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

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