mirror of https://github.com/davisking/dlib.git
303 lines
16 KiB
Python
Executable File
303 lines
16 KiB
Python
Executable File
#!/usr/bin/python
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# The contents of this file are in the public domain. See LICENSE_FOR_EXAMPLE_PROGRAMS.txt
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#
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# This is an example illustrating the use of the structural SVM solver from the dlib C++
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# Library. Therefore, this example teaches you the central ideas needed to setup a
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# structural SVM model for your machine learning problems. To illustrate the process, we
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# use dlib's structural SVM solver to learn the parameters of a simple multi-class
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# classifier. We first discuss the multi-class classifier model and then walk through
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# using the structural SVM tools to find the parameters of this classification model.
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#
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# COMPILING THE DLIB PYTHON INTERFACE
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# Dlib comes with a compiled python interface for python 2.7 on MS Windows. If
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# you are using another python version or operating system then you need to
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# compile the dlib python interface before you can use this file. To do this,
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# run compile_dlib_python_module.bat. This should work on any operating system
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# so long as you have CMake and boost-python installed. On Ubuntu, this can be
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# done easily by running the command: sudo apt-get install libboost-python-dev cmake
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import dlib
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def main():
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# In this example, we have three types of samples: class 0, 1, or 2. That is, each of
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# our sample vectors falls into one of three classes. To keep this example very
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# simple, each sample vector is zero everywhere except at one place. The non-zero
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# dimension of each vector determines the class of the vector. So for example, the
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# first element of samples has a class of 1 because samples[0][1] is the only non-zero
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# element of samples[0].
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samples = [[0,2,0], [1,0,0], [0,4,0], [0,0,3]];
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# Since we want to use a machine learning method to learn a 3-class classifier we need
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# to record the labels of our samples. Here samples[i] has a class label of labels[i].
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labels = [1,0,1,2]
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# Now that we have some training data we can tell the structural SVM to learn the
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# parameters of our 3-class classifier model. The details of this will be explained
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# later. For now, just note that it finds the weights (i.e. a vector of real valued
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# parameters) such that predict_label(weights, sample) always returns the correct label
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# for a sample vector.
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problem = three_class_classifier_problem(samples, labels)
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weights = dlib.solve_structural_svm_problem(problem)
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# Print the weights and then evaluate predict_label() on each of our training samples.
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# Note that the correct label is predicted for each sample.
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print weights
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for i in range(len(samples)):
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print "predicted label for sample[{0}]: {1}".format(i, predict_label(weights, samples[i]))
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def predict_label(weights, sample):
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"""Given the 9-dimensional weight vector which defines a 3 class classifier, predict the
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class of the given 3-dimensional sample vector. Therefore, the output of this
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function is either 0, 1, or 2 (i.e. one of the three possible labels)."""
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# Our 3-class classifier model can be thought of as containing 3 separate linear
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# classifiers. So to predict the class of a sample vector we evaluate each of these
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# three classifiers and then whatever classifier has the largest output "wins" and
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# predicts the label of the sample. This is the popular one-vs-all multi-class
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# classifier model.
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#
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# Keeping this in mind, the code below simply pulls the three separate weight vectors
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# out of weights and then evaluates each against sample. The individual classifier
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# scores are stored in scores and the highest scoring index is returned as the label.
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w0 = weights[0:3]
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w1 = weights[3:6]
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w2 = weights[6:9]
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scores = [dot(w0, sample), dot(w1,sample), dot(w2, sample)]
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max_scoring_label = scores.index(max(scores))
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return max_scoring_label
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def dot(a, b):
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"Compute the dot product between the two vectors a and b."
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return sum(i*j for i,j in zip(a,b))
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###########################################################################################
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class three_class_classifier_problem:
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# Now we arrive at the meat of this example program. To use the
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# dlib.solve_structural_svm_problem() routine you need to define an object which tells
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# the structural SVM solver what to do for your problem. In this example, this is done
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# by defining the three_class_classifier_problem object. Before we get into the
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# details, we first discuss some background information on structural SVMs.
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#
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# A structural SVM is a supervised machine learning method for learning to predict
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# complex outputs. This is contrasted with a binary classifier which makes only simple
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# yes/no predictions. A structural SVM, on the other hand, can learn to predict
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# complex outputs such as entire parse trees or DNA sequence alignments. To do this,
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# it learns a function F(x,y) which measures how well a particular data sample x
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# matches a label y, where a label is potentially a complex thing like a parse tree.
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# However, to keep this example program simple we use only a 3 category label output.
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#
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# At test time, the best label for a new x is given by the y which maximizes F(x,y).
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# To put this into the context of the current example, F(x,y) computes the score for a
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# given sample and class label. The predicted class label is therefore whatever value
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# of y which makes F(x,y) the biggest. This is exactly what predict_label() does.
