 keyword arguments to pass as keywords arguments to
            `func`.

        Returns
        -------
        Series or DataFrame
            Result of applying ``func`` along the given axis of the
            DataFrame.

        See Also
        --------
        DataFrame.map: For elementwise operations.
        DataFrame.aggregate: Only perform aggregating type operations.
        DataFrame.transform: Only perform transforming type operations.

        Notes
        -----
        Functions that mutate the passed object can produce unexpected
        behavior or errors and are not supported. See :ref:`gotchas.udf-mutation`
        for more details.

        Examples
        --------
        >>> df = pd.DataFrame([[4, 9]] * 3, columns=['A', 'B'])
        >>> df
           A  B
        0  4  9
        1  4  9
        2  4  9

        Using a numpy universal function (in this case the same as
        ``np.sqrt(df)``):

        >>> df.apply(np.sqrt)
             A    B
        0  2.0  3.0
        1  2.0  3.0
        2  2.0  3.0

        Using a reducing function on either axis

        >>> df.apply(np.sum, axis=0)
        A    12
        B    27
        dtype: int64

        >>> df.apply(np.sum, axis=1)
        0    13
        1    13
        2    13
        dtype: int64

        Returning a list-like will result in a Series

        >>> df.apply(lambda x: [1, 2], axis=1)
        0    [1, 2]
        1    [1, 2]
        2    [1, 2]
        dtype: object

        Passing ``result_type='expand'`` will expand list-like results
        to columns of a Dataframe

        >>> df.apply(lambda x: [1, 2], axis=1, result_type='expand')
           0  1
        0  1  2
        1  1  2
        2  1  2

        Returning a Series inside the function is similar to passing
        ``result_type='expand'``. The resulting column names
        will be the Series index.

        >>> df.apply(lambda x: pd.Series([1, 2], index=['foo', 'bar']), axis=1)
           foo  bar
        0    1    2
        1    1    2
        2    1    2

        Passing ``result_type='broadcast'`` will ensure the same shape
        result, whether list-like or scalar is returned by the function,
        and broadcast it along the axis. The resulting column names will
        be the originals.

        >>> df.apply(lambda x: [1, 2], axis=1, result_type='broadcast')
           A  B
        0  1  2
        1  1  2
        2  1  2
        r