How could I union or concatenate a static dataframe with only one row to a stream dataframe with around 500 rows in spark. it is somehow put a stamp or mark (add a row to each table) for each streaming dataframe.
Adding more, my streaming data has no timestamp and I'm wondering if I can use foreach() or foreachBatch() or not?
I really appreciate it if you can help me.
Related
I have a scenario where i have dataset with date column and later i use the dataset in iteration to save the dataset into multiple partition files in parquet format. I do iterate the date list and while writing to parquet format with that date partition folder i do filter the dataset with date.
I was able to write for certain iterations but after that its failing with Spark out of memory exceptions.
Whats the best way to optimise this to persist the data with OOM.
dataset = dataset with some transformations
for date in date-list
pd.write_part_file("part-data-file", dataset.filter(archive_date==date))
The code looks like above.
I am new to databricks notebooks and dataframes. I have a requirement to load few columns(out of many) in a table of around 14million records into a dataframe. once the table is loaded, I need to create a new column based on values present in two columns.
I want to write the logic for the new column along with the select command while loading the table into dataframe.
Ex:
df = spark.read.table(tableName)
.select(columnsList)
.withColumn('newColumnName', 'logic')
will it have any performance impact? is it better to first load the table for the few columns into the df and then perform the column manipulation on the loaded df?
does the table data gets loaded all at once or row by row into the df? if row by row, then by including column manipulation logic while reading the table, am I causing any performance degradation?
Thanks in advance!!
This really depends on the underlying format of the table - is it backed by Parquet or Delta, or it's an interface to the actual database, etc. In general, Spark is trying to read only necessary data, and if, for example, Parquet is used (or Delta), then it's easier because it's column-oriented file format, so data for each column is placed together.
Regarding the question on the reading - Spark is lazy by default, so even if you put df = spark.read.table(....) as separate variable, then add .select, and then add .withColumn, it won't do anything until you call some action, for example .count, or write your results. Until that time, Spark will just check that table exists, your operations are correct, etc. You can always call .explain on the resulting dataframe to see how Spark will perform operations.
P.S. I recommend to grab a free copy of the Learning Spark, 2ed that is provided by Databricks - it will provide you a foundation for development of the code for Spark/Databricks
I'd like to understand how structured streaming treats new data coming.
If more rows arrive at the same time, spark append them to the input streaming dataframe, right?
If I have a withColumn and apply a pandas_udf, the function is called once per each row, or only one time and the rows are passed to the pandas_udf?
Let's say something like this:
dfInt = spark \
.readStream \
.load() \
.withColumn("prediction", predict( (F.struct([col(x) for x in (features)]))))
If more rows arrive at the same time, they are processed together or once per each?=
There is the chance to limit this to only one row per time?
If more rows arrive at the same time, spark append them to the input streaming dataframe, right?
Let's talk Micro-Batch Execution Engine only, right? That's what you most likely use in streaming queries.
Structured Streaming queries the streaming sources in a streaming query using Source.getBatch (DataSource API V1):
getBatch(start: Option[Offset], end: Offset): DataFrame
Returns the data that is between the offsets (start, end]. When start is None, then the batch should begin with the first record.
Whatever the source returns in a DataFrame is the data to be processed in a micro-batch.
If I have a withColumn and apply a pandas_udf, the function is called once per each row
Always. That's how user-defined functions work in Spark SQL.
or only one time and the rows are passed to the pandas_udf?
This says:
Pandas UDFs are user defined functions that are executed by Spark using Arrow to transfer data and Pandas to work with the data.
The Python function should take pandas.Series as inputs and return a pandas.Series of the same length. Internally, Spark will execute a Pandas UDF by splitting columns into batches and calling the function for each batch as a subset of the data, then concatenating the results together.
If more rows arrive at the same time, they are processed together or once per each?
If "arrive" means "part of a single DataFrame", then "they are processed together", but one row at a time (per the UDF contract).
There is the chance to limit this to only one row per time?
You don't have to. It's as such by design. One row at a time only.
In python or R, there are ways to slice DataFrame using index.
For example, in pandas:
df.iloc[5:10,:]
Is there a similar way in pyspark to slice data based on location of rows?
Short Answer
If you already have an index column (suppose it was called 'id') you can filter using pyspark.sql.Column.between:
from pyspark.sql.functions import col
df.where(col("id").between(5, 10))
If you don't already have an index column, you can add one yourself and then use the code above. You should have some ordering built in to your data based on some other columns (orderBy("someColumn")).
Full Explanation
No it is not easily possible to slice a Spark DataFrame by index, unless the index is already present as a column.
Spark DataFrames are inherently unordered and do not support random access. (There is no concept of a built-in index as there is in pandas). Each row is treated as an independent collection of structured data, and that is what allows for distributed parallel processing. Thus, any executor can take any chunk of the data and process it without regard for the order of the rows.
Now obviously it is possible to perform operations that do involve ordering (lead, lag, etc), but these will be slower because it requires spark to shuffle data between the executors. (The shuffling of data is typically one of the slowest components of a spark job.)
Related/Futher Reading
PySpark DataFrames - way to enumerate without converting to Pandas?
PySpark - get row number for each row in a group
how to add Row id in pySpark dataframes
You can convert your spark dataframe to koalas dataframe.
Koalas is a dataframe by Databricks to give an almost pandas like interface to spark dataframe. See here https://pypi.org/project/koalas/
import databricks.koalas as ks
kdf = ks.DataFrame(your_spark_df)
kdf[0:500] # your indexes here
I am a newbie in Spark.I want to write the dataframe data into hive table. Hive table is partitioned on mutliple column. Through, Hivemetastore client I am getting the partition column and passing that as a variable in partitionby clause in write method of dataframe.
var1="country","state" (Getting the partiton column names of hive table)
dataframe1.write.partitionBy(s"$var1").mode("overwrite").save(s"$hive_warehouse/$dbname.db/$temp_table/")
When I am executing the above code,it is giving me error partiton "country","state" does not exists.
I think it is taking "country","state" as a string.
Can you please help me out.
The partitionBy function takes a varargs not a list. You can use this as
dataframe1.write.partitionBy("country","state").mode("overwrite").save(s"$hive_warehouse/$dbname.db/$temp_table/")
Or in scala you can convert a list into a varargs like
val columns = Seq("country","state")
dataframe1.write.partitionBy(columns:_*).mode("overwrite").save(s"$hive_warehouse/$dbname.db/$temp_table/")