Just a hint for everybody reading that:
the functions above do not compute the divergence of a vector field. they sum the derivatives of a scalar field A:
result = dA/dx + dA/dy
in contrast to a vector field (with three dimensional example):
result = sum dAi/dxi = dAx/dx + dAy/dy + dAz/dz
Vote down for all! It is mathematically simply wrong.
Cheers!
Answer from Polly on Stack OverflowJust a hint for everybody reading that:
the functions above do not compute the divergence of a vector field. they sum the derivatives of a scalar field A:
result = dA/dx + dA/dy
in contrast to a vector field (with three dimensional example):
result = sum dAi/dxi = dAx/dx + dAy/dy + dAz/dz
Vote down for all! It is mathematically simply wrong.
Cheers!
import numpy as np
def divergence(field):
"return the divergence of a n-D field"
return np.sum(np.gradient(field),axis=0)
try something like:
dqu_dx, dqu_dy = np.gradient(qu, [dx, dy])
dqv_dx, dqv_dy = np.gradient(qv, [dx, dy])
you can not assign to any operation in python; any of those are syntax errors:
a + b = 3
a * b = 7
# or, in your case:
a / b = 9
UPDATE
following Pinetwig's comment: a/b is not a valid identifier name; it is (the return value of) an operator.
Try removing the [dx, dy].
[dqu_dx, dqu_dy] = np.gradient(qu)
[dqv_dx, dqv_dy] = np.gradient(qv)
Also to point out if you are recreating plots. Gradient changed in numpy between 1.82 and 1.9. This had an effect for recreating matlab plots in python as 1.82 was the matlab method. I am not sure how this relates to GrADs. Here is the wording for both.
1.82
"The gradient is computed using central differences in the interior and first differences at the boundaries. The returned gradient hence has the same shape as the input array."
1.9
"The gradient is computed using second order accurate central differences in the interior and either first differences or second order accurate one-sides (forward or backwards) differences at the boundaries. The returned gradient hence has the same shape as the input array."
The gradient function for 1.82 is here.
def gradient(f, *varargs):
"""
Return the gradient of an N-dimensional array.
The gradient is computed using central differences in the interior
and first differences at the boundaries. The returned gradient hence has
the same shape as the input array.
Parameters
----------
f : array_like
An N-dimensional array containing samples of a scalar function.
`*varargs` : scalars
0, 1, or N scalars specifying the sample distances in each direction,
that is: `dx`, `dy`, `dz`, ... The default distance is 1.
Returns
-------
gradient : ndarray
N arrays of the same shape as `f` giving the derivative of `f` with
respect to each dimension.
Examples
--------
>>> x = np.array([1, 2, 4, 7, 11, 16], dtype=np.float)
>>> np.gradient(x)
array([ 1. , 1.5, 2.5, 3.5, 4.5, 5. ])
>>> np.gradient(x, 2)
array([ 0.5 , 0.75, 1.25, 1.75, 2.25, 2.5 ])
>>> np.gradient(np.array([[1, 2, 6], [3, 4, 5]], dtype=np.float))
[array([[ 2., 2., -1.],
[ 2., 2., -1.]]),
array([[ 1. , 2.5, 4. ],
[ 1. , 1. , 1. ]])]
"""
f = np.asanyarray(f)
N = len(f.shape) # number of dimensions
n = len(varargs)
if n == 0:
dx = [1.0]*N
elif n == 1:
dx = [varargs[0]]*N
elif n == N:
dx = list(varargs)
else:
raise SyntaxError(
"invalid number of arguments")
# use central differences on interior and first differences on endpoints
outvals = []
# create slice objects --- initially all are [:, :, ..., :]
slice1 = [slice(None)]*N
slice2 = [slice(None)]*N
slice3 = [slice(None)]*N
otype = f.dtype.char
if otype not in ['f', 'd', 'F', 'D', 'm', 'M']:
otype = 'd'
# Difference of datetime64 elements results in timedelta64
if otype == 'M' :
# Need to use the full dtype name because it contains unit information
otype = f.dtype.name.replace('datetime', 'timedelta')
elif otype == 'm' :
# Needs to keep the specific units, can't be a general unit
otype = f.dtype
for axis in range(N):
# select out appropriate parts for this dimension
out = np.empty_like(f, dtype=otype)
slice1[axis] = slice(1, -1)
slice2[axis] = slice(2, None)
slice3[axis] = slice(None, -2)
# 1D equivalent -- out[1:-1] = (f[2:] - f[:-2])/2.0
out[slice1] = (f[slice2] - f[slice3])/2.0
slice1[axis] = 0
slice2[axis] = 1
slice3[axis] = 0
# 1D equivalent -- out[0] = (f[1] - f[0])
out[slice1] = (f[slice2] - f[slice3])
slice1[axis] = -1
slice2[axis] = -1
slice3[axis] = -2
# 1D equivalent -- out[-1] = (f[-1] - f[-2])
out[slice1] = (f[slice2] - f[slice3])
# divide by step size
outvals.append(out / dx[axis])
# reset the slice object in this dimension to ":"
slice1[axis] = slice(None)
slice2[axis] = slice(None)
slice3[axis] = slice(None)
if N == 1:
return outvals[0]
else:
return outvals
First of all, sklearn.metrics.mutual_info_score implements mutual information for evaluating clustering results, not pure Kullback-Leibler divergence!
This is equal to the Kullback-Leibler divergence of the joint distribution with the product distribution of the marginals.
KL divergence (and any other such measure) expects the input data to have a sum of 1. Otherwise, they are not proper probability distributions. If your data does not have a sum of 1, most likely it is usually not proper to use KL divergence! (In some cases, it may be admissible to have a sum of less than 1, e.g. in the case of missing data.)
Also note that it is common to use base 2 logarithms. This only yields a constant scaling factor in difference, but base 2 logarithms are easier to interpret and have a more intuitive scale (0 to 1 instead of 0 to log2=0.69314..., measuring the information in bits instead of nats).
> sklearn.metrics.mutual_info_score([0,1],[1,0])
0.69314718055994529
as we can clearly see, the MI result of sklearn is scaled using natural logarithms instead of log2. This is an unfortunate choice, as explained above.
Kullback-Leibler divergence is fragile, unfortunately. On above example it is not well-defined: KL([0,1],[1,0]) causes a division by zero, and tends to infinity. It is also asymmetric.
Scipy's entropy function will calculate KL divergence if feed two vectors p and q, each representing a probability distribution. If the two vectors aren't pdfs, it will normalize then first.
Mutual information is related to, but not the same as KL Divergence.
"This weighted mutual information is a form of weighted KL-Divergence, which is known to take negative values for some inputs, and there are examples where the weighted mutual information also takes negative values"
» pip install divergence