  ...         "A": [1, 1, 2, 2],
    ...         "B": [1, 2, 3, 4],
    ...         "C": [0.362838, 0.227877, 1.267767, -0.562860],
    ...     }
    ... )

    >>> df
       A  B         C
    0  1  1  0.362838
    1  1  2  0.227877
    2  2  3  1.267767
    3  2  4 -0.562860

    The aggregation is for each column.

    >>> df.groupby('A').agg('min')
       B         C
    A
    1  1  0.227877
    2  3 -0.562860

    Multiple aggregations

    >>> df.groupby('A').agg(['min', 'max'])
        B             C
      min max       min       max
    A
    1   1   2  0.227877  0.362838
    2   3   4 -0.562860  1.267767

    Select a column for aggregation

    >>> df.groupby('A').B.agg(['min', 'max'])
       min  max
    A
    1    1    2
    2    3    4

    User-defined function for aggregation

    >>> df.groupby('A').agg(lambda x: sum(x) + 2)
        B	       C
    A
    1	5	2.590715
    2	9	2.704907

    Different aggregations per column

    >>> df.groupby('A').agg({'B': ['min', 'max'], 'C': 'sum'})
        B             C
      min max       sum
    A
    1   1   2  0.590715
    2   3   4  0.704907

    To control the output names with different aggregations per column,
    pandas supports "named aggregation"

    >>> df.groupby("A").agg(
    ...     b_min=pd.NamedAgg(column="B", aggfunc="min"),
    ...     c_sum=pd.NamedAgg(column="C", aggfunc="sum"))
       b_min     c_sum
    A
    1      1  0.590715
    2      3  0.704907

    - The keywords are the *output* column names
    - The values are tuples whose first element is the column to select
      and the second element is the aggregation to apply to that column.
      Pandas provides the ``pandas.NamedAgg`` namedtuple with the fields
      ``['column', 'aggfunc']`` to make it clearer what the arguments are.
      As usual, the aggregation can be a callable or a string alias.

    See :ref:`groupby.aggregate.named` for more.

    .. versionchanged:: 1.3.0

        The resulting dtype will reflect the return value of the aggregating function.

    >>> df.groupby("A")[["B"]].agg(lambda x: x.astype(float).min())
          B
    A
    1   1.0
    2   3.0
    r'