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Estimating entropy and mutual information with scikit-learn: visit https://github.com/mutualinfo/mutual_info
'''
Non-parametric computation of entropy and mutual-information
Adapted by G Varoquaux for code created by R Brette, itself
from several papers (see in the code).
This code is maintained at https://github.com/mutualinfo/mutual_info
Please download the latest code there, to have improvements and
bug fixes.
These computations rely on nearest-neighbor statistics
'''
import numpy as np
from scipy.special import gamma,psi
from scipy import ndimage
from scipy.linalg import det
from numpy import pi
from sklearn.neighbors import NearestNeighbors
__all__=['entropy', 'mutual_information', 'entropy_gaussian']
EPS = np.finfo(float).eps
def nearest_distances(X, k=1):
'''
X = array(N,M)
N = number of points
M = number of dimensions
returns the distance to the kth nearest neighbor for every point in X
'''
knn = NearestNeighbors(n_neighbors=k + 1)
knn.fit(X)
d, _ = knn.kneighbors(X) # the first nearest neighbor is itself
return d[:, -1] # returns the distance to the kth nearest neighbor
def entropy_gaussian(C):
'''
Entropy of a gaussian variable with covariance matrix C
'''
if np.isscalar(C): # C is the variance
return .5*(1 + np.log(2*pi)) + .5*np.log(C)
else:
n = C.shape[0] # dimension
return .5*n*(1 + np.log(2*pi)) + .5*np.log(abs(det(C)))
def entropy(X, k=1):
''' Returns the entropy of the X.
Parameters
===========
X : array-like, shape (n_samples, n_features)
The data the entropy of which is computed
k : int, optional
number of nearest neighbors for density estimation
Notes
======
Kozachenko, L. F. & Leonenko, N. N. 1987 Sample estimate of entropy
of a random vector. Probl. Inf. Transm. 23, 95-101.
See also: Evans, D. 2008 A computationally efficient estimator for
mutual information, Proc. R. Soc. A 464 (2093), 1203-1215.
and:
Kraskov A, Stogbauer H, Grassberger P. (2004). Estimating mutual
information. Phys Rev E 69(6 Pt 2):066138.
'''
# Distance to kth nearest neighbor
r = nearest_distances(X, k) # squared distances
n, d = X.shape
volume_unit_ball = (pi**(.5*d)) / gamma(.5*d + 1)
'''
F. Perez-Cruz, (2008). Estimation of Information Theoretic Measures
for Continuous Random Variables. Advances in Neural Information
Processing Systems 21 (NIPS). Vancouver (Canada), December.
return d*mean(log(r))+log(volume_unit_ball)+log(n-1)-log(k)
'''
return (d*np.mean(np.log(r + np.finfo(X.dtype).eps))
+ np.log(volume_unit_ball) + psi(n) - psi(k))
def mutual_information(variables, k=1):
'''
Returns the mutual information between any number of variables.
Each variable is a matrix X = array(n_samples, n_features)
where
n = number of samples
dx,dy = number of dimensions
Optionally, the following keyword argument can be specified:
k = number of nearest neighbors for density estimation
Example: mutual_information((X, Y)), mutual_information((X, Y, Z), k=5)
'''
if len(variables) < 2:
raise AttributeError(
"Mutual information must involve at least 2 variables")
all_vars = np.hstack(variables)
return (sum([entropy(X, k=k) for X in variables])
- entropy(all_vars, k=k))
def mutual_information_2d(x, y, sigma=1, normalized=False):
"""
Computes (normalized) mutual information between two 1D variate from a
joint histogram.
Parameters
----------
x : 1D array
first variable
y : 1D array
second variable
sigma: float
sigma for Gaussian smoothing of the joint histogram
Returns
-------
nmi: float
the computed similariy measure
"""
bins = (256, 256)
jh = np.histogram2d(x, y, bins=bins)[0]
