dlib/python_examples/svm_struct.py

#!/usr/bin/python
# The contents of this file are in the public domain. See LICENSE_FOR_EXAMPLE_PROGRAMS.txt
#
# This is an example illustrating the use of the structural SVM solver from the dlib C++
# Library.  Therefore, this example teaches you the central ideas needed to setup a
# structural SVM model for your machine learning problems.  To illustrate the process, we
# use dlib's structural SVM solver to learn the parameters of a simple multi-class
# classifier.  We first discuss the multi-class classifier model and then walk through
# using the structural svm tools to find the parameters of this classification model.   
#
# COMPILING THE DLIB PYTHON INTERFACE
#   Dlib comes with a compiled python interface for python 2.7 on MS Windows.  If
#   you are using another python version or operating system then you need to
#   compile the dlib python interface before you can use this file.  To do this,
#   run compile_dlib_python_module.bat.  This should work on any operating system
#   so long as you have CMake and boost-python installed.  On Ubuntu, this can be
#   done easily by running the command:  sudo apt-get install libboost-python-dev cmake


import dlib


def main():
    # In this example, we have three types of samples: class 0, 1, or 2.  That is, each of
    # our sample vectors falls into one of three classes.  To keep this example very
    # simple, each sample vector is zero everywhere except at one place.  The non-zero
    # dimension of each vector determines the class of the vector.  So for example, the
    # first element of samples has a class of 1 because samples[0][1] is the only non-zero
    # element of samples[0].   
    samples = [[0,2,0], [1,0,0], [0,4,0], [0,0,3]];
    # Since we want to use a machine learning method to learn a 3-class classifier we need
    # to record the labels of our samples.  Here samples[i] has a class label of labels[i].
    labels =  [1,0,1,2]

    # Now that we have some training data we can tell the structural SVM to learn the
    # parameters of our 3-class classifier model.  The details of this will be explained
    # later.  For now, just note that it finds the weights (i.e. a vector of real valued
    # parameters) such that predict_label(weights, sample) always returns the correct label
    # for a sample vector. 
    problem = three_class_classifier_problem(samples, labels)
    weights = dlib.solve_structural_svm_problem(problem)

    # Print the weights and then evaluate predict_label() on each of our training samples.
    # Note that the correct label is predicted for each sample.
    print weights
    for i in range(len(samples)):
        print "predicted label for sample[{0}]: {1}".format(i, predict_label(weights, samples[i]))

def predict_label(weights, sample):
    """Given the 9-dimensional weight vector which defines a 3 class classifier, predict the
    class of the given 3-dimensional sample vector.   Therefore, the output of this
    function is either 0, 1, or 2 (i.e. one of the three possible labels)."""

    # Our 3-class classifier model can be thought of as containing 3 separate linear
    # classifiers.  So to predict the class of a sample vector we evaluate each of these
    # three classifiers and then whatever classifier has the largest output "wins" and
    # predicts the label of the sample.  This is the popular one-vs-all multi-class
    # classifier model.  
    #
    # Keeping this in mind, the code below simply pulls the three separate weight vectors
    # out of weights and then evaluates each against sample.  The individual classifier
    # scores are stored in scores and the highest scoring index is returned as the label.
    w0 = weights[0:3]
    w1 = weights[3:6]
    w2 = weights[6:9]
    scores = [dot(w0, sample), dot(w1,sample), dot(w2, sample)]
    max_scoring_label = scores.index(max(scores))
    return max_scoring_label

def dot(a, b):
    "Compute the dot product between the two vectors a and b."
    return sum(i*j for i,j in zip(a,b))

###########################################################################################

