Intellipaat Back

Explore Courses Blog Tutorials Interview Questions
0 votes
2 views
in Big Data Hadoop & Spark by (11.4k points)

I am trying to create a new column of lists in Pyspark using a groupby aggregation on existing set of columns. An example input data frame is provided below:

------------------------
id | date        | value
------------------------
1  |2014-01-03   | 10
1  |2014-01-04   | 5
1  |2014-01-05   | 15
1  |2014-01-06   | 20
2  |2014-02-10   | 100  
2  |2014-03-11   | 500
2  |2014-04-15   | 1500


The expected output is:

id | value_list
------------------------
1  | [10, 5, 15, 20]
2  | [100, 500, 1500]


The values within a list are sorted by the date.

I tried using collect_list as follows:

from pyspark.sql import functions as F
ordered_df = input_df.orderBy(['id','date'],ascending = True)
grouped_df = ordered_df.groupby("id").agg(F.collect_list("value"))

 

But collect_list doesn't guarantee order even if I sort the input data frame by date before aggregation.

Could someone help on how to do aggregation by preserving the order based on a second (date) variable?

1 Answer

0 votes
by (32.3k points)

from pyspark.sql import functions as F

from pyspark.sql import Window

w = Window.partitionBy('id').orderBy('date')

sorted_list_df = input_df.withColumn(

            'sorted_list', F.collect_list('value').over(w)

        )\

        .groupBy('id')\

        .agg(F.max('sorted_list').alias('sorted_list'))

Window examples provided by users often don't really explain what is going on so let me dissect it for you.

As you know, using collect_list together with groupBy will result in an unordered list of values. This is because depending on how your data is partitioned, Spark will append values to your list as soon as it finds a row in the group. The order then depends on how Spark plans your aggregation over the executors.

A Window function allows you to control that situation, grouping rows by a certain value so you can perform an operation over each of the resultant groups:

w = Window.partitionBy('id').orderBy('date')

  • partitionBy - you want groups/partitions of rows with the same id

  • orderBy - you want each row in the group to be sorted by date

Once you have defined the scope of your Window - "rows with the same id, sorted by date" -, you can use it to perform an operation over it, in this case, a collect_list:

F.collect_list('value').over(w)

At this point you created a new column sorted_list with an ordered list of values, sorted by date, but you still have duplicated rows per id. To trim out the duplicated rows you want to groupBy id and keep the max value in for each group:

.groupBy('id')\

.agg(F.max('sorted_list').alias('sorted_list'))

Related questions

1.2k questions

2.7k answers

501 comments

693 users

Browse Categories

...