snowflake.snowpark.functions.grouping¶
- snowflake.snowpark.functions.grouping(*cols: Union[Column, str]) Column [source]¶
Describes which of a list of expressions are grouped in a row produced by a GROUP BY query.
grouping_id()
is an alias ofgrouping()
.- 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").alias("count_c"), grouping("a").alias("ga"), grouping("b").alias("gb"), grouping("a", "b").alias("gab")]).sort("a", "b").collect() [Row(A=None, B=2, COUNT_C=1, GA=1, GB=0, GAB=2), Row(A=None, B=5, COUNT_C=1, GA=1, GB=0, GAB=2), Row(A=1, B=None, COUNT_C=1, GA=0, GB=1, GAB=1), Row(A=1, B=2, COUNT_C=1, GA=0, GB=0, GAB=0), Row(A=4, B=None, COUNT_C=1, GA=0, GB=1, GAB=1), Row(A=4, B=5, COUNT_C=1, GA=0, GB=0, GAB=0)]