If my goal is to collect distinct values in a column as a list, is there a performance difference or pros/cons using either of these?
df.select(column).distinct().collect()...
vs
df.select(collect_set(column)).first()...
collect_set is an aggregator function and requires a groupBy in the beginning. When there is no grouping provided it will take entire data as 1 big group.
1. collect_set
df.select(collect_set(column)).first()...
This will send all data of column column to a single node which will perform collect_set operation (removing duplicates). If your data size is big then it will swamp the single executor where all data goes.
2. distinct
df.select(column).distinct().collect()...
This will partition all data of column column based on its value (called partition key), no. of partitions will be the value of spark.sql.shuffle.partitions (say 200). So 200 tasks will execute to remove duplicates, 1 for each partition key. Then only dedup data will be sent to the driver for .collect() operation. This will fail if your data after removing duplicates is huge, will cause driver to go out of memory.
TLDR:
.distinct is better than .collect_set for your specific need
Related
I have a case where i am trying to write some results using dataframe write into S3 using the below query with input_table_1 size is 13 Gb and input_table_2 as 1 Mb
input_table_1 has columns account, membership and
input_table_2 has columns role, id , membership_id, quantity, start_date
SELECT
/*+ BROADCASTJOIN(input_table_2) */
account,
role,
id,
quantity,
cast(start_date AS string) AS start_date
FROM
input_table_1
INNER JOIN
input_table_2
ON array_contains(input_table_1.membership, input_table_2.membership_id)
where membership array contains list of member_ids
This dataset write using Spark dataframe is generating around 1.1TiB of data in S3 with around 700 billion records.
We identified that there are duplicates and used dataframe.distinct.write.parquet("s3path") to remove the duplicates . The record count is reduced to almost 1/3rd of the previous total count with around 200 billion rows but we observed that the output size in S3 is now 17.2 TiB .
I am very confused how this can happen.
I have used the following spark conf settings
spark.sql.shuffle.partitions=20000
I have tried to do a coalesce and write to s3 but it did not work.
Please suggest if this is expected and when can be done ?
There's two sides to this:
1) Physical translation of distinct in Spark
The Spark catalyst optimiser turns a distinct operation into an aggregation by means of the ReplaceDeduplicateWithAggregate rule (Note: in the execution plan distinct is named Deduplicate).
This basically means df.distinct() on all columns is translated into a groupBy on all columns with an empty aggregation:
df.groupBy(df.columns:_*).agg(Map.empty).
Spark uses a HashPartitioner when shuffling data for a groupBy on respective columns. Since the groupBy clause in your case contains all columns (well, implicitly, but it does), you're more or less randomly shuffling data to different nodes in the cluster.
Increasing spark.sql.shuffle.partitions in this case is not going to help.
Now on to the 2nd side, why does this affect the size of your parquet files so much?
2) Compression in parquet files
Parquet is a columnar format, will say your data is organised in columns rather than row by row. This allows for powerful compression if data is adequately laid-out & ordered. E.g. if a column contains the same value for a number of consecutive rows, it is enough to write that value just once and make a note of the number of repetitions (a strategy called run length encoding). But Parquet also uses various other compression strategies.
Unfortunately, data ends up pretty randomly in your case after shuffling to remove duplicates. The original partitioning of input_table_1 was much better fitted.
Solutions
There's no single answer how to solve this, but here's a few pointers I'd suggest doing next:
What's causing the duplicates? Could these be removed upstream? Or is there a problem with the join condition causing duplicates?
A simple solution is to just repartition the dataset after distinct to match the partitioning of your input data. Adding a secondary sorting (sortWithinPartition) is likely going to give you even better compression. However, this comes at the cost of an additional shuffle!
As #matt-andruff pointed out below, you can also achieve this in SQL using cluster by. Obviously, that also requires you to move the distinct keyword into your SQL statement.
Write your own deduplication algorithm as Spark Aggregator and group / shuffle the data just once in a meaningful way.
Here is an example of it. The cell 44 output shows the count of distinct keys but when i find the partition size in cell 45 then it combines together 3 and 5. also on saving the size of one partition is still zero. Any help would be appreciated.
By default, Spark applies a HashPartitioner to the values of the column overint. Apparently, the values 3 and 5 fall into the same partition after being hashed.
You may want to choose the RangePartitioner. Or, in case you need full flexibility, you could also write your custome Partitioner class. However, this is only available on the RDD API and not the Structured API.
