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jensenshannon(p, q, base=None, *, axis=0, keepdims=False)

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.

Notes

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Parameters

p : (N,) array_like

left probability vector

q : (N,) array_like

right probability vector

base : double, 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.

axis : int, optional

Axis along which the Jensen-Shannon distances are computed. The default is 0.

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keepdims : bool, optional

If this is set to :None:None:`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.

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Returns

js : double or ndarray

The Jensen-Shannon distances between p and q along the :None:None:`axis`.

Compute the Jensen-Shannon distance (metric) between two probability arrays. This is the square root of the Jensen-Shannon divergence.

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])
See :

Back References

The following pages refer to to this document either explicitly or contain code examples using this.

scipy.spatial.distance.jensenshannon

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GitHub : /scipy/spatial/distance.py#1220
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