snowflake.snowpark.functions.grouping

snowflake.snowpark.functions.grouping(*cols: Union[Column, str]) → Column[source] (https://github.com/snowflakedb/snowpark-python/blob/v1.26.0/snowpark-python/src/snowflake/snowpark/functions.py#L1643-L1663)

Describes which of a list of expressions are grouped in a row produced by a GROUP BY query.

grouping_id() is an alias of grouping().

Example::
>>> from snowflake.snowpark import GroupingSets
>>> df = session.create_dataframe([[1, 2, 3], [4, 5, 6]],schema=["a", "b", "c"])
>>> grouping_sets = GroupingSets([col("a")], [col("b")], [col("a"), col("b")])
>>> df.group_by_grouping_sets(grouping_sets).agg([count("c"), grouping("a"), grouping("b"), grouping("a", "b")]).collect()
[Row(A=1, B=2, COUNT(C)=1, GROUPING(A)=0, GROUPING(B)=0, GROUPING(A, B)=0), Row(A=4, B=5, COUNT(C)=1, GROUPING(A)=0, GROUPING(B)=0, GROUPING(A, B)=0), Row(A=1, B=None, COUNT(C)=1, GROUPING(A)=0, GROUPING(B)=1, GROUPING(A, B)=1), Row(A=4, B=None, COUNT(C)=1, GROUPING(A)=0, GROUPING(B)=1, GROUPING(A, B)=1), Row(A=None, B=2, COUNT(C)=1, GROUPING(A)=1, GROUPING(B)=0, GROUPING(A, B)=2), Row(A=None, B=5, COUNT(C)=1, GROUPING(A)=1, GROUPING(B)=0, GROUPING(A, B)=2)]
Copy
Language: English