# smooth the jh with a gaussian filter of given sigma
ndimage.gaussian_filter(jh, sigma=sigma, mode='constant',
output=jh)
# compute marginal histograms
jh = jh + EPS
sh = np.sum(jh)
jh = jh / sh
s1 = np.sum(jh, axis=0).reshape((-1, jh.shape[0]))
s2 = np.sum(jh, axis=1).reshape((jh.shape[1], -1))
# Normalised Mutual Information of:
# Studholme, jhill & jhawkes (1998).
# "A normalized entropy measure of 3-D medical image alignment".
# in Proc. Medical Imaging 1998, vol. 3338, San Diego, CA, pp. 132-143.
if normalized:
mi = ((np.sum(s1 * np.log(s1)) + np.sum(s2 * np.log(s2)))
/ np.sum(jh * np.log(jh))) - 1
else:
mi = ( np.sum(jh * np.log(jh)) - np.sum(s1 * np.log(s1))
- np.sum(s2 * np.log(s2)))
return mi
###############################################################################
# Tests
def test_entropy():
# Testing against correlated Gaussian variables
# (analytical results are known)
# Entropy of a 3-dimensional gaussian variable
rng = np.random.RandomState(0)
n = 50000
d = 3
P = np.array([[1, 0, 0], [0, 1, .5], [0, 0, 1]])
C = np.dot(P, P.T)
Y = rng.randn(d, n)
X = np.dot(P, Y)
H_th = entropy_gaussian(C)
H_est = entropy(X.T, k=5)
# Our estimated entropy should always be less that the actual one
# (entropy estimation undershoots) but not too much
np.testing.assert_array_less(H_est, H_th)
np.testing.assert_array_less(.9*H_th, H_est)
def test_mutual_information():
# Mutual information between two correlated gaussian variables
# Entropy of a 2-dimensional gaussian variable
n = 50000
rng = np.random.RandomState(0)
#P = np.random.randn(2, 2)
P = np.array([[1, 0], [0.5, 1]])
C = np.dot(P, P.T)
U = rng.randn(2, n)
Z = np.dot(P, U).T
X = Z[:, 0]
X = X.reshape(len(X), 1)
Y = Z[:, 1]
Y = Y.reshape(len(Y), 1)
# in bits
MI_est = mutual_information((X, Y), k=5)
MI_th = (entropy_gaussian(C[0, 0])
+ entropy_gaussian(C[1, 1])
- entropy_gaussian(C)
)
# Our estimator should undershoot once again: it will undershoot more
# for the 2D estimation that for the 1D estimation
print((MI_est, MI_th))
np.testing.assert_array_less(MI_est, MI_th)
np.testing.assert_array_less(MI_th, MI_est + .3)
def test_degenerate():
# Test that our estimators are well-behaved with regards to
# degenerate solutions
rng = np.random.RandomState(0)
x = rng.randn(50000)
X = np.c_[x, x]
assert np.isfinite(entropy(X))
assert np.isfinite(mutual_information((x[:, np.newaxis],
x[:, np.newaxis])))
assert 2.9 < mutual_information_2d(x, x) < 3.1
def test_mutual_information_2d():
# Mutual information between two correlated gaussian variables
# Entropy of a 2-dimensional gaussian variable
n = 50000
rng = np.random.RandomState(0)
#P = np.random.randn(2, 2)
P = np.array([[1, 0], [.9, .1]])
C = np.dot(P, P.T)
U = rng.randn(2, n)
Z = np.dot(P, U).T
X = Z[:, 0]
X = X.reshape(len(X), 1)
Y = Z[:, 1]
Y = Y.reshape(len(Y), 1)
# in bits
MI_est = mutual_information_2d(X.ravel(), Y.ravel())
MI_th = (entropy_gaussian(C[0, 0])
+ entropy_gaussian(C[1, 1])
- entropy_gaussian(C)
)
print((MI_est, MI_th))
# Our estimator should undershoot once again: it will undershoot more
# for the 2D estimation that for the 1D estimation
np.testing.assert_array_less(MI_est, MI_th)
np.testing.assert_array_less(MI_th, MI_est + .2)
if __name__ == '__main__':
# Run our tests
test_entropy()
test_mutual_information()
test_degenerate()
test_mutual_information_2d()
@Striderl
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Thank you for your work. Just one quick comment about the nearest_distances function.
Shouldn't line 31 be "knn = NearestNeighbors(n_neighbors=k+1)" ?