[New to Spark] Language - Scala
As per docs, RangePartitioner sorts and divides the elements into chunks and distributes the chunks to different machines. How would it work for below example.
Let's say we have a dataframe with 2 columns and one column (say 'A') has continuous values from 1 to 1000. There is another dataframe with same schema but the corresponding column has only 4 values 30, 250, 500, 900. (These could be any values, randomly selected from 1 to 1000)
If I partition both using RangePartitioner,
df_a.partitionByRange($"A")
df_b.partitionByRange($"A")
how will the data from both the dataframes be distributed across nodes ?
Assuming that the number of partitions is 5.
Also, if I know that second DataFrame has less number of values then will reducing number of partitions for it make any difference ?
What I am struggling to understand is that how Spark maps one partition of df_a to a partition of df_b and how it sends (if it does) both those partitions to same machine for processing.
A very detailed explanation of how RangePartitioner works internally is described here
Specific to your question, RangePartitioner samples the RDD at runtime, collects the statistics, and only then are the ranges (limits) evaluated. Note that there are 2 parameters here - ranges (logical), and partitions (physical). The number of partitions can be affected by many factors - number of input files, inherited number from parent RDD, 'spark.sql.shuffle.partitions' in case of shuffling, etc. The ranges evaluated according to the sampling. In any case, RangePartitioner ensures every range is contained in single partition.
how will the data from both the dataframes be distributed across nodes ? how Spark maps one partition of df_a to a partition of df_b
I assume you implicitly mean joining 'A' and 'B', otherwise the question does not make any sense. In that case, Spark would make sure to match partitions with ranges on both DataFrames, according to their statistics.
I've done this in PySpark:
Created a DataFrame using a SELECT statement to get asset data ordered by asset serial number and then time.
Used DataFrame.map() to convert the DataFrame to an RDD.
Used RDD.combineByKey() to collate all the data for each asset, using the asset's serial number as the key.
Question: Can I be certain that the data for each asset will still be sorted in time order in the RDD resulting from the last step?
Time order is crucial for me (I need to calculate statistics over a moving time window across the data for each asset). When RDD.combineByKey() combines data from different nodes in the Spark cluster for a given key, is any order in that key's data retained? Or is the data from the different nodes combined in no particular order for a given key?
Can I be certain that the data for each asset will still be sorted in time order in the RDD resulting from the last step?
You cannot. When you apply sort across multiple dimensions (data ordered by asset serial number and then time) records for a single asset can be spread across multiple partitions. combineByKey will require a shuffle and the order in which these parts are combined is not guaranteed.
You can try with repartition and sortWithinPartitions (or its equivalent on RDDs):
df.repartition("asset").sortWithinPartitions("time")
or
df.repartition("asset").sortWithinPartitions("asset", "time")
or window functions with frame definition as follows:
w = Window.partitionBy("asset").orderBy("time")
In Spark >= 2.0 window functions can be used with UserDefinedFunctions so if you're fine with writing your own SQL extensions in Scala you can skip conversion to RDD completely.
What is the Big-O time complexity for Apache Spark RDD sortByKey?
I am trying to assign row numbers to an RDD based on a particular order.
Say I have a {K,V} pair RDD and I wish to perform an order by key using
myRDD.sortByKey(true).zipWithIndex
What is the time complexity for this operation, in big-O form?
And what is happening under-the-covers? Bubble sort? I hope not! My dataset is very large and runs across partitions, so I'm curious whether the sortByKey function is optimal, or does some kind of intermediate data structure within a partition and then something else across partitions to optimize message passing, or what.
A quick look at the code shows that a RangePartitioner is being used under the covers. The docs say:
partitions sortable records by range into roughly
* equal ranges. The ranges are determined by sampling the content of the RDD passed in
So in essence your data is sampled (O[n]), then only the unique sample keys (m) are sorted are sorted (O[m log(m)]) and ranges of keys determined, then the entire data is shuffled around (O[n], but costly), then the data sorted internally for the range of keys received on a given partition (O[p log[p)).
zipWithIndex probably uses local sizes to assign numbers, using the partition number, so it is likely that partition metadata is stored for this effect:
Zips this RDD with its element indices. The ordering is first based on the partition index
* and then the ordering of items within each partition. So the first item in the first
* partition gets index 0, and the last item in the last partition receives the largest index.