def adjustedMutualInformation(cluster1, cluster2):
'''
Using the scikit-learn algorithms, calculate the adjusted mutual
information for two clusterings. Assume cluster1 is the
reference/ground truth clustering.
The adjusted MI accounts for higher scores by chance, particularly
in the case where a larger number of clusters leads to a higher MI.
AMI(U, V) = [MI(U, V) - E(MI(U, V))] / [max(H(U), H(V)) - E(MI(U, V))]
where E(MI(U, V)) is the expected mutual information given the
number of clusters and H(U), H(V) are the entropies of clusterings
U and V.
'''
cont = contingency(cluster1, cluster2)
mi = mutualInformation(cluster1, cluster2)
sample_size = float(sum([len(cluster1[x]) for x in cluster1.keys()]))
# Given the number of samples, what is the expected number
# of overlaps that would occur by chance?
emi = supervised.expected_mutual_information(cont, sample_size)
# calculate the entropy for each clustering
h_clust1, h_clust2 = entropy(cluster1), entropy(cluster2)
if abs(h_clust1) == 0.0:
h_clust1 = 0.0
else:
pass
if abs(h_clust2) == 0.0:
h_clust2 = 0.0
else:
pass
ami = (mi - emi) / (max(h_clust1, h_clust2) - emi)
# bug: entropy will return -0 in some instances
# make sure this is actually 0 else ami will return None
# instead of 0.0
if np.isnan(ami):
ami = np.nan_to_num(ami)
else:
pass
return ami
def adjustedMutualInformation(cluster1, cluster2):
'''
Using the scikit-learn algorithms, calculate the adjusted mutual
information for two clusterings. Assume cluster1 is the
reference/ground truth clustering.
The adjusted MI accounts for higher scores by chance, particularly
in the case where a larger number of clusters leads to a higher MI.
AMI(U, V) = [MI(U, V) - E(MI(U, V))] / [max(H(U), H(V)) - E(MI(U, V))]
where E(MI(U, V)) is the expected mutual information given the
number of clusters and H(U), H(V) are the entropies of clusterings
U and V.
'''
cont = contingency(cluster1, cluster2)
mi = mutualInformation(cluster1, cluster2)
sample_size = float(sum([len(cluster1[x]) for x in cluster1.keys()]))
# Given the number of samples, what is the expected number
# of overlaps that would occur by chance?
emi = supervised.expected_mutual_information(cont, sample_size)
# calculate the entropy for each clustering
h_clust1, h_clust2 = entropy(cluster1), entropy(cluster2)
if abs(h_clust1) == 0.0:
h_clust1 = 0.0
else:
pass
if abs(h_clust2) == 0.0:
h_clust2 = 0.0
else:
pass
ami = (mi - emi) / (max(h_clust1, h_clust2) - emi)
# bug: entropy will return -0 in some instances
# make sure this is actually 0 else ami will return None
# instead of 0.0
if np.isnan(ami):
ami = np.nan_to_num(ami)
else:
pass
return ami
def _ami(ab_cts, average_method='arithmetic'):
"""Adjusted mutual information between two discrete categorical random variables
based on counts observed and provided in ab_cts.
Code adapted directly from scikit learn AMI to
accomodate having counts/contingency table instead of rows/instances:
https://scikit-learn.org/stable/modules/generated/sklearn.metrics.adjusted_mutual_info_score.html
Parameters
----------
ab_cts : np.ndarray [len(a_classes) x len(b_classes)
Counts for each combination of classes in random variables a and b
organized in a rectangular array.
average_method : str
See sklearn documentation for details
Returns
-------
ami : float
Adjusted mutual information score for variables a and b"""
a_freq = np.sum(ab_cts, axis=1)
a_freq = a_freq / np.sum(a_freq)
b_freq = np.sum(ab_cts, axis=0)
b_freq = b_freq / np.sum(b_freq)
n_samples = np.sum(ab_cts)
""" Calculate the MI for the two clusterings
contingency is a joint count distribution [a_classes x b_classes]"""
mi = mutual_info_score(None, None, contingency=ab_cts)
"""Calculate the expected value for the mutual information"""
emi = expected_mutual_information(ab_cts, n_samples)
"""Calculate entropy"""
h_true, h_pred = _entropy(a_freq), _entropy(b_freq)
normalizer = _generalized_average(h_true, h_pred, average_method)
denominator = normalizer - emi
if denominator < 0:
denominator = min(denominator, -np.finfo('float64').eps)
else:
denominator = max(denominator, np.finfo('float64').eps)
ami = (mi - emi) / denominator
return ami