Complete and hand in this completed worksheet (including its outputs and any supporting code outside of the worksheet) with your assignment submission. For more details see the assignments page on the course website.
The kNN classifier consists of two stages:
In this exercise you will implement these steps and understand the basic Image Classification pipeline, cross-validation, and gain proficiency in writing efficient, vectorized code.
In [2]:
# Run some setup code for this notebook.
import sys
import os
sys.path.append('..')
import random
import numpy as np
from algorithms.data_utils import load_CIFAR10
import matplotlib.pyplot as plt
# This is a bit of magic to make matplotlib figures appear inline in the notebook
# rather than in a new window.
%matplotlib inline
plt.rcParams['figure.figsize'] = (10.0, 8.0) # set default size of plots
plt.rcParams['image.interpolation'] = 'nearest'
plt.rcParams['image.cmap'] = 'gray'
# Some more magic so that the notebook will reload external python modules;
# see http://stackoverflow.com/questions/1907993/autoreload-of-modules-in-ipython
%load_ext autoreload
%autoreload 2
In [4]:
# Load the raw CIFAR-10 data.
cifar10_dir = '../datasets/cifar-10-batches-py'
X_train, y_train, X_test, y_test = load_CIFAR10(cifar10_dir)
# As a sanity check, we print out the size of the training and test data.
print 'Training data shape: ', X_train.shape
print 'Training labels shape: ', y_train.shape
print 'Test data shape: ', X_test.shape
print 'Test labels shape: ', y_test.shape
In [5]:
plt.imshow(X_train[1].astype('uint8'))
Out[5]:
In [6]:
# Visualize some examples from the dataset.
# We show a few examples of training images from each class.
classes = ['plane', 'car', 'bird', 'cat', 'deer', 'dog', 'frog', 'horse', 'ship', 'truck']
num_classes = len(classes)
samples_per_class = 7
for y, cls in enumerate(classes):
idxs = np.flatnonzero(y_train == y)
idxs = np.random.choice(idxs, samples_per_class, replace=False)
for i, idx in enumerate(idxs):
plt_idx = i * num_classes + y + 1
plt.subplot(samples_per_class, num_classes, plt_idx)
plt.imshow(X_train[idx].astype('uint8'))
plt.axis('off')
if i == 0:
plt.title(cls)
plt.show()
In [7]:
# Subsample the data for more efficient code execution in this exercise
num_training = 5000
mask = range(num_training)
X_train = X_train[mask]
y_train = y_train[mask]
num_test = 500
mask = range(num_test)
X_test = X_test[mask]
y_test = y_test[mask]
In [8]:
# Reshape the image data into rows
X_train = np.reshape(X_train, (X_train.shape[0], -1))
X_test = np.reshape(X_test, (X_test.shape[0], -1))
print X_train.shape, X_test.shape
print(X_train[0])
In [11]:
from algorithms.k_nearest_neighbor import KNearestNeighbor
# Create a kNN classifier instance.
# Remember that training a kNN classifier is a loop:
# the Classifier simply remembers the data and does no further processing
classifier = KNearestNeighbor()
classifier.train(X_train, y_train)
We would now like to classify the test data with the kNN classifier. Recall that we can break down this process into two steps:
Lets begin with computing the distance matrix between all training and test examples. For example, if there are Ntr training examples and Nte test examples, this stage should result in a Nte x Ntr matrix where each element (i,j) is the distance between the i-th test and j-th train example.
First, open algorithms/k_nearest_neighbor.py and implement the function compute_distances_two_loops that uses a (very inefficient) double loop over all pairs of (test, train) examples and computes the distance matrix one element at a time.
In [12]:
# Open cs231n/classifiers/k_nearest_neighbor.py and implement
# compute_distances_two_loops.
# Test your implementation:
dists = classifier.compute_distances_two_loops(X_test)
print dists.shape
In [13]:
# We can visualize the distance matrix: each row is a single test example and
# its distances to training examples
plt.imshow(dists, interpolation='none')
plt.show()
In [14]:
# Now implement the function predict_labels and run the code below:
# We use k = 1 (which is Nearest Neighbor).
y_test_pred = classifier.predict_labels(dists, k=1)
# Compute and print the fraction of correctly predicted examples
num_correct = np.sum(y_test_pred == y_test)
accuracy = float(num_correct) / num_test
print 'Got %d / %d correct => accuracy: %f' % (num_correct, num_test, accuracy)
You should expect to see approximately 27% accuracy. Now lets try out a larger k, say k = 5:
In [15]:
y_test_pred = classifier.predict_labels(dists, k=5)
num_correct = np.sum(y_test_pred == y_test)
accuracy = float(num_correct) / num_test
print 'Got %d / %d correct => accuracy: %f' % (num_correct, num_test, accuracy)
You should expect to see a slightly better performance than with k = 1.
