mirror of https://github.com/davisking/dlib.git
344 lines
17 KiB
Python
Executable File
344 lines
17 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
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# the dlib C++ Library. Therefore, this example teaches you the central ideas
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# needed to setup a structural SVM model for your machine learning problems. To
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# illustrate the process, we use dlib's structural SVM solver to learn the
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# parameters of a simple multi-class classifier. We first discuss the
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# multi-class classifier model and then walk through using the structural SVM
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# tools to find the parameters of this classification model. As an aside,
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# dlib's C++ interface to the structural SVM solver is threaded. So on a
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# multi-core computer it is significantly faster than using the python
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# interface. So consider using the C++ interface instead if you find that
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# running it in python is slow.
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#
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#
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# COMPILING/INSTALLING THE DLIB PYTHON INTERFACE
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# You can install dlib using the command:
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# pip install dlib
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#
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# Alternatively, if you want to compile dlib yourself then go into the dlib
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# root folder and run:
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# python setup.py install
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# or
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# python setup.py install --yes USE_AVX_INSTRUCTIONS
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# if you have a CPU that supports AVX instructions, since this makes some
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# things run faster.
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#
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# Compiling dlib should work on any operating system so long as you have
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# CMake and boost-python installed. On Ubuntu, this can be done easily by
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# running the command:
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# sudo apt-get install libboost-python-dev cmake
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#
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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
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# is, each of our sample vectors falls into one of three classes. To keep
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# this example very simple, each sample vector is zero everywhere except at
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# one place. The non-zero dimension of each vector determines the class of
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# the vector. So for example, the first element of samples has a class of 1
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# because samples[0][1] is the only non-zero 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
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# classifier we need to record the labels of our samples. Here samples[i]
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# 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
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# learn the parameters of our 3-class classifier model. The details of this
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# will be explained later. For now, just note that it finds the weights
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# (i.e. a vector of real valued parameters) such that predict_label(weights,
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# sample) always returns the correct label for a sample vector.
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problem = ThreeClassClassifierProblem(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
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# training samples. Note that the correct label is predicted for each
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# sample.
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print(weights)
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for k, s in enumerate(samples):
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print("Predicted label for sample[{0}]: {1}".format(
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k, predict_label(weights, s)))
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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,
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predict the class of the given 3-dimensional sample vector. Therefore, the
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output of this function is either 0, 1, or 2 (i.e. one of the three possible
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labels)."""
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# Our 3-class classifier model can be thought of as containing 3 separate
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# linear classifiers. So to predict the class of a sample vector we
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# evaluate each of these three classifiers and then whatever classifier has
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# the largest output "wins" and predicts the label of the sample. This is
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# the popular one-vs-all multi-class classifier model.
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# Keeping this in mind, the code below simply pulls the three separate
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# weight vectors out of weights and then evaluates each against sample. The
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# individual classifier scores are stored in scores and the highest scoring
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# 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 ThreeClassClassifierProblem:
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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
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# which tells the structural SVM solver what to do for your problem. In
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# this example, this is done by defining the ThreeClassClassifierProblem
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# object. Before we get into the details, we first discuss some background
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# information on structural SVMs.
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#
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# A structural SVM is a supervised machine learning method for learning to
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# predict complex outputs. This is contrasted with a binary classifier
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# which makes only simple yes/no predictions. A structural SVM, on the
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# other hand, can learn to predict complex outputs such as entire parse
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# trees or DNA sequence alignments. To do this, it learns a function F(x,y)
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# which measures how well a particular data sample x matches a label y,
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# where a label is potentially a complex thing like a parse tree. However,
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# 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
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# maximizes F(x,y). To put this into the context of the current example,
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# F(x,y) computes the score for a given sample and class label. The
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# predicted class label is therefore whatever value of y which makes F(x,y)
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# the biggest. This is exactly what predict_label() does. That is, it
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# 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
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# parameter space of F(x,y) for the set of parameters that make the
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# following inequality true as often 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
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# gives the highest score to the correct output. To define the structural
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# SVM optimization 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
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# of weights called w)
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# - let LOSS(idx,y) == the loss incurred for predicting that the
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# idx-th training sample has a label of y. Note that LOSS()
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# should always be >= 0 and should become exactly 0 when y is the
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# correct label for the idx-th sample. Moreover, it should notionally
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# 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) and
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# sample_risk(i,w) == max over all
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# Y: LOSS(i,Y) + F(x_i,Y|w) - F(x_i,y_i|w) 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
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# you would make when predicting the label of the i-th sample using
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# parameters w. That is, it is zero only when the correct label would be
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# predicted and grows larger the more "wrong" the predicted output becomes.
