
Related
I have a pyspark dataframe which connect to oracle database and read a table which has 3 million records. I need to write this dataframe to azure eventhub.
Below is the sample pyspark datframe write to eventhub code.
df.select("body") \
.write\
.format("eventhubs") \
.options(**ehconf) \
.save()
How to split my pyspark dataframe into 10 parts equally (300k records/ dataframe) ?
So that I can send iterate each of these 10 pyspark dataframes to eventhub.
You can specify the number of partitions by
df.select('body').coalesce(10).write
OR
df.select('body').repartition(10).write
coalesce only decrease the number of partitions while repartition can
increase or decrease. repartition will do full shuffle, while
coalesce(2) will keep partition 1 & 2 as it is and move anything in
other partitions to fit into partition 1 & 2.
So, if your current partitions are higher than 10, use coalesce.
(You can check the num of partitions by df.rdd.getNumPartitions())
Let's say I have 2 RDDs, each of them being partitioned and each of them persisted.
Now I'm calling zipPartitions in order to iterate over each pair of RDD's partition.
Is there a way to ensure minimal data transfer? That is to say, can I ensure that partition0 of RDD1 was persisted at the same location as partition0 of RDD2 and so on?
I'm using spark for processing large files, I have 12 partitions.
I have rdd1 and rdd2 i make a join between them, than select (rdd3).
My problem is, i consulted that the last partition is too big than other partitions, from partition 1 to partitions 11 45000 recodrs but the partition 12 9100000 recodrs.
so i divided 9100000 / 45000 =~ 203. i repartition my rdd3 into 214(203+11)
but i last partition still too big.
How i can balance the size of my partitions ?
My i write my own custom partitioner?
I have rdd1 and rdd2 i make a join between them
join is the most expensive operation is Spark. To be able to join by key, you have to shuffle values, and if keys are not uniformly distributed, you get described behavior. Custom partitioner won't help you in that case.
I'd consider adjusting the logic, so it doesn't require a full join.
For the following join between two DataFrames in Spark 1.6.0
val df0Rep = df0.repartition(32, col("a")).cache
val df1Rep = df1.repartition(32, col("a")).cache
val dfJoin = df0Rep.join(df1Rep, "a")
println(dfJoin.count)
Does this join not only co-partitioned but also co-located? I know that for RDDs if using the same partitioner and shuffled in the same operation, the join would be co-located. But what about dataframes? Thank you.
[https://medium.com/#achilleus/https-medium-com-joins-in-apache-spark-part-3-1d40c1e51e1c]
According to the article link provided above Sort-Merge join is the default join, would like to add important point
For Ideal performance of Sort-Merge join, it is important that all
rows having the same value for the join key are available in the same
partition. This warrants for the infamous partition exchange(shuffle)
between executors. Collocated partitions can avoid unnecessary data
shuffle. Data needs to be evenly distributed n the join keys. The
number of join keys is unique enough so that they can be equally
distributed across the cluster to achieve the max parallelism from the
available partitions
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.