cs231 assignment1--KNN
这次作业的主要的难点：
实现训练集和测试集之间距离的方法，主要是用向量形式来实现的部分
交叉检验的部分
import numpy as np
from collections import Counter
class KNearestNeighbor(object):
""" a kNN classifier with L2 distance """
def __init__(self):
pass
def train(self, X, y):
"""
Train the classifier. For k-nearest neighbors this is just
memorizing the training data.
Inputs:
- X: A numpy array of shape (num_train, D) containing the training data
consisting of num_train samples each of dimension D.
- y: A numpy array of shape (N,) containing the training labels, where
y[i] is the label for X[i].
"""
self.X_train = X
self.y_train = y
def predict(self, X, k=1, num_loops=0):
"""
Predict labels for test data using this classifier.
Inputs:
- X: A numpy array of shape (num_test, D) containing test data consisting
of num_test samples each of dimension D.
- k: The number of nearest neighbors that vote for the predicted labels.
- num_loops: Determines which implementation to use to compute distances
between training points and testing points.
Returns:
- y: A numpy array of shape (num_test,) containing predicted labels for the
test data, where y[i] is the predicted label for the test point X[i].
"""
if num_loops == 0:
dists = self.compute_distances_no_loops(X)
elif num_loops == 1:
dists = self.compute_distances_one_loop(X)
elif num_loops == 2:
dists = self.compute_distances_two_loops(X)
else:
raise ValueError('Invalid value %d for num_loops' % num_loops)
return self.predict_labels(dists, k=k)
#两个循环部分 ，实现比较简单
def compute_distances_two_loops(self, X):
"""
Compute the distance between each test point in X and each training point
in self.X_train using a nested loop over both the training data and the
test data.
Inputs:
- X: A numpy array of shape (num_test, D) containing test data.
Returns:
- dists: A numpy array of shape (num_test, num_train) where dists[i, j]
is the Euclidean distance between the ith test point and the jth training
point.
"""
num_test = X.shape[0]
num_train = self.X_train.shape[0]
dists = np.zeros((num_test, num_train))
for i in xrange(num_test):
for j in xrange(num_train):
dists[i, j] = np.sqrt(np.sum(np.square(self.X_train[j,:]- X[i, :])))
#####################################################################
# TODO: #
# Compute the l2 distance between the ith test point and the jth #
# training point, and store the result in dists[i, j]. You should #
# not use a loop over dimension. #
#####################################################################
pass
#####################################################################
# END OF YOUR CODE #
#####################################################################
return dists
#一个循环实现部分，主要是通过numpy广播机制来实现
def compute_distances_one_loop(self, X):
"""
Compute the distance between each test point in X and each training point
in self.X_train using a single loop over the test data.
Input / Output: Same as compute_distances_two_loops
"""
num_test = X.shape[0]
num_train = self.X_train.shape[0]
dists = np.zeros((num_test, num_train))
for i in xrange(num_test):
#######################################################################
# TODO: #
# Compute the l2 distance between the ith test point and all training #
# points, and store the result in dists[i, :]. #
#######################################################################
pass
dists[i,:] = np.sqrt(np.sum(np.square(self.X_train-X[i,:]),axis=1))
#X_train中每一行都减去X[i,:]
#######################################################################
# END OF YOUR CODE #
#######################################################################
return dists
#两个循环部分
def compute_distances_no_loops(self, X):
"""
Compute the distance between each test point in X and each training point
in self.X_train using no explicit loops.
Input / Output: Same as compute_distances_two_loops
"""
num_test = X.shape[0]
num_train = self.X_train.shape[0]
dists = np.zeros((num_test, num_train))
#########################################################################
# TODO: #
# Compute the l2 distance between all test points and all training #
# points without using any explicit loops, and store the result in #
# dists. #
# #
# You should implement this function using only basic array operations; #
# in particular you should not use functions from scipy. #
# #
# HINT: Try to formulate the l2 distance using matrix multiplication #
# and two broadcast sums. #
#########################################################################
pass
#注意点：numpy中dot才是点乘，而*预算则是所有相对应的元素进行相乘
#此处实现的原理，计算距离方法：（X1-X2）^2，将其展开，X1^2+X2^2-2X1*X2
#下面就是分别计算 X1^2,X2^2,2X1*X2
# X1*X2部分 500*5000
M = np.dot(X,self.X_train.T)
# X1^2 部分 5000*1
sqtr = np.sum(np.square(self.X_train),axis=1,keepdims=True)
# X2^2 部分 500*1
sqte = np.sum(np.square(X),axis=1,keepdims=True)
# 三部分相加，其中sqte+np.matrix(sqtr).T 500*1 + 1*5000 广播机制实现
dists = np.sqrt(sqte+np.matrix(sqtr).T-2*M)
#########################################################################
# END OF YOUR CODE #
#########################################################################
return dists
def predict_labels(self, dists, k=1):
"""
Given a matrix of distances between test points and training points,
predict a label for each test point.