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# That is, it computes F(x,0), F(x,1), and F(x,2) and then reports which label has the
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# biggest value.
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#
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# At a high level, a structural SVM can be thought of as searching the parameter space
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# of F(x,y) for the set of parameters that make the following inequality true as often
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# as possible:
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# F(x_i,y_i) > max{over all incorrect labels of x_i} F(x_i, y_incorrect)
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# That is, it seeks to find the parameter vector such that F(x,y) always gives the
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# highest score to the correct output. To define the structural SVM optimization
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# problem precisely, we first introduce some notation:
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# - let PSI(x,y) == the joint feature vector for input x and a label y.
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# - let F(x,y|w) == dot(w,PSI(x,y)).
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# (we use the | notation to emphasize that F() has the parameter vector of
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# weights called w)
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# - let LOSS(idx,y) == the loss incurred for predicting that the idx-th training
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# sample has a label of y. Note that LOSS() should always be >= 0 and should
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# become exactly 0 when y is the correct label for the idx-th sample. Moreover,
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# it should notionally indicate how bad it is to predict y for the idx'th sample.
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# - let x_i == the i-th training sample.
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# - let y_i == the correct label for the i-th training sample.
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# - The number of data samples is N.
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#
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# Then the optimization problem solved by a structural SVM using
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# dlib.solve_structural_svm_problem() is the following:
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# Minimize: h(w) == 0.5*dot(w,w) + C*R(w)
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#
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# Where R(w) == sum from i=1 to N: 1/N * sample_risk(i,w)
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# and sample_risk(i,w) == max over all Y: LOSS(i,Y) + F(x_i,Y|w) - F(x_i,y_i|w)
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# and C > 0
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#
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# You can think of the sample_risk(i,w) as measuring the degree of error you would make
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# when predicting the label of the i-th sample using parameters w. That is, it is zero
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# only when the correct label would be predicted and grows larger the more "wrong" the
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# predicted output becomes. Therefore, the objective function is minimizing a balance
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# between making the weights small (typically this reduces overfitting) and fitting the
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# training data. The degree to which you try to fit the data is controlled by the C
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# parameter.
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#
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# For a more detailed introduction to structured support vector machines you should
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# consult the following paper:
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# Predicting Structured Objects with Support Vector Machines by
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# Thorsten Joachims, Thomas Hofmann, Yisong Yue, and Chun-nam Yu
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#
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# Finally, we come back to the code. To use dlib.solve_structural_svm_problem() you
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# need to provide the things discussed above. This is the value of C, the number of
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# training samples, the dimensionality of PSI(), as well as methods for calculating the
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# loss values and PSI() vectors. You will also need to write code that can compute:
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# max over all Y: LOSS(i,Y) + F(x_i,Y|w). To summarize, the
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# three_class_classifier_problem class is required to have the following fields:
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# - C
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# - num_samples
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# - num_dimensions
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# - get_truth_joint_feature_vector()
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# - separation_oracle()
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C = 1
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# There are also a number of optional arguments:
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# epsilon is the stopping tolerance. The optimizer will run until R(w) is within
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# epsilon of its optimal value. If you don't set this then it defaults to 0.001.
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#epsilon = 1e-13
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# Uncomment this and the optimizer will print its progress to standard out. You will
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# be able to see things like the current risk gap. The optimizer continues until the
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# risk gap is below epsilon.
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#be_verbose = True
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# If you want to require that the learned weights are all non-negative then set this
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# field to True.
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#learns_nonnegative_weights = True
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# The optimizer uses an internal cache to avoid unnecessary calls to your
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# separation_oracle() routine. This parameter controls the size of that cache. Bigger
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# values use more RAM and might make the optimizer run faster. You can also disable it
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# by setting it to 0 which is good to do when your separation_oracle is very fast. If
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# If you don't call this function it defaults to a value of 5.
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#max_cache_size = 20
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def __init__(self, samples, labels):
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# dlib.solve_structural_svm_problem() expects the class to have num_samples and
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# num_dimensions fields. These fields should contain the number of training
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# samples and the dimensionality of the PSI feature vector respectively.
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self.num_samples = len(samples)
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self.num_dimensions = len(samples[0])*3
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self.samples = samples
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self.labels = labels
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def make_psi(self, x, label):
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"""Compute PSI(x,label)."""
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# All we are doing here is taking x, which is a 3 dimensional sample vector in this
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# example program, and putting it into one of 3 places in a 9 dimensional PSI
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# vector, which we then return. So this function returns PSI(x,label). To see why
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# we setup PSI like this, recall how predict_label() works. It takes in a 9
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# dimensional weight vector and breaks the vector into 3 pieces. Each piece then
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# defines a different classifier and we use them in a one-vs-all manner to predict
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# the label. So now that we are in the structural SVM code we have to define the
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# PSI vector to correspond to this usage. That is, we need to setup PSI so that
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# argmax_y dot(weights,PSI(x,y)) == predict_label(weights,x). This is how we tell
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# the structural SVM solver what kind of problem we are trying to solve.
