#!/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()