Inputs:
- dists: A numpy array of shape (num_test, num_train) where dists[i, j]
gives the distance betwen the ith test point and the jth training point.
Returns:
- y: A numpy array of shape (num_test,) containing predicted labels for the
test data, where y[i] is the predicted label for the test point X[i].
"""
num_test = dists.shape[0]
y_pred = np.zeros(num_test)
for i in xrange(num_test):
# A list of length k storing the labels of the k nearest neighbors to
# the ith test point.
closest_y = []
"""
closest_y = self.y_train[np.argsort(dists[i:])[:k]]
y_pred[i] = np.argmax(np.bincount(np.matrix(closest_y)))
"""
labels = self.y_train[np.argsort(dists[i, :])].flatten()
closest_y = labels[0:k]
c = Counter(closest_y)
y_pred[i] = c.most_common(1)[0][0]
#########################################################################
# TODO: #
# Use the distance matrix to find the k nearest neighbors of the ith #
# testing point, and use self.y_train to find the labels of these #
# neighbors. Store these labels in closest_y. #
# Hint: Look up the function numpy.argsort. #
#########################################################################
pass
#########################################################################
# TODO: #
# Now that you have found the labels of the k nearest neighbors, you #
# need to find the most common label in the list closest_y of labels. #
# Store this label in y_pred[i]. Break ties by choosing the smaller #
# label. #
#########################################################################
pass
#########################################################################
# END OF YOUR CODE #
#########################################################################
return y_pred
# Run some setup code for this notebook.
import random
import numpy as np
from cs231n.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
# 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()
# 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]
# 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
from cs231n.classifiers import KNearestNeighbor
# Create a kNN classifier instance.
# Remember that training a kNN classifier is a noop:
# the Classifier simply remembers the data and does no further processing
classifier = KNearestNeighbor()
classifier.train(X_train, y_train)
# 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
# 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()
# 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)
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)
# 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'
# 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'
# 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
Cross-validation
We have implemented the k-Nearest Neighbor classifier but we set the value k = 5 arbitrarily.
We will now determine the best value of this hyperparameter with cross-validation.
# num_folds = 5
k_choices = [1, 3, 5, 8, 10, 12, 15, 20, 50, 100]
X_train_folds = []
y_train_folds = []
################################################################################
# TODO: #
# Split up the training data into folds. After splitting, X_train_folds and #
# y_train_folds should each be lists of length num_folds, where #
# y_train_folds[i] is the label vector for the points in X_train_folds[i]. #
# Hint: Look up the numpy array_split function. #
################################################################################
pass
X_train_folds = np.array_split(X_train,num_folds,axis=0)
y_train_folds = np.array_split(y_train,num_folds,axis=0)
################################################################################
# END OF YOUR CODE #
################################################################################
# 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.
# 交叉检验部分，一开始完全没有实现的思路，查阅各种资料后，才发现是才训练数据分成5部分
# 5个部分轮流当测试集，并将5次的结果取平均值作为最终结果
k_to_accuracies = {}
classifier = KNearestNeighbor()
for k in k_choices:
k_to_accuracies[k] = []
for j in xrange(num_folds):
X_tr = X_train_folds[0:j]+X_train_folds[(j+1):num_folds]
X_tr = np.reshape(X_tr,(X_train.shape[0]*(num_folds-1)/num_folds,-1))
y_tr = y_train_folds[0:j]+y_train_folds[(j+1):num_folds]
y_tr = np.reshape(y_tr,(X_train.shape[0]*(num_folds-1)/num_folds,-1))
X_te = np.reshape( X_train_folds[j], (X_train.shape[0]/num_folds,-1))
y_te = y_train_folds[j]
classifier.train(X_tr,y_tr)
y_test_pred = classifier.predict(X_te,k,0)
num_correct = sum(y_te==y_test_pred)
num_test1 = np.shape(X_te)[0]
accuracy = float(num_correct)/num_test1
k_to_accuracies[k].append(accuracy)
################################################################################
# TODO: #
# Perform k-fold cross validation to find the best value of k. For each #
# possible value of k, run the k-nearest-neighbor algorithm num_folds times, #
# where in each case you use all but one of the folds as training data and the #
# last fold as a validation set. Store the accuracies for all fold and all #
# values of k in the k_to_accuracies dictionary. #
################################################################################
pass
################################################################################
# END OF YOUR CODE #
################################################################################
# Print out the computed accuracies
for k in sorted(k_to_accuracies):
for accuracy in k_to_accuracies[k]:
print 'k = %d, accuracy = %f' % (k, accuracy)
# 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()
# 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 = 7
num_test=500
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)