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@stulacy
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"""
MAUCpy
~~~~~~
Contains two equations from Hand and Till's 2001 paper on a multi-class
approach to the AUC. The a_value() function is the probabilistic approximation
of the AUC found in equation 3, while MAUC() is the pairwise averaging of this
value for each of the classes. This is equation 7 in their paper.
"""
def a_value(probabilities, zero_label=0, one_label=1):
"""
Approximates the AUC by the method described in Hand and Till 2001,
equation 3.
NB: The class labels should be in the set [0,n-1] where n = # of classes.
The class probability should be at the index of its label in the
probability list.
I.e. With 3 classes the labels should be 0, 1, 2. The class probability
for class '1' will be found in index 1 in the class probability list
wrapped inside the zipped list with the labels.
Args:
probabilities (list): A zipped list of the labels and the
class probabilities in the form (m = # data instances):
[(label1, [p(x1c1), p(x1c2), ... p(x1cn)]),
(label2, [p(x2c1), p(x2c2), ... p(x2cn)])
...
(labelm, [p(xmc1), p(xmc2), ... (pxmcn)])
]
zero_label (optional, int): The label to use as the class '0'.
Must be an integer, see above for details.
one_label (optional, int): The label to use as the class '1'.
Must be an integer, see above for details.
Returns:
The A-value as a floating point.
"""
# Obtain a list of the probabilities for the specified zero label class
expanded_points = []
for instance in probabilities:
if instance[0] == zero_label or instance[0] == one_label:
expanded_points.append((instance[0], instance[1][zero_label]))
sorted_ranks = sorted(expanded_points, key=lambda x: x[1])
n0, n1, sum_ranks = 0, 0, 0
# Iterate through ranks and increment counters for overall count and ranks of class 0
for index, point in enumerate(sorted_ranks):
if point[0] == zero_label:
n0 += 1
sum_ranks += index + 1 # Add 1 as ranks are one-based
elif point[0] == one_label:
n1 += 1
else:
pass # Not interested in this class
return (sum_ranks - (n0*(n0+1)/2.0)) / float(n0 * n1) # Eqn 3
def MAUC(data, num_classes):
"""
Calculates the MAUC over a set of multi-class probabilities and
their labels. This is equation 7 in Hand and Till's 2001 paper.
NB: The class labels should be in the set [0,n-1] where n = # of classes.
The class probability should be at the index of its label in the
probability list.
I.e. With 3 classes the labels should be 0, 1, 2. The class probability
for class '1' will be found in index 1 in the class probability list
wrapped inside the zipped list with the labels.
Args:
data (list): A zipped list (NOT A GENERATOR) of the labels and the
class probabilities in the form (m = # data instances):
[(label1, [p(x1c1), p(x1c2), ... p(x1cn)]),
(label2, [p(x2c1), p(x2c2), ... p(x2cn)])
...
(labelm, [p(xmc1), p(xmc2), ... (pxmcn)])
]
num_classes (int): The number of classes in the dataset.
Returns:
The MAUC as a floating point value.
"""
# Find all pairwise comparisons of labels
class_pairs = [x for x in itertools.combinations(xrange(num_classes), 2)]
# Have to take average of A value with both classes acting as label 0 as this
# gives different outputs for more than 2 classes
sum_avals = 0
for pairing in class_pairs:
sum_avals += (a_value(data, zero_label=pairing[0], one_label=pairing[1]) +
a_value(data, zero_label=pairing[1], one_label=pairing[0])) / 2.0
return sum_avals * (2 / float(num_classes * (num_classes-1))) # Eqn 7
@stulacy
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stulacy commented Apr 14, 2015

An implementation of the MAUC measure from Hand & Till's 2001 paper "A Simple Generalisation of the Area Under the ROC Curve for Multiple Class Classification Problems". It has been implemented in R in the HandTill2001 package, as well as pROC but this is the first (that I can tell) readily available Python implementation.

@muhammedniyas
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My doubt is whether finding MAUC measure depends on the classifier you are using? Can you please clarify it?

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