numpy.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: |
a : array_like Input array or object that can be converted to an array. axis : int, optional Axis along which the medians are computed. The default (None) is to compute the median along a flattened version of the array. out : ndarray, 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_input : bool, 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 keepdims : bool, 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. |
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Returns: |
median : ndarray 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 |
See also
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.
>>> 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. 5.], mask = False, fill_value = 1e+20)
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https://docs.scipy.org/doc/numpy-1.12.0/reference/generated/numpy.ma.median.html