# 2. 具体实现部分

## 2.1 knn 调用程序

### 2.1.1 简单说明

1. knn 算法原理非常简单， 我们之前也总结过一次： http://blog.csdn.net/zhyh1435589631/article/details/53875182
2. 这个算法需要 对每个输入的测试数据计算他与所有的训练集数据之间的距离 （可以是 曼哈顿距离 L1， 欧式距离 L2）， 然后挑选出其中距离最小的k个值作为 选民， 并根据他们的党派进行投票， 这是一种典型的少数服从多数的方法

### 2.1.2 knn 调用程序 代码分析

#### 2.1.2.1 data_utils 载入数据集

1. 这里选用的数据集是 cifar-10 数据集 http://www.cs.toronto.edu/~kriz/cifar.html
2. 载入代码：
输出相应的训练集和测试集数据 Xtr, Ytr, Xte, Yte
def load_CIFAR_batch(filename):""" load single batch of cifar """with open(filename, 'rb') as f:datadict = pickle.load(f)X = datadict['data']Y = datadict['labels']X = X.reshape(10000, 3, 32, 32).transpose(0,2,3,1).astype("float")Y = np.array(Y)return X, Ydef load_CIFAR10(ROOT):""" load all of cifar """xs = []ys = []for b in range(1,6):f = os.path.join(ROOT, 'data_batch_%d' % (b, ))X, Y = load_CIFAR_batch(f)xs.append(X)ys.append(Y)    Xtr = np.concatenate(xs)Ytr = np.concatenate(ys)del X, YXte, Yte = load_CIFAR_batch(os.path.join(ROOT, 'test_batch'))return Xtr, Ytr, Xte, Yte

#### 2.1.2.2 载入数据集的调用

# Run some setup code for this notebook.import random
import numpy as np
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;
cifar10_dir = 'cs231n/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

Training data shape:  (50000L, 32L, 32L, 3L)
Training labels shape:  (50000L,)
Test data shape:  (10000L, 32L, 32L, 3L)
Test labels shape:  (10000L,)

#### 2.1.2.3 显示数据集的一部分信息

# 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 + 1plt.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()

#### 2.1.2.4 调整数据集大小

# Subsample the data for more efficient code execution in this exercise
num_training = 5000
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

(5000L, 3072L) (500L, 3072L)

#### 2.1.2.5 使用KNN进行训练

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
dists = classifier.compute_distances_two_loops(X_test)
print dists.shape# 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)

Got 137 / 500 correct => accuracy: 0.274000

#### 2.1.2.6 修改 k 参数

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)

Got 145 / 500 correct => accuracy: 0.290000

#### 2.1.2.7 验证其他两种实现方式

# 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'

#### 2.1.2.8 查看三种实现方法的使用时间

# 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 timetic = time.time()f(*args)toc = time.time()return toc - tictwo_loop_time = time_function(classifier.compute_distances_two_loops, X_test)
print 'Two loop version took %f seconds' % two_loop_timeone_loop_time = time_function(classifier.compute_distances_one_loop, X_test)
print 'One loop version took %f seconds' % one_loop_timeno_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

Two loop version took 46.657000 seconds
One loop version took 109.456000 seconds
No loop version took 1.205000 seconds

### 2.2.3 knn 本质实现部分 代码分析

#### 2.2.3.1 KNearestNeighbor 类整体分析

1. 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 trainpredict 即可得到想要的预测数据
2. 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):passdef 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):...

#### 2.2.3.2 compute_distances_two_loops

d2(I1,I2)=p(Ip1Ip2)2d2(I1,I2)=∑p(I1p−I2p)2

def compute_distances_two_loops(self, X):"""Compute the distance between each test point in X and each training pointin 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 trainingpoint."""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.dot(X[i] - self.X_train[j], X[i] - self.X_train[j]))######################################################################                       END OF YOUR CODE                            ######################################################################return dists

#### 2.2.3.3 compute_distances_one_loop

def compute_distances_one_loop(self, X):"""Compute the distance between each test point in X and each training pointin 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(X[i] - self.X_train), axis = 1))########################################################################                         END OF YOUR CODE                            ########################################################################return dists

