snowflake.snowpark.modin.plugin.extensions.groupby_overrides.DataFrameGroupBy.indices¶

property DataFrameGroupBy.indices: dict[collections.abc.Hashable, numpy.ndarray[Any, numpy.dtype[numpy.int64]]][source] (https://github.com/snowflakedb/snowpark-python/blob/v1.26.0/snowpark-python/src/snowflake/snowpark/modin/plugin/extensions/groupby_overrides.py#L211-L223)¶

Get a dictionary mapping group key to row positions.

Returns:

Dict {group name -> group positions}.

Return type:

Dict[Any, np.array]

Examples

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>>> df = pd.DataFrame({'A': [1, 1, 2, 1, 2],
...                    'B': [np.nan, 2, 3, 4, 5],
...                    'C': [1, 2, 1, 1, 2]})

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Groupby one column and get the positions of each member of each group.

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>>> df.groupby('A').indices
{1: array([0, 1, 3]), 2: array([2, 4])}

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Group the same dataframe with a different index. The result is the same because the row positions for each group are the same.

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>>> df.set_index('B').groupby('A').indices
{1: array([0, 1, 3]), 2: array([2, 4])}

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Notes

Beware that the return value is a python dictionary, so evaluating this property will trigger evaluation of the pandas dataframe and will materialize data that could be as large as the size of the grouping columns.