#!/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. As an aside, # dlib's C++ interface to the structural SVM solver is threaded. So on a # multi-core computer it is significantly faster than using the python # interface. So consider using the C++ interface instead if you find that # running it in python is slow. # # # COMPILING/INSTALLING THE DLIB PYTHON INTERFACE # You can install dlib using the command: # pip install dlib # # Alternatively, if you want to compile dlib yourself then go into the dlib # root folder and run: # python setup.py install # or # python setup.py install --yes USE_AVX_INSTRUCTIONS # if you have a CPU that supports AVX instructions, since this makes some # things run faster. # # Compiling dlib should work on any operating system so long as you have # CMake installed. On Ubuntu, this can be done easily by running the # command: # sudo apt-get install 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 = ThreeClassClassifierProblem(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 k, s in enumerate(samples): print("Predicted label for sample[{0}]: {1}".format( k, predict_label(weights, s))) 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 ThreeClassClassifierProblem: # 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 ThreeClassClassifierProblem # 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 which 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 # ThreeClassClassifierProblem 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. If If you don't call this # function it defaults to a value of 5. # 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()