class three_class_classifier_problem:
    # Now we arrive at the meat of this example program.  To use the
    # dlib.solve_structural_svm_problem() routine you need to define an object which tells
    # the structural SVM solver what to do for your problem.  In this example, this is done
    # by defining the three_class_classifier_problem object.  Before we get into the
    # details, we first discuss some background information on structural SVMs.  
    # 
    # A structural SVM is a supervised machine learning method for learning to predict
    # complex outputs.  This is contrasted with a binary classifier which makes only simple
    # yes/no predictions.  A structural SVM, on the other hand, can learn to predict
    # complex outputs such as entire parse trees or DNA sequence alignments.  To do this,
    # it learns a function F(x,y) which measures how well a particular data sample x
    # matches a label y, where a label is potentially a complex thing like a parse tree.
    # However, to keep this example program simple we use only a 3 category label output. 
    #
    # At test time, the best label for a new x is given by the y which maximizes F(x,y).
    # To put this into the context of the current example, F(x,y) computes the score for a
    # given sample and class label.  The predicted class label is therefore whatever value
    # of y makes F(x,y) the biggest.  This is exactly what predict_label() does.  That is,
    # it computes F(x,0), F(x,1), and F(x,2) and then reports which label has the biggest
    # value.
    #
    # At a high level, a structural SVM can be thought of as searching the parameter space
    # of F(x,y) for the set of parameters that make the following inequality true as often
    # as possible:
    #     F(x_i,y_i) > max{over all incorrect labels of x_i} F(x_i, y_incorrect)
    # That is, it seeks to find the parameter vector such that F(x,y) always gives the
    # highest score to the correct output.  To define the structural SVM optimization
    # problem precisely, we first introduce some notation:
    #     - let PSI(x,y)    == the joint feature vector for input x and a label y.
    #     - let F(x,y|w)    == dot(w,PSI(x,y)).  
    #       (we use the | notation to emphasize that F() has the parameter vector of
    #       weights called w)
    #     - let LOSS(idx,y) == the loss incurred for predicting that the idx-th training 
    #       sample has a label of y.  Note that LOSS() should always be >= 0 and should
    #       become exactly 0 when y is the correct label for the idx-th sample.  Moreover,
    #       it should notionally indicate how bad it is to predict y for the idx'th sample.
    #     - let x_i == the i-th training sample.
    #     - let y_i == the correct label for the i-th training sample.
    #     - The number of data samples is N.
    #
    # Then the optimization problem solved by a structural SVM using
    # dlib.solve_structural_svm_problem() is the following:
    #     Minimize: h(w) == 0.5*dot(w,w) + C*R(w)
    #
    #     Where R(w) == sum from i=1 to N: 1/N * sample_risk(i,w)
    #     and sample_risk(i,w) == max over all Y: LOSS(i,Y) + F(x_i,Y|w) - F(x_i,y_i|w)
    #     and C > 0
    #
    # You can think of the sample_risk(i,w) as measuring the degree of error you would make
    # when predicting the label of the i-th sample using parameters w.  That is, it is zero
    # only when the correct label would be predicted and grows larger the more "wrong" the
    # predicted output becomes.  Therefore, the objective function is minimizing a balance
    # between making the weights small (typically this reduces overfitting) and fitting the
    # training data.  The degree to which you try to fit the data is controlled by the C
    # parameter.
    #
    # For a more detailed introduction to structured support vector machines you should
    # consult the following paper: 
    #     Predicting Structured Objects with Support Vector Machines by 
    #     Thorsten Joachims, Thomas Hofmann, Yisong Yue, and Chun-nam Yu
    #

    # Finally, we come back to the code.  To use dlib.solve_structural_svm_problem() you
    # need to provide the things discussed above.  This is the value of C, the number of
    # training samples, the dimensionality of PSI(), as well as methods for calculating the
    # loss values and PSI() vectors.  You will also need to write code that can compute:
    # max over all Y: LOSS(i,Y) + F(x_i,Y|w).  To summarize, the
    # three_class_classifier_problem class is required to have the following fields:
    #   - C
    #   - num_samples
    #   - num_dimensions
    #   - get_truth_joint_feature_vector()
    #   - separation_oracle()

    C = 1

    # There are also a number of optional arguments:
    # epsilon is the stopping tolerance.  The optimizer will run until R(w) is within
    # epsilon of its optimal value. If you don't set this then it defaults to 0.001.
    #epsilon = 1e-13                     

    # Uncomment this and the optimizer will print its progress to standard out.  You will
    # be able to see things like the current risk gap.  The optimizer continues until the
    # risk gap is below epsilon.
    #be_verbose = True

    # If you want to require that the learned weights are all non-negative then set this
    # field to True.
    #learns_nonnegative_weights = True

    # The optimizer uses an internal cache to avoid unnecessary calls to your
    # separation_oracle() routine.  This parameter controls the size of that cache.  Bigger
    # values use more RAM and might make the optimizer run faster.  You can also disable it
    # by setting it to 0 which is good to do when your separation_oracle is very fast.
    #max_cache_size = 20


    def __init__(self, samples, labels):
        # dlib.solve_structural_svm_problem() expects the class to have num_samples and
        # num_dimensions fields.  These fields should contain the number of training
        # samples and the dimensionality of the PSI feature vector respectively.
        self.num_samples = len(samples)
        self.num_dimensions = len(samples[0])*3

        self.samples = samples
        self.labels = labels


    def make_psi(self, x, label):
        """Compute PSI(x,label)."""
        # All we are doing here is taking x, which is a 3 dimensional sample vector in this
        # example program, and putting it into one of 3 places in a 9 dimensional PSI
        # vector, which we then return.  So this function returns PSI(x,label).  To see why
        # we setup PSI like this, recall how predict_label() works.  It takes in a 9
        # dimensional weight vector and breaks the vector into 3 pieces.  Each piece then
        # defines a different classifier and we use them in a one-vs-all manner to predict
        # the label.  So now that we are in the structural SVM code we have to define the
        # PSI vector to correspond to this usage.  That is, we need to setup PSI so that
        # argmax_y dot(weights,PSI(x,y)) == predict_label(weights,x).  This is how we tell
        # the structural SVM solver what kind of problem we are trying to solve.
        #
        # It's worth emphasizing that the single biggest step in using a structural SVM is
        # deciding how you want to represent PSI(x,label).  It is always a vector, but
        # deciding what to put into it to solve your problem is often not a trivial task.
        # Part of the difficulty is that you need an efficient method for finding the label
        # that makes dot(w,PSI(x,label)) the biggest.  Sometimes this is easy, but often
        # finding the max scoring label turns into a difficult combinatorial optimization
        # problem.  So you need to pick a PSI that doesn't make the label maximization step
        # intractable but also still well models your problem.  

