CNN for Visual Recognition (assignment1_Q1)
参考:http://cs231n.github.io/assignment1/
Q1: k-Nearest Neighbor classifier (30 points)
import numpy as np
from matplotlib.cbook import todate 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(X[i,:] - self.X_train[j,:])))
#####################################################################
# 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. #
#########################################################################
tDot = np.multiply(np.dot(X, self.X_train.T), -2)
t1 = np.sum(np.square(X), axis=1, keepdims=True)
t2 = np.sum(np.square(self.X_train), axis=1)
tDot = np.add(t1, tDot)
tDot = np.add(tDot, t2)
dists = np.sqrt(tDot)
#########################################################################
# 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. #
#########################################################################
# pass
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
输出:
Two loop version took 55.817642 seconds
One loop version took 49.692089 seconds
No loop version took 1.267753 seconds
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