作業網址：http://cs231n.github.io/assignment1/
k_nearest_neighbor.py：
import numpy as np
class KNearestNeighbor:
""" 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.
Input:
X - A num_train x dimension array where each row is a training point.
y - A vector of length num_train, 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.
Input:
X - A num_test x dimension array where each row is a test point.
k - The number of nearest neighbors that vote for predicted label
num_loops - Determines which method to use to compute distances
between training points and test points.
Output:
y - A vector of length num_test, 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)
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.
Input:
X - An num_test x dimension array where each row is a test point.
Output:
dists - A num_test x num_train array where dists[i, j] is the 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):
#####################################################################
# TODO: #
# Compute the l2 distance between the ith test point and the jth #
# training point, and store the result in dists[i, j] #
#####################################################################
dists[i][j] = np.sqrt(np.sum(np.square(self.X_train[j,:] - X[i,:])))
#####################################################################
# END OF YOUR CODE #
#####################################################################
return dists
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, :]. #
#######################################################################
dists[i,:] = np.sqrt(np.sum(np.square(self.X_train-X[i,:]),axis = 1))
#######################################################################
# END OF YOUR CODE #
#######################################################################
return dists
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. #
# HINT: Try to formulate the l2 distance using matrix multiplication #
# and two broadcast sums. #
#########################################################################
dists = np.multiply(np.dot(X,self.X_train.T),-2)
sq1 = np.sum(np.square(X),axis=1,keepdims = True)
sq2 = np.sum(np.square(self.X_train),axis=1)
dists = np.add(dists,sq1)
dists = np.add(dists,sq2)
dists = np.sqrt(dists)
#########################################################################
# 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.
Input:
dists - A num_test x num_train array where dists[i, j] gives the distance
between the ith test point and the jth training point.
Output:
y - A vector of length num_test where y[i] is the predicted label for the
ith test point.
"""
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 = []
#########################################################################
# TODO: #
# Use the distance matrix to find the k nearest neighbors of the ith #
# training 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. #
#########################################################################
closest_y = self.y_train[np.argsort(dists[i,:])[:k]]
#########################################################################
# 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. #
#########################################################################
y_pred[i] = np.argmax(np.bincount(closest_y))
#########################################################################
# END OF YOUR CODE #
#########################################################################
return y_pred
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. #
################################################################################
X_train_folds = np.split(X_train,5)
y_train_folds = np.split(y_train,5)
################################################################################
# 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.
k_to_accuracies = {}
################################################################################
# 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. #
################################################################################
for k in k_choices:
k_to_accuracies[k] = np.zeros(5)
for i in range(5):
X_tr = X_train_folds[0:i] + X_train_folds[(i+1):5]
X_tr = np.reshape(X_tr, (X_train.shape[0]*4/5, -1))
y_tr = y_train_folds[0:i] + y_train_folds[(i+1):5]
y_tr = np.reshape(y_tr, (X_train.shape[0]*4/5 ))
X_tt = np.reshape(X_train_folds[i], (X_train.shape[0]*1/5, -1))
classifier.train(X_tr,y_tr)
y_test_pred = classifier.predict(X_tt, k)
num_correct = np.sum(y_test_pred == y_train_folds[i])
accuracy = float(num_correct) / num_test
k_to_accuracies[k][i] = accuracy
################################################################################
# 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)
（初次使用python，不熟悉操作，代碼質量無法保證，僅供參考！）
