applying cache() and count() to Spark Dataframe in Databricks is very slow [pyspark]

Question

I have a spark dataframe in Databricks cluster with 5 million rows. And what I want is to cache this spark dataframe and then apply .count() so for the next operations to run extremely fast. I have done it in the past with 20,000 rows and it works. However, in my trial to do this I came into the following paradox:

Dataframe creation

Step 1: Read 8 millions rows from Azure Data Lake storage account

read_avro_data=spark.read.format("avro").load(list_of_paths) #list_of_paths[0]='abfss://storage_container_name@storage_account_name.dfs.core.windows.net/folder_1/folder_2/0/2020/06/02/00/00/27.avro'
avro_decoded=read_avro_data.withColumn('Body_decoded', sql_function.decode(read_avro_data.Body, charset="UTF-8")).select("Body_decoded")
datalake_spark_dataframe=datalake_spark_dataframe.union(avro_decoded.withColumn("Body_decoded", sql_function.from_json("Body_decoded", schema)).select(*['Body_decoded.{}'.format(x) for x in columns_selected]))

datalake_spark_dataframe.printSchema()
"root
 |-- id: string (nullable = true)
 |-- BatteryPercentage: float (nullable = true)
 |-- SensorConnected: integer (nullable = false)
 |-- TemperatureOutside: float (nullable = true)
 |-- ReceivedOn: string (nullable = true)"

datalake_spark_dataframe.rdd.getNumPartitions() # 635 partitions

This dataframe has 8 million rows. With 8 million rows my application runs pretty good, but I wanted to stress test my application in a big-data environment. Because 8 million rows is not Big-Data. Thus I replicated my 8 millions rows Spark Dataframe 287 times ~ 2.2 billion rows. To make the replication I did the following:

Step 2: Replicate the 8 million rows dataframe

datalake_spark_dataframe_new=datalake_spark_dataframe
for i in range(287):
  print(i)
  datalake_spark_dataframe_new=datalake_spark_dataframe_new.union(datalake_spark_dataframe)
  print("done on iteration: {0}".format(i))

datalake_spark_dataframe_new.rdd.getNumPartitions() #182880

Having the final 2.2 billion rows dataframe, I made a time-window GroupBy of my data, ending up with some millions of rows. I have written approximately that the grouped dataset has 5 million rows in the top of my question.

Step 3: GroupBy the 2.2 billion rows dataframe by a time window of 6 hours & Apply the .cache() and .count()

%sql set spark.sql.shuffle.partitions=100

import pyspark.sql.functions as sql_function
from pyspark.sql.types import StructType, StructField, StringType, IntegerType, FloatType, BooleanType, DateType, DoubleType, ArrayType

datalake_spark_dataframe_downsampled=datalake_spark_dataframe_new.withColumn(timestamp_column, sql_function.to_timestamp(timestamp_column, "yyyy-MM-dd HH:mm"))
datalake_spark_dataframe_downsampled=datalake_spark_dataframe_downsampled.groupBy("id", sql_function.window("ReceivedOn","{0} minutes".format(time_interval)))\
                                                                         .agg(
                                                                              sql_function.mean("BatteryPercentage").alias("BatteryPercentage"),
                                                                              sql_function.mean("SensorConnected").alias("OuterSensorConnected"),
                                                                              sql_function.mean("TemperatureOutside").alias("averageTemperatureOutside"))

columns_to_drop=['window']
datalake_spark_dataframe_downsampled=datalake_spark_dataframe_downsampled.drop(*columns_to_drop)

# From 2.2 billion rows down to 5 million rows after the GroupBy...
datalake_spark_dataframe_downsampled.repartition(100)
datalake_spark_dataframe_downsampled.cache()
datalake_spark_dataframe_downsampled.count() # job execution takes for ever

datalake_spark_dataframe_downsampled.rdd.getNumPartitions() #100 after re-partition

Spark UI before showing the .count()

Spark UI during the count execution

When I apply the following commands to my spark Dataframe it takes more than 3 hours to complete this task, which in the end fails.

I want to add that before and after the re-partitioning, the job had the same behavior in time execution. So, I did the re-partitioning in case the default values were making the job to run very slow. Thus, I was keep adding partitions in case the job would execute faster.

%sql set spark.sql.shuffle.partitions=1000000
datalake_spark_dataframe_downsampled.repartition(1000000)

datalake_spark_dataframe_downsampled.cache()
datalake_spark_dataframe_downsampled.count()

Below is the output of the spark job:

The error I get:

My cluster resources:

As you can see it's not a matter of RAM or CPU cores, as I have plenty of them. Why the job splits only to 5 stages even after I apply re-partitioning? and How can I split the jobs so the .cache() and .count() commands run faster based on my 48 vCPU cores?

Screenshots provided per job execution Execution on 80 million rows (8m * 10 iterations = 80m rows)

user_3807020 · Answer

I had the similiar issue in the past while iterating through for loop as my iteration is dynamic depending on input combination.

I resolved the performance issue by persisting data (you can try to persist in ADLS2 or if in case On-Prem then HDFS / Hive Tables) on each iteration. In next Iteration again read from that location, union and again overwrite the same location. There is a network lag and not efficient. Still it brought down execution time by 10x.

Possible reason could be Spark Lineage (I believe for every iteration it does all previous iteration again and again). Persisting data with overwrite avoids that. I tried cache() and other options as well but did not help me.

Edited #1 Try something like this

datalake_spark_dataframe_new=datalake_spark_dataframe
datalake_spark_dataframe.write.mode("overwrite").option("header", "true").format("parquet").save("abfss://")
for i in range(287):
  print(i)
  datalake_spark_dataframe_new=spark.read.parquet("abfss://")
  datalake_spark_dataframe_new.union(datalake_spark_dataframe).write.mode("overwrite").option("header", "true").format("parquet").save("abfss://")
  print("done on iteration: {0}".format(i))

Edited #2 This should be more efficient than previous edition,

for i in range(287):
  print(i)
  datalake_spark_dataframe.write.mode("append").option("header", "true").format("parquet").save("abfss://")
  print("done on iteration: {0}".format(i))

datalake_spark_dataframe_new=spark.read.parquet("abfss://")

user_8593414 · Answer

I think you have used very huge shuffle partition number 1000000 that why it is taking more time to complete job.

I will follow below logic to calculate shuffle partition based on data size. for example

Say 5 millions of data is comes around 20 GB of data.

shuffle stage input = 20 GB

so total number of shuffle partitions are 20000MB/200MB = 100,

let assume only 50 cores in cluster in that case shuffle partition value is 50 or 200 cores in cluster in that case shuffle partition value will be 200.

Choosing high value as shuffle partition value there will be lot of shuffling data & hence task will take more time to complete or sometime it might fail.

spark.sql.shuffle.partitions=50 // 50 or 100 for better option.

applying cache() and count() to Spark Dataframe in Databricks is very slow [pyspark]

2 Answers