OrgLion-ML
/
dlib
mirror of https://github.com/davisking/dlib.git
1
0
Fork 0
Code Issues Releases Wiki Activity

updated example program

This commit is contained in:
Davis King 2013-07-07 16:37:18 -04:00
parent d134e2ec94
commit aa46752d89
1 changed files with 223 additions and 32 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

255
python_examples/svm_struct.py
View File

@ -2,9 +2,11 @@
# 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. This example will briefly introduce it and then walk through an example showing
# how to use it to create a simple multi-class classifier.
#
# 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
@ -17,46 +19,244 @@
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[{}]: {}".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:
C = 10
be_verbose = True
epsilon = 0.0001
# 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() also expects the class to have num_samples
# and num_dimensions fields. These fields are expected to 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, vector, label):
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.
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(vector)
dims = len(x)
if (label == 0):
for i in range(0,dims):
psi[i] = vector[i]
psi[i] = x[i]
elif (label == 1):
for i in range(dims,2*dims):
psi[i] = vector[i-dims]
psi[i] = x[i-dims]
else: # the label must be 2
for i in range(2*dims,3*dims):
psi[i] = vector[i-2*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 be more precise, separation_oracle() 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 = samples[idx]
samp = self.samples[idx]
dims = len(samp)
scores = [0,0,0]
# compute scores for each of the three classifiers
@ -64,39 +264,30 @@ class three_class_classifier_problem:
scores[1] = dot(current_solution[dims:2*dims], samp)
scores[2] = dot(current_solution[2*dims:3*dims], samp)
# Add in the loss-augmentation
if (labels[idx] != 0):
# 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().
if (self.labels[idx] != 0):
scores[0] += 1
if (labels[idx] != 1):
if (self.labels[idx] != 1):
scores[1] += 1
if (labels[idx] != 2):
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))
if (max_scoring_label == labels[idx]):
# We incur a loss of 1 if we don't predict the correct label and a loss of 0 if we
# get the right answer.
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
samples = [[0,0,1], [0,1,0], [1,0,0]];
labels = [0,1,2]
problem = three_class_classifier_problem(samples, labels)
weights = dlib.solve_structural_svm_problem(problem)
print weights
w1 = weights[0:3]
w2 = weights[3:6]
w3 = weights[6:9]
print "scores for class 1 sample: ", dot(w1, samples[0]), dot(w2,samples[0]), dot(w3, samples[0])
print "scores for class 2 sample: ", dot(w1, samples[1]), dot(w2,samples[1]), dot(w3, samples[1])
print "scores for class 3 sample: ", dot(w1, samples[2]), dot(w2,samples[2]), dot(w3, samples[2])
if __name__ == "__main__":
main()