--g

I agree. according to the original paper, http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.422.5607&rep=rep1&type=pdf, this should calculate the Euclidean distances to the k-nn of xi in X \xi.

@GaelVaroquaux
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What if I want the MI of two sets that don't have the same amount of samples? like X.shape = (30, 10) and Y.shape = (1,10). My impression is that the np.hstack in line 103 is not enough.

I don't really see how it would be possible to compute non-parametric mutual information between two sets where there is no correspondence between the samples. By construction to estimate MI one needs this correspondence.

@GaelVaroquaux
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Shouldn't line 31 be "knn = NearestNeighbors(n_neighbors=k+1)" ?

I agree that using n_neighbors=k gives the k-1 nearest neighbor, as the first is the point itself. The code was wrong. Maybe that error was introduced in an evolution of scikit-learn. I've fixed it.

@Sandy4321
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Hello friends
can be similarity for categorical numbers used as mutual information
per
https://towardsdatascience.com/the-search-for-categorical-correlation-a1cf7f1888c9
https://github.com/shakedzy/dython
https://en.wikipedia.org/wiki/Uncertainty_coefficient
https://stackoverflow.com/questions/20892799/using-pandas-calculate-cram%C3%A9rs-coefficient-matrix
https://stackoverflow.com/questions/46498455/categorical-features-correlation/46498792#46498792

thanks
PS
it would be very kind of you to share some links to understand
why
https://scikit-learn.org/stable/modules/generated/sklearn.feature_selection.mutual_info_classif.html#sklearn.feature_selection.mutual_info_classif
Kozachenko, N. N. Leonenko,
is good for feature selection
why it impossible just calculate similarity/mutual information for each feature and target?
Then if similarity/mutual information for given feature and target is high then this feature is good to use??

@cottrell
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cottrell commented Jan 22, 2020

I've noticed that scaling samples changes the output. I suspect this is a small bug in the scaling somewhere. Will post back if I trace it down.

Compare the following:

In [126]: np.random.seed(12)
     ...: for i in range(20):
     ...:     print(mi.mutual_information([np.random.randn(100, 1), 1 * np.random.randn(100, 1)], k=10))
     ...:
-0.020132950207217615
0.0007764531823655219
-0.10097362008313171
0.008234538277922088
-0.019989602665604345
-0.07712501107587677
0.02317718497197019
-0.01624389693603323
-0.037344436692854366
0.04950941718866808
-0.05049871422680052
-0.028624985257037494
-0.004130417096971595
-0.09652205195754915
0.07427403221566875
-0.0040393263534936885
-0.058616537991666995
0.09200872016468775
0.04512428420750236
-0.07361872725115903

In [127]: np.random.seed(12)
     ...: for i in range(20):
     ...:     print(mi.mutual_information([np.random.randn(100, 1), 100 * np.random.randn(100, 1)], k=10))
     ...:
-2.0021553153700378
-1.7289150378782132
-1.938608386083576
-1.999900260449417
-1.7849647677622498
-1.8527331847233786
-2.0258220171874264
-2.00130782835749
-2.017433687721022
-1.8716648526808015
-2.159052646410464
-2.090519859781173
-2.1565500325401965
-1.9789886099442322
-1.9400487388101713
-2.004367515829771
-1.9796955922060349
-2.016857734129518
-2.1187142194709843
-1.9679714350429087

UPDATE: I think this is just the result of using a differential entropy throughout. Probably (not sure yet) but using some copula style normalization (see pd.rank for example) is a decent way of getting something invariant to some transformations. Might destroy some structure depending on the use case though.