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# Therefore, the objective function is minimizing a balance between making
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# 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
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# by the C parameter.
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#
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# For a more detailed introduction to structured support vector machines
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# you should 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
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# dlib.solve_structural_svm_problem() you need to provide the things
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# discussed above. This is the value of C, the number of training samples,
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# the dimensionality of PSI(), as well as methods for calculating the loss
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# values and PSI() vectors. You will also need to write code that can
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# compute:
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# max over all Y: LOSS(i,Y) + F(x_i,Y|w). To summarize, the
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# ThreeClassClassifierProblem class is required to have the following
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# 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
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# within epsilon of its optimal value. If you don't set this then it
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# 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
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# out. You will be able to see things like the current risk gap. The
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# 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
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# then set this 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
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# cache. Bigger values use more RAM and might make the optimizer run
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# faster. You can also disable it by setting it to 0 which is good to do
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# when your separation_oracle is very fast. If If you don't call this
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# 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
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# num_samples and num_dimensions fields. These fields should contain
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# the number of training samples and the dimensionality of the PSI
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# 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
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# vector in this example program, and putting it into one of 3 places in
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# a 9 dimensional PSI vector, which we then return. So this function
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# returns PSI(x,label). To see why we setup PSI like this, recall how
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# predict_label() works. It takes in a 9 dimensional weight vector and
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# breaks the vector into 3 pieces. Each piece then defines a different
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# classifier and we use them in a one-vs-all manner to predict the
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# label. So now that we are in the structural SVM code we have to
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# define the PSI vector to correspond to this usage. That is, we need
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# to setup PSI so that argmax_y dot(weights,PSI(x,y)) ==
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# predict_label(weights,x). This is how we tell the structural SVM
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# 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
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# structural SVM is deciding how you want to represent PSI(x,label). It
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# is always a vector, but deciding what to put into it to solve your
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# problem is often not a trivial task. Part of the difficulty is that
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# you need an efficient method for finding the label that makes
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# 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
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# optimization problem. So you need to pick a PSI that doesn't make the
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# label maximization step intractable but also still well models your
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# problem.
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#
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# Create a dense vector object (note that you can also use unsorted
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# sparse vectors (i.e. dlib.sparse_vector objects) to represent your
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# PSI vector. This is useful if you have very high dimensional PSI
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# vectors that are mostly zeros. In the context of this example, you
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# would simply return a dlib.sparse_vector at the end of make_psi() and
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# 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
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# are 0 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
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# PSI() vector for the idx-th training sample when it has its true label.
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# 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.
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# dlib.solve_structural_svm_problem() will call separation_oracle() many
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# times during the optimization. Each time it will give it the current
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# value of the parameter weights and the separation_oracle() is supposed to
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# find the label that most violates the structural SVM objective function
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# for the idx-th sample. Then the separation oracle reports the
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# corresponding PSI vector and loss value. To state this more precisely,
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# the separation_oracle() member function 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.
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# 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
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# 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
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# idx-th sample has a label of y. Note that LOSS()
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# should always be >= 0 and should become exactly 0 when y is the
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# 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
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# LOSS(idx,y) + F(X,y) in the separate oracle, not just F(X,y) as we
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# normally would in predict_label(). Therefore, we must add in this
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# extra amount to account for the loss-augmentation. For our simple
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# multi-class classifier, we incur a loss of 1 if we don't predict the
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# 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
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# label. 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
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# 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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