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snowflake.snowpark.functions.countDistinct¶

snowflake.snowpark.functions.countDistinct(*cols: Union[Column, str]) → Column[source]¶

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()
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|"RESULT"  |
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|2         |
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>>> #  The result should be 2 for {[1,2],[2,3]} since the rest are either duplicate or NULL records
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