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#
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# It's worth emphasizing that the single biggest step in using a structural SVM is
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# deciding how you want to represent PSI(x,label). It is always a vector, but
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# deciding what to put into it to solve your problem is often not a trivial task.
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# Part of the difficulty is that you need an efficient method for finding the label
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# that makes dot(w,PSI(x,label)) the biggest. Sometimes this is easy, but often
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# finding the max scoring label turns into a difficult combinatorial optimization
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# problem. So you need to pick a PSI that doesn't make the label maximization step
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# intractable but also still well models your problem.
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# Create a dense vector object (note that you can also use unsorted sparse vectors
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# (i.e. dlib.sparse_vector objects) to represent your PSI vector. This is useful
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# if you have very high dimensional PSI vectors that are mostly zeros. In the
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# context of this example, you would simply return a dlib.sparse_vector at the end
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# of make_psi() and the rest of the example would still work properly. ).
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psi = dlib.vector()
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# Set it to have 9 dimensions. Note that the elements of the vector are 0
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# initialized.
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psi.resize(self.num_dimensions)
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dims = len(x)
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if (label == 0):
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for i in range(0,dims):
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psi[i] = x[i]
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elif (label == 1):
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for i in range(dims,2*dims):
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psi[i] = x[i-dims]
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else: # the label must be 2
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for i in range(2*dims,3*dims):
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psi[i] = x[i-2*dims]
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return psi
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# Now we get to the two member functions that are directly called by
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# dlib.solve_structural_svm_problem().
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#
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# In get_truth_joint_feature_vector(), all you have to do is return the PSI() vector
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# for the idx-th training sample when it has its true label. So here it returns
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# PSI(self.samples[idx], self.labels[idx]).
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def get_truth_joint_feature_vector(self, idx):
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return self.make_psi(self.samples[idx], self.labels[idx])
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# separation_oracle() is more interesting. dlib.solve_structural_svm_problem() will
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# call separation_oracle() many times during the optimization. Each time it will give
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# it the current value of the parameter weights and the separation_oracle() is supposed
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# to find the label that most violates the structural SVM objective function for the
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# idx-th sample. Then the separation oracle reports the corresponding PSI vector and
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# loss value. To state this more precisely, the separation_oracle() member function
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# has the following contract:
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# requires
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# - 0 <= idx < self.num_samples
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# - len(current_solution) == self.num_dimensions
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# ensures
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# - runs the separation oracle on the idx-th sample. We define this as follows:
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# - let X == the idx-th training sample.
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# - let PSI(X,y) == the joint feature vector for input X and an arbitrary label y.
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# - let F(X,y) == dot(current_solution,PSI(X,y)).
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# - let LOSS(idx,y) == the loss incurred for predicting that the idx-th sample
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# has a label of y. Note that LOSS() should always be >= 0 and should
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# become exactly 0 when y is the correct label for the idx-th sample.
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#
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# Then the separation oracle finds a Y such that:
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# Y = argmax over all y: LOSS(idx,y) + F(X,y)
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# (i.e. It finds the label which maximizes the above expression.)
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#
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# Finally, separation_oracle() returns LOSS(idx,Y),PSI(X,Y)
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def separation_oracle(self, idx, current_solution):
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samp = self.samples[idx]
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dims = len(samp)
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scores = [0,0,0]
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# compute scores for each of the three classifiers
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scores[0] = dot(current_solution[0:dims], samp)
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scores[1] = dot(current_solution[dims:2*dims], samp)
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scores[2] = dot(current_solution[2*dims:3*dims], samp)
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# Add in the loss-augmentation. Recall that we maximize LOSS(idx,y) + F(X,y) in
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# the separate oracle, not just F(X,y) as we normally would in predict_label().
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# Therefore, we must add in this extra amount to account for the loss-augmentation.
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# For our simple multi-class classifier, we incur a loss of 1 if we don't predict
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# the correct label and a loss of 0 if we get the right label.
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if (self.labels[idx] != 0):
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scores[0] += 1
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if (self.labels[idx] != 1):
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scores[1] += 1
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if (self.labels[idx] != 2):
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scores[2] += 1
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# Now figure out which classifier has the largest loss-augmented score.
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max_scoring_label = scores.index(max(scores))
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# And finally record the loss that was associated with that predicted label.
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# Again, the loss is 1 if the label is incorrect and 0 otherwise.
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if (max_scoring_label == self.labels[idx]):
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loss = 0
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else:
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loss = 1
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# Finally, return the loss and PSI vector corresponding to the label we just found.
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psi = self.make_psi(samp, max_scoring_label)
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return loss,psi
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if __name__ == "__main__":
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main()
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