In [16]:
# Now lets speed up distance matrix computation by using partial vectorization
# with one loop. Implement the function compute_distances_one_loop and run the
# code below:
dists_one = classifier.compute_distances_one_loop(X_test)
# To ensure that our vectorized implementation is correct, we make sure that it
# agrees with the naive implementation. There are many ways to decide whether
# two matrices are similar; one of the simplest is the Frobenius norm. In case
# you haven't seen it before, the Frobenius norm of two matrices is the square
# root of the squared sum of differences of all elements; in other words, reshape
# the matrices into vectors and compute the Euclidean distance between them.
difference = np.linalg.norm(dists - dists_one, ord='fro')
print 'Difference was: %f' % (difference, )
if difference < 0.001:
print 'Good! The distance matrices are the same'
else:
print 'Uh-oh! The distance matrices are different'
In [17]:
# Now implement the fully vectorized version inside compute_distances_no_loops
# and run the code
dists_two = classifier.compute_distances_no_loops(X_test)
# check that the distance matrix agrees with the one we computed before:
difference = np.linalg.norm(dists - dists_two, ord='fro')
print 'Difference was: %f' % (difference, )
if difference < 0.001:
print 'Good! The distance matrices are the same'
else:
print 'Uh-oh! The distance matrices are different'
In [18]:
# Let's compare how fast the implementations are
def time_function(f, *args):
"""
Call a function f with args and return the time (in seconds) that it took to execute.
"""
import time
tic = time.time()
f(*args)
toc = time.time()
return toc - tic
two_loop_time = time_function(classifier.compute_distances_two_loops, X_test)
print 'Two loop version took %f seconds' % two_loop_time
one_loop_time = time_function(classifier.compute_distances_one_loop, X_test)
print 'One loop version took %f seconds' % one_loop_time
no_loop_time = time_function(classifier.compute_distances_no_loops, X_test)
print 'No loop version took %f seconds' % no_loop_time
# you should see significantly faster performance with the fully vectorized implementation
In [19]:
print(X_train.shape)
# classifier = KNearestNeighbor()
# classifier.train(X_train, y_train)
# dists = classifier.compute_distances_no_loops(X_test)
# y_test_pred = classifier.predict_labels(dists, k=5)
# num_correct = np.sum(y_test_pred == y_test)
# accuracy = float(num_correct) / num_test
# print 'Got %d / %d correct => accuracy: %f' % (num_correct, num_test, accuracy)
In [21]:
num_folds = 5
k_choices = [1, 3, 5, 8, 10, 12, 15, 20, 50, 100]
n = len(y_train)
cross_valid_classifier = KNearestNeighbor()
def shuffle_examples(data, labels):
assert len(data) == len(labels)
p = np.random.permutation(len(data))
return data[p], labels[p]
X_train, y_train = shuffle_examples(X_train, y_train)
# A dictionary holding the accuracies for different values of k that we find
# when running cross-validation. After running cross-validation,
# k_to_accuracies[k] should be a list of length num_folds giving the different
# accuracy values that we found when using that value of k.
k_to_accuracies = {}
for i, k_choice in enumerate(k_choices):
accuracy = []
for j in xrange(num_folds):
start = (n*j)/num_folds
end = (n*(j+1))/num_folds-1
Xvalidation_set = X_train[start:end+1]
Yvalidation_set = y_train[start:end+1]
Xtraining_set = np.concatenate((X_train[0:start], X_train[end+1:n]), axis=0)
ytraining_set = np.concatenate((y_train[0:start], y_train[end+1:n]), axis=0)
cross_valid_classifier.train(Xtraining_set, ytraining_set)
dists = cross_valid_classifier.compute_distances_no_loops(Xvalidation_set)
y_test_pred = cross_valid_classifier.predict_labels(dists, k=k_choice)
num_correct = np.sum(y_test_pred == Yvalidation_set)
accuracy.append( float(num_correct) / len(y_test_pred) )
k_to_accuracies[k_choice] = accuracy
In [22]:
for k in sorted(k_to_accuracies):
for accuracy in k_to_accuracies[k]:
print 'k = %d, accuracy = %f' % (k, accuracy)
In [52]:
# plot the raw observations
for k in k_choices:
accuracies = k_to_accuracies[k]
plt.scatter([k] * len(accuracies), accuracies)
# plot the trend line with error bars that correspond to standard deviation
accuracies_mean = np.array([np.mean(v) for k,v in sorted(k_to_accuracies.items())])
accuracies_std = np.array([np.std(v) for k,v in sorted(k_to_accuracies.items())])
plt.errorbar(k_choices, accuracies_mean, yerr=accuracies_std)
plt.title('Cross-validation on k')
plt.xlabel('k')
plt.ylabel('Cross-validation accuracy')
plt.show()
Choose k = 10
In [24]:
# Based on the cross-validation results above, choose the best value for k,
# retrain the classifier using all the training data, and test it on the test
# data. You should be able to get above 28% accuracy on the test data.
best_k = 10
classifier = KNearestNeighbor()
classifier.train(X_train, y_train)
y_test_pred = classifier.predict(X_test, k=best_k)
# Compute and display the accuracy
num_correct = np.sum(y_test_pred == y_test)
accuracy = float(num_correct) / num_test
print 'Got %d / %d correct => accuracy: %f' % (num_correct, num_test, accuracy)
In [ ]: