CS231n-assignment1
参考文章：http://www.cnblogs.com/daihengchen/p/5754383.html
KNN
参考文章：http://blog.csdn.net/zhyh1435589631/article/details/54236643
knn 本质实现部分 代码分析
2.2.3.1 KNearestNeighbor 类整体分析
- 本质上， 这是一个类， 有多个成员函数构成， 用户调用的时候， 只需要调用
train和predict即可得到想要的预测数据 - 其中，
compute_distances_two_loops,compute_distances_one_loop,compute_distances_no_loops分别是用来实现需要预测的数据集X和 原始记录的训练集self.X_train之间的距离关系， 并通过predict_labels进行KNN预测
class KNearestNeighbor(object):
""" a kNN classifier with L2 distance """
def __init__(self):
pass
def train(self, X, y):
...
def predict(self, X, k=1, num_loops=0):
...
def compute_distances_two_loops(self, X):
...
def compute_distances_one_loop(self, X):
...
def compute_distances_no_loops(self, X):
...
def getNormMatrix(self, x, lines_num):
...
def predict_labels(self, dists, k=1):
...
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2.2.3.2 compute_distances_two_loops
这个函数主要通过两层 for 循环对计算测试集与训练集数据之间的欧式距离
d2(I1,I2)=∑p(Ip1−Ip2)2−−−−−−−−−−−−√
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):
#####################################################################
# 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. #
#####################################################################
dists[i][j] = np.sqrt(np.sum(np.square(self.X_train[j,:] - X[i,:])))
#####################################################################
# END OF YOUR CODE #
#####################################################################
return dists
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2.2.3.3 compute_distances_one_loop
本质上这里填入的代码和 上一节中的是一致的， 只是多了一个 axis = 1 指定方向
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
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2.2.3.4 compute_distances_no_loops
- 这部分公式虽然短小， 但是需要一定的数学功底， 参考文章：http://blog.csdn.net/geekmanong/article/details/51524402
- 我们记测试集矩阵 为 P 大小为 M×D , 训练集矩阵 为 C 大小为 N×D
- 记 Pi 是 P 的第 i 行， 同理 Cj 是 C 的 第 j 行：
Pi=[Pi1Pi2⋯PiD]Cj=[Cj1Cj2⋯CjD] - 我们先来计算一下 Pi 和 Cj 之间的距离
d(Pi,Cj)=(Pi1−Cj1)2+(Pi2−Cj2)2+⋯+(PiD−CjD)2−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√=(P2i1+P2i2+⋯+P2iD)+(C2j1+C2j2+⋯+C2jD)−2∗(Pi1Cj1+Pi2Cj2+⋯+PiDCjD)−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√=||Pi||2+||Cj||2−2∗PiC′j−−−−−−−−−−−−−−−−−−−−√ - 我们可以推广得到，结果矩阵的每行元素为：
Res(i)=(||Pi||2||Pi||2⋯||Pi||2)+(||C1||2||C2||2⋯||CN||2)−2∗Pi(C′1C′2⋯C′N)−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√=(||Pi||2||Pi||2⋯||Pi||2)+(||C1||2||C2||2⋯||CN||2)−2∗PiC′−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√ - 继而， 结果矩阵为：
Res=⎛⎝⎜⎜⎜⎜⎜⎜||P1||2||P2||2⋮||PM||2||P1||2||P2||2⋮||PM||2⋯⋯⋱⋯||P1||2||P2||2⋮||PM||2⎞⎠⎟⎟⎟⎟⎟⎟+⎛⎝⎜⎜⎜⎜⎜⎜||C1||2||C1||2⋮||C1||2||C2||2||C2||2⋮||C2||2⋯⋯⋱⋯||CN||2||CN||2⋮||CN||2⎞⎠⎟⎟⎟⎟⎟⎟−2PC′−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−⎷=⎛⎝⎜⎜⎜⎜⎜||P1||2||P2||2⋮||PM||2⎞⎠⎟⎟⎟⎟⎟M×1∗(11⋯1)1×N+⎛⎝⎜⎜⎜⎜11⋮1⎞⎠⎟⎟⎟⎟M×1∗(||C1||2||C2||2⋯||CN||2)1×N−2PM×DC′N×D−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−⎷ - 转换为python 代码如下：
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. #
#########################################################################
sq1=np.sum(np.square(X),axis=1)
sq2=np.sum(np.square(self.X_train),axis=1)
s=np.dot(X,self.X_train.T)
dist = np.sqrt(sq1+sq2.T-2*s) #dists = np.sqrt(self.getNormMatrix(X, num_train).T + self.getNormMatrix(self.X_train, num_test) - 2 * np.dot(X, self.X_train.T))
#########################################################################
# END OF YOUR CODE #
#########################################################################
return dists
def getNormMatrix(self, x, lines_num):
"""
Get a lines_num x size(x, 1) matrix
"""
return np.ones((lines_num, 1)) * np.sum(np.square(x), axis = 1)
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2.2.3.5 predict_labels
根据计算得到的距离关系， 挑选 K 个数据组成选民， 进行党派选举
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 = []
#########################################################################
# 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. #
#########################################################################
kids = np.argsort(dists[i])
closest_y = self.y_train[kids[: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_pred2.2.4 cross-validation 代码分析
- 交叉验证实际上是将数据的训练集进行拆分， 分成多个组， 构成多个训练和测试集， 来筛选较好的超参数
- 如图所示， 可以分为 5组数据， （分别将 fold 1, 2 .. 5 作为验证集， 将剩余的数据作为训练集， 训练得到超参数）
2.2.4.1 筛选不同的k
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.array_split(X_train, num_folds)
y_train_folds = np.array_split(y_train, num_folds)
################################################################################
# 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(num_folds)
for i in range(num_folds):
Xtr = np.array(X_train_folds[:i] + X_train_folds[i+1:])
ytr = np.array(y_train_folds[:i] + y_train_folds[i+1:])
Xte = np.array(X_train_folds[i])
yte = np.array(y_train_folds[i])
Xtr = np.reshape(Xtr, (X_train.shape[0] * 4 / 5, -1))
ytr = np.reshape(ytr, (y_train.shape[0] * 4 / 5, -1))
Xte = np.reshape(Xte, (X_train.shape[0] / 5, -1))
yte = np.reshape(yte, (y_train.shape[0] / 5, -1))
classifier.train(Xtr, ytr)
yte_pred = classifier.predict(Xte, k)
yte_pred = np.reshape(yte_pred, (yte_pred.shape[0], -1))
num_correct = np.sum(yte_pred == yte)
accuracy = float(num_correct) / len(yte)
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)