Also, there are few forks kicking around now. I think I've got one that fixed some small issue with the case of mutual information with itself.

https://gist.github.com/cottrell/770b811c03c846386b19e9c4bd18772f/revisions

I'm wondering if we should try to coral everyone together and make a repo. Viewing diffs is a little less convenient from the gist. Though I like the single discussion thread.

@GaelVaroquaux
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GaelVaroquaux commented Jan 23, 2020 via email

@AbdolvakilFazli
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Thank you for the work,

Assume that I have a node with some features(f1,f2,f3). And another nodes from features (f1',f2',f3'). how can I find the mutual information of the two?

@Sandy4321
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Good question
Really, how it will be?

@maajdl
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maajdl commented Apr 22, 2020

If I use k = [1,2,3,4,5,6] and then I test mutual_info_2d(k,k), I get as a result: 1.7456376329342476.
Shouldn't it be 1 when I test the mutual information between two identical vectors?

I would assume the natural result would be -6*(1/6)*log(1/6)=log(6) = 1.79175946923 .
This is assuming each value has the same probability of 1/6 .
Looks good given the that it is an estimation.

@thismartian
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I noted two errors in this code (based on the reference paper from Krasov), and this also fixed the negative MI values I was getting.

  1. volume_unit_ball is missing a small expression. It should be : (pi**(d/2)/gamma(d/2 +1)/2**d)
  2. volume_unit_ball values for every variable need to be multiplied to each other before calculating the log. So line 75, volume_unit_ball = (pi**(.5d)) / gamma(.5d + 1) should actually be replaced by something like this:

volume_unit_ball = 1
for dim in range(1, d+1):
volume_unit_ball = volume_unit_ball * (pi**(dim/2)/gamma(dim/2 +1)/2**dim)

@cottrell
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cottrell commented Feb 5, 2021

@thismartion I haven't yet found it in another library where it should probably sit. Have created this and added you and @GaelVaroquaux . You should have admin privileges if I did it right.

@GaelVaroquaux
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Creating a library for collective maintenance was really the right move. I'll update the gist to point there.

@cottrell
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cottrell commented Feb 5, 2021

volume_unit_ball = 1
for dim in range(1, d+1):
volume_unit_ball = volume_unit_ball * (pi**(dim/2)/gamma(dim/2 +1)/2**dim)

I've copy pasta'd your change into this branch. https://github.com/mutualinfo/mutual_info/tree/thismartian

There is some test failure but I haven't looked at it yet. I suspect if you still have it in your head it is an easy test change. I'm not ever sure what is correct since it's been a few months since I've used this stuff.

@thismartian
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Hi, cottrell. Thanks a lot for this. I figured out what the problem is. The distance used to calculate the entropy should be 2x the distance to the nearest neighbor. Not sure I'm doing it right but I don't seem to have the permission to make changes to the file, perhaps you could try this: in the entropy function: return d * np.mean(np.log(2*r + np.finfo(X.dtype).eps)) + np.log(volume_unit_ball) + psi(n) - psi(k)

@cottrell
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cottrell commented Feb 6, 2021

Hi, cottrell. Thanks a lot for this. I figured out what the problem is. The distance used to calculate the entropy should be 2x the distance to the nearest neighbor. Not sure I'm doing it right but I don't seem to have the permission to make changes to the file, perhaps you could try this: in the entropy function: return d * np.mean(np.log(2*r + np.finfo(X.dtype).eps)) + np.log(volume_unit_ball) + psi(n) - psi(k)

Hmmm, I can past it in and try but let's try to fix the permissions. Having a look now ...