        # Create a dense vector object (note that you can also use unsorted sparse vectors
        # (i.e.  dlib.sparse_vector objects) to represent your PSI vector.  This is useful
        # if you have very high dimensional PSI vectors that are mostly zeros.  In the
        # context of this example, you would simply return a dlib.sparse_vector at the end
        # of make_psi() and the rest of the example would still work properly. ).
        psi = dlib.vector()
        # Set it to have 9 dimensions.  Note that the elements of the vector are 0
        # initialized.
        psi.resize(self.num_dimensions)
        dims = len(x)
        if (label == 0):
            for i in range(0,dims):
                psi[i] = x[i]
        elif (label == 1):
            for i in range(dims,2*dims):
                psi[i] = x[i-dims]
        else: # the label must be 2
            for i in range(2*dims,3*dims):
                psi[i] = x[i-2*dims]
        return psi


    # Now we get to the two member functions that are directly called by
    # dlib.solve_structural_svm_problem().  
    #
    # In get_truth_joint_feature_vector(), all you have to do is return the PSI() vector
    # for the idx-th training sample when it has its true label.  So here it returns
    # PSI(self.samples[idx], self.labels[idx]).
    def get_truth_joint_feature_vector(self, idx):
        return self.make_psi(self.samples[idx], self.labels[idx])


    # separation_oracle() is more interesting.  dlib.solve_structural_svm_problem() will
    # call separation_oracle() many times during the optimization.  Each time it will give
    # it the current value of the parameter weights and the separation_oracle() is supposed
    # to find the label that most violates the structural SVM objective function for the
    # idx-th sample.  Then the separation oracle reports the corresponding PSI vector and
    # loss value.  To state this more precisely, the separation_oracle() member function
    # has the following contract:
    #   requires
    #       - 0 <= idx < self.num_samples 
    #       - len(current_solution) == self.num_dimensions 
    #   ensures
    #       - runs the separation oracle on the idx-th sample.  We define this as follows: 
    #           - let X           == the idx-th training sample.
    #           - let PSI(X,y)    == the joint feature vector for input X and an arbitrary label y.
    #           - let F(X,y)      == dot(current_solution,PSI(X,y)).  
    #           - let LOSS(idx,y) == the loss incurred for predicting that the idx-th sample
    #             has a label of y.  Note that LOSS() should always be >= 0 and should
    #             become exactly 0 when y is the correct label for the idx-th sample.
    #  
    #               Then the separation oracle finds a Y such that: 
    #                   Y = argmax over all y: LOSS(idx,y) + F(X,y) 
    #                   (i.e. It finds the label which maximizes the above expression.)
    #  
    #               Finally, separation_oracle() returns LOSS(idx,Y),PSI(X,Y) 
    def separation_oracle(self, idx, current_solution):
        samp = self.samples[idx]
        dims = len(samp)
        scores = [0,0,0]
        # compute scores for each of the three classifiers
        scores[0] = dot(current_solution[0:dims], samp)
        scores[1] = dot(current_solution[dims:2*dims], samp)
        scores[2] = dot(current_solution[2*dims:3*dims], samp)

        # Add in the loss-augmentation.  Recall that we maximize LOSS(idx,y) + F(X,y) in
        # the separate oracle, not just F(X,y) as we normally would in predict_label().
        # Therefore, we must add in this extra amount to account for the loss-augmentation.
        # For our simple multi-class classifier, we incur a loss of 1 if we don't predict
        # the correct label and a loss of 0 if we get the right label.
        if (self.labels[idx] != 0): 
            scores[0] += 1
        if (self.labels[idx] != 1): 
            scores[1] += 1
        if (self.labels[idx] != 2): 
            scores[2] += 1

        # Now figure out which classifier has the largest loss-augmented score.
        max_scoring_label = scores.index(max(scores))
        # And finally record the loss that was associated with that predicted label.
        # Again, the loss is 1 if the label is incorrect and 0 otherwise.
        if (max_scoring_label == self.labels[idx]):
            loss = 0
        else:
            loss = 1

        # Finally, return the loss and PSI vector corresponding to the label we just found.
        psi = self.make_psi(samp, max_scoring_label)
        return loss,psi



if __name__ == "__main__":
    main()