pyspark collect_set or collect_list with groupby

ListGroup BySetPysparkCollect

List Problem Overview


How can I use collect_set or collect_list on a dataframe after groupby. for example: df.groupby('key').collect_set('values'). I get an error: AttributeError: 'GroupedData' object has no attribute 'collect_set'

List Solutions


Solution 1 - List

You need to use agg. Example:

from pyspark import SparkContext
from pyspark.sql import HiveContext
from pyspark.sql import functions as F

sc = SparkContext("local")

sqlContext = HiveContext(sc)

df = sqlContext.createDataFrame([
    ("a", None, None),
    ("a", "code1", None),
    ("a", "code2", "name2"),
], ["id", "code", "name"])

df.show()

+---+-----+-----+
| id| code| name|
+---+-----+-----+
|  a| null| null|
|  a|code1| null|
|  a|code2|name2|
+---+-----+-----+

Note in the above you have to create a HiveContext. See https://stackoverflow.com/a/35529093/690430 for dealing with different Spark versions.

(df
  .groupby("id")
  .agg(F.collect_set("code"),
       F.collect_list("name"))
  .show())

+---+-----------------+------------------+
| id|collect_set(code)|collect_list(name)|
+---+-----------------+------------------+
|  a|   [code1, code2]|           [name2]|
+---+-----------------+------------------+

Solution 2 - List

If your dataframe is large, you can try using pandas udf(GROUPED_AGG) to avoid memory error. It is also much faster.

>Grouped aggregate Pandas UDFs are similar to Spark aggregate functions. Grouped aggregate Pandas UDFs are used with groupBy().agg() and pyspark.sql.Window. It defines an aggregation from one or more pandas.Series to a scalar value, where each pandas.Series represents a column within the group or window. pandas udf

example:

import pyspark.sql.functions as F

@F.pandas_udf('string', F.PandasUDFType.GROUPED_AGG)
def collect_list(name):
    return ', '.join(name)

grouped_df = df.groupby('id').agg(collect_list(df["name"]).alias('names'))

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionHanan ShteingartView Question on Stackoverflow
Solution 1 - ListKamil SindiView Answer on Stackoverflow
Solution 2 - ListAllenView Answer on Stackoverflow