snowflake.snowpark.functions.countDistinct¶
- snowflake.snowpark.functions.countDistinct(*cols: Union[Column, str]) Column[source] (https://github.com/snowflakedb/snowpark-python/blob/v1.16.0/src/snowflake/snowpark/functions.py#L707-L726)¶
Returns either the number of non-NULL distinct records for the specified columns, or the total number of the distinct records.
Example
>>> df = session.create_dataframe([[1, 2], [1, 2], [3, None], [2, 3], [3, None], [4, None]], schema=["a", "b"]) >>> df.select(count_distinct(col("a"), col("b")).alias("result")).show() ------------ |"RESULT" | ------------ |2 | ------------ >>> # The result should be 2 for {[1,2],[2,3]} since the rest are either duplicate or NULL records