scipy.spatial.distance.jensenshannon¶
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scipy.spatial.distance.jensenshannon(p, q, base=None, *, axis=0, keepdims=False)[source]¶ Compute the Jensen-Shannon distance (metric) between two probability arrays. This is the square root of the Jensen-Shannon divergence.
The Jensen-Shannon distance between two probability vectors p and q is defined as,
\[\sqrt{\frac{D(p \parallel m) + D(q \parallel m)}{2}}\]where \(m\) is the pointwise mean of \(p\) and \(q\) and \(D\) is the Kullback-Leibler divergence.
This routine will normalize p and q if they don’t sum to 1.0.
- Parameters
- p(N,) array_like
left probability vector
- q(N,) array_like
right probability vector
- basedouble, optional
the base of the logarithm used to compute the output if not given, then the routine uses the default base of scipy.stats.entropy.
- axisint, optional
Axis along which the Jensen-Shannon distances are computed. The default is 0.
New in version 1.7.0.
- keepdimsbool, optional
If this is set to True, the reduced axes are left in the result as dimensions with size one. With this option, the result will broadcast correctly against the input array. Default is False.
New in version 1.7.0.
- Returns
- jsdouble or ndarray
The Jensen-Shannon distances between p and q along the axis.
Notes
New in version 1.2.0.
Examples
>>> from scipy.spatial import distance >>> distance.jensenshannon([1.0, 0.0, 0.0], [0.0, 1.0, 0.0], 2.0) 1.0 >>> distance.jensenshannon([1.0, 0.0], [0.5, 0.5]) 0.46450140402245893 >>> distance.jensenshannon([1.0, 0.0, 0.0], [1.0, 0.0, 0.0]) 0.0 >>> a = np.array([[1, 2, 3, 4], ... [5, 6, 7, 8], ... [9, 10, 11, 12]]) >>> b = np.array([[13, 14, 15, 16], ... [17, 18, 19, 20], ... [21, 22, 23, 24]]) >>> distance.jensenshannon(a, b, axis=0) array([0.1954288, 0.1447697, 0.1138377, 0.0927636]) >>> distance.jensenshannon(a, b, axis=1) array([0.1402339, 0.0399106, 0.0201815])