@cottrell
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cottrell commented Feb 6, 2021

Ok, I think I've now added @thismartian and @GaelVaroquaux as admins on the org and the repo now. For some reason it only sets you up as members initially when I punched them in. Let me know if you can't access it.

@cottrell
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cottrell commented Feb 6, 2021

@thismartian The tests now pass after I pasted the change in. It's still just on that branch if you want to review it. If there is an obvious test that would have failed before your change and passed after, maybe we can add that?

@thismartian
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@cottrell the changes look fine. I think the entropy test was failing (as expected). Is that what you mean? Sorry it's not clear to me...

@cottrell
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@cottrell the changes look fine. I think the entropy test was failing (as expected). Is that what you mean? Sorry it's not clear to me...

No I mean before the change, no tests were failing but you noticed something was wrong. What is a missing test that would have shown something was wrong?

@thismartian
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thismartian commented Feb 11, 2021

Ah, for me the main indication was just that for some random combinations the mutual info was negative (as you had pointed out earlier too). I think in the test itself, if you change the covariance matrix to unit variance, it shows that the test failed. Honestly, I am not sure I understand how the covariance matrix is generated in the tests (I need to read about it). What I would do is: define a correlation coefficient (r), standard deviation (s), and based on these two it's possible to have a unit variance, zero mean covariance matrix:

`

s = 0.6
r = 0.9

C= np.array([[0.5*s**2, r*0.5*s**2], [r*0.5*s**2, 0.5*s**2]]) 

mean = [0,0]

x,y = np.random.multivariate_normal(mean, C, size = n).T

X = x.reshape(len(x), 1)

Y = y.reshape(len(y), 1) `

@thismartian
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I was probably not clear, let me try to rephrase:

  • In the mutual info test written by @GaelVaroquaux, the covariance matrix does not have a unit variance. If you change it to a unit variance matrix, the test fails. The gaussian reference used in the paper is based on a zero mean, unit variance covariance matrix.
  • I don't have an expertise in statistics so I don't really understand the approach used in the code to generate the correlated XY pair. I have instead generated the XY pair based on a numpy function (previous comment), which I understand better.

@cottrell
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Thanks @thismartian I will take a look a try to jam a test in shortly.

@thismartian
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thismartian commented Mar 3, 2021 via email

@cottrell
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cottrell commented Mar 3, 2021

@thismartian Yeah I can't remember where I left it. I think I put some new tests in and have a branch there. I think I was looking at differential entropy in general and how it sort of has a dimensionality issue. The branch should run and we could just merge. Likely I'll dig into it more next time I am looking for entropy of continuous random variables.

@Sandy4321
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friends
lets finish this great task?

@thismartian
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Yes, I let cottrell manage the merge after reviewing thismartian branch.

@cottrell
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friends
lets finish this great task?

I can merge it in. There isn't a ton of testing or anything because it's just the gist copied over so we can break and fix as we go along. I should probably set up my notifications for that org ... am not noticing these often enough.

@cottrell
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friends
lets finish this great task?

I can merge it in. There isn't a ton of testing or anything because it's just the gist copied over so we can break and fix as we go along. I should probably set up my notifications for that org ... am not noticing these often enough.

I think my reply got dropped. It is all merged in now. I had some minor refactors and added some tests. I can't remember the details but was trying to make in shift and scale invariant.

@EtienneCmb
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Hi everyone, thanks for the shared code, very interesting.

I've a question regarding the estimation of the entropy for gaussian variables. What is the reason for taking the absolute value of the determinant? Is it for numerical reasons?

Thank you in advance,

@cottrell
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Hi everyone, thanks for the shared code, very interesting.

I've a question regarding the estimation of the entropy for gaussian variables. What is the reason for taking the absolute value of the determinant? Is it for numerical reasons?

Thank you in advance,

That is in the definition of the pdf basically. Plus there is a log there so you need an abs.

You can also think of it as a scale transformation (change of variables, law of the unconsious statistician) on gaussian with identity covariance.

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