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diff --git a/ml/dlib/python_examples/svm_struct.py b/ml/dlib/python_examples/svm_struct.py new file mode 100755 index 000000000..7f0004ccd --- /dev/null +++ b/ml/dlib/python_examples/svm_struct.py @@ -0,0 +1,343 @@ +#!/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() |