numpy.ma.median

ma.median(a, axis=None, out=None, overwrite_input=False, keepdims=False)[source]

Compute the median along the specified axis.

Returns the median of the array elements.

Parameters
aarray_like

Input array or object that can be converted to an array.

axisint, optional

Axis along which the medians are computed. The default (None) is to compute the median along a flattened version of the array.

outndarray, optional

Alternative output array in which to place the result. It must have the same shape and buffer length as the expected output but the type will be cast if necessary.

overwrite_inputbool, optional

If True, then allow use of memory of input array (a) for calculations. The input array will be modified by the call to median. This will save memory when you do not need to preserve the contents of the input array. Treat the input as undefined, but it will probably be fully or partially sorted. Default is False. Note that, if overwrite_input is True, and the input is not already an ndarray, an error will be raised.

keepdimsbool, optional

If this is set to True, the axes which are reduced are left in the result as dimensions with size one. With this option, the result will broadcast correctly against the input array.

New in version 1.10.0.

Returns
medianndarray

A new array holding the result is returned unless out is specified, in which case a reference to out is returned. Return data-type is float64 for integers and floats smaller than float64, or the input data-type, otherwise.

See also

mean

Notes

Given a vector V with N non masked values, the median of V is the middle value of a sorted copy of V (Vs) - i.e. Vs[(N-1)/2], when N is odd, or {Vs[N/2 - 1] + Vs[N/2]}/2 when N is even.

Examples

>>> x = np.ma.array(np.arange(8), mask=[0]*4 + [1]*4)
>>> np.ma.median(x)
1.5
>>> x = np.ma.array(np.arange(10).reshape(2, 5), mask=[0]*6 + [1]*4)
>>> np.ma.median(x)
2.5
>>> np.ma.median(x, axis=-1, overwrite_input=True)
masked_array(data=[2.0, 5.0],
             mask=[False, False],
       fill_value=1e+20)