pandas.DataFrame.median#

DataFrame.median(*, axis=0, skipna=True, numeric_only=False, **kwargs)[source]#

Return the median of the values over the requested axis.

Parameters:

axis{index (0), columns (1)}

Axis for the function to be applied on. For Series this parameter is unused and defaults to 0.

For DataFrames, specifying axis=None will apply the aggregation across both axes.

Added in version 2.0.0.

skipnabool, default True

Exclude NA/null values when computing the result.

numeric_onlybool, default False

Include only float, int, boolean columns.

**kwargs

Additional keyword arguments to be passed to the function.

Returns:

Series or scalar: Value containing the calculation referenced in the description.

See also

Series.sum: Return the sum.
Series.min: Return the minimum.
Series.max: Return the maximum.
Series.idxmin: Return the index of the minimum.
Series.idxmax: Return the index of the maximum.
DataFrame.sum: Return the sum over the requested axis.
DataFrame.min: Return the minimum over the requested axis.
DataFrame.max: Return the maximum over the requested axis.
DataFrame.idxmin: Return the index of the minimum over the requested axis.
DataFrame.idxmax: Return the index of the maximum over the requested axis.

Examples

>>> s = pd.Series([1, 2, 3])
>>> s.median()
2.0

With a DataFrame

>>> df = pd.DataFrame({'a': [1, 2], 'b': [2, 3]}, index=['tiger', 'zebra'])
>>> df
       a   b
tiger  1   2
zebra  2   3
>>> df.median()
a   1.5
b   2.5
dtype: float64

Using axis=1

>>> df.median(axis=1)
tiger   1.5
zebra   2.5
dtype: float64

In this case, numeric_only should be set to True to avoid getting an error.

>>> df = pd.DataFrame({'a': [1, 2], 'b': ['T', 'Z']},
...                   index=['tiger', 'zebra'])
>>> df.median(numeric_only=True)
a   1.5
dtype: float64