snowflake.snowpark.functions.grouping_id¶

snowflake.snowpark.functions.grouping_id(*cols: Union[Column, str]) → Column[source] (https://github.com/snowflakedb/snowpark-python/blob/v1.30.0/snowpark-python/src/snowflake/snowpark/functions.py#L1785-L1805)¶

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::

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>>> 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)]

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