#### 2.2.3.4 compute_distances_no_loops

1. 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
2. 我们记测试集矩阵 为 PP$P$ 大小为 $M×D$$M \times D$ , 训练集矩阵 为 CC$C$ 大小为 $N×D$$N \times D$
3. PiPi$P_i$PP$P$ 的第 $i$$i$ 行， 同理 CjCj$C_j$CC$C$ 的 第 $j$$j$ 行：
Pi=[Pi1Pi2PiD]Cj=[Cj1Cj2CjD]Pi=[Pi1Pi2⋯PiD]Cj=[Cj1Cj2⋯CjD]
4. 我们先来计算一下 PiPi$P_i$CjCj$C_j$ 之间的距离
d(Pi,Cj)=(Pi1Pj1)2+(Pi2Pj2)2++(PiDPjD)2=(P2i1+P2i2++P2iD)+(C2j1+C2j2++C2jD)2(Pi1Cj1+Pi2Cj2++PiDCjD)=||Pi||2+||Cj||22PiCjd(Pi,Cj)=(Pi1−Pj1)2+(Pi2−Pj2)2+⋯+(PiD−PjD)2=(Pi12+Pi22+⋯+PiD2)+(Cj12+Cj22+⋯+CjD2)−2∗(Pi1Cj1+Pi2Cj2+⋯+PiDCjD)=||Pi||2+||Cj||2−2∗PiCj′
5. 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)2Pi(C1C2CN)=(||Pi||2||Pi||2||Pi||2)+(||C1||2||C2||2||CN||2)2PiCRes(i)=(||Pi||2||Pi||2⋯||Pi||2)+(||C1||2||C2||2⋯||CN||2)−2∗Pi(C1′C2′⋯CN′)=(||Pi||2||Pi||2⋯||Pi||2)+(||C1||2||C2||2⋯||CN||2)−2∗PiC′
6. 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||22PC=||P1||2||P2||2||PM||2M×1(111)1×N+111M×1(||C1||2||C2||2||CN||2)1×N2PM×DCN×DRes=(||P1||2||P1||2⋯||P1||2||P2||2||P2||2⋯||P2||2⋮⋮⋱⋮||PM||2||PM||2⋯||PM||2)+(||C1||2||C2||2⋯||CN||2||C1||2||C2||2⋯||CN||2⋮⋮⋱⋮||C1||2||C2||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×DCN×D′
7. 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 pointin 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.                                         ##########################################################################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 distsdef 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) 

#### 2.2.3.5 predict_labels

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 thetest 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.                                                                ##########################################################################count = 0label = 0for j in closest_y:tmp = 0for kk in closest_y:tmp += (kk == j)if tmp > count:count = tmplabel = jy_pred[i] = label#y_pred[i] = np.argmax(np.bincount(closest_y))##########################################################################                           END OF YOUR CODE                            # #########################################################################return y_pred

#### 2.2.3.6 predict

1. 这里主要做了两个步骤：
1. 计算欧式距离
2. KNN 统计预测信息
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 consistingof 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 distancesbetween training points and testing points.Returns:- y: A numpy array of shape (num_test,) containing predicted labels for thetest 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)

### 2.2.4 cross-validation 代码分析

1. 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
2. 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)

k = 1, accuracy = 0.263000
k = 1, accuracy = 0.257000
k = 1, accuracy = 0.264000
k = 1, accuracy = 0.278000
k = 1, accuracy = 0.266000
k = 3, accuracy = 0.257000
k = 3, accuracy = 0.263000
k = 3, accuracy = 0.273000
k = 3, accuracy = 0.282000
k = 3, accuracy = 0.270000
k = 5, accuracy = 0.265000
k = 5, accuracy = 0.275000
k = 5, accuracy = 0.295000
k = 5, accuracy = 0.298000
k = 5, accuracy = 0.284000
k = 8, accuracy = 0.272000
k = 8, accuracy = 0.295000
k = 8, accuracy = 0.284000
k = 8, accuracy = 0.298000
k = 8, accuracy = 0.290000
k = 10, accuracy = 0.272000
k = 10, accuracy = 0.303000
k = 10, accuracy = 0.289000
k = 10, accuracy = 0.292000
k = 10, accuracy = 0.285000
k = 12, accuracy = 0.271000
k = 12, accuracy = 0.305000
k = 12, accuracy = 0.285000
k = 12, accuracy = 0.289000
k = 12, accuracy = 0.281000
k = 15, accuracy = 0.260000
k = 15, accuracy = 0.302000
k = 15, accuracy = 0.292000
k = 15, accuracy = 0.292000
k = 15, accuracy = 0.285000
k = 20, accuracy = 0.268000
k = 20, accuracy = 0.293000
k = 20, accuracy = 0.291000
k = 20, accuracy = 0.287000
k = 20, accuracy = 0.286000
k = 50, accuracy = 0.273000
k = 50, accuracy = 0.291000
k = 50, accuracy = 0.274000
k = 50, accuracy = 0.267000
k = 50, accuracy = 0.273000
k = 100, accuracy = 0.261000
k = 100, accuracy = 0.272000
k = 100, accuracy = 0.267000
k = 100, accuracy = 0.260000
k = 100, accuracy = 0.267000

#### 2.2.4.2 图形化显示

# 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()

#### 2.2.4.3 选取最好的k 进行训练

# 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 = 8classifier = 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)

Got 147 / 500 correct => accuracy: 0.294000

# 3. 参考资料

1. 寒老师博客 http://blog.csdn.net/han_xiaoyang/article/details/49949535
2. cs231n 课程主页 http://vision.stanford.edu/teaching/cs231n/syllabus.html
3. github 主页 http://cs231n.github.io/
4. http://www.cnblogs.com/daihengchen/p/5754383.html
5. https://docs.scipy.org/doc/numpy/reference/generated/numpy.sqrt.html
6. https://docs.scipy.org/doc/numpy-1.10.1/reference/generated/numpy.sum.html
7. https://docs.scipy.org/doc/numpy/reference/generated/numpy.ndarray.shape.html
8. https://docs.scipy.org/doc/numpy/reference/generated/numpy.reshape.html
9. enumerate ： http://blog.chinaunix.net/uid-27040911-id-3429751.html
10. https://docs.scipy.org/doc/numpy-dev/reference/generated/numpy.random.choice.html
11. https://docs.scipy.org/doc/numpy/reference/generated/numpy.flatnonzero.html
12. https://docs.scipy.org/doc/numpy/reference/generated/numpy.sum.html
13. https://docs.scipy.org/doc/numpy/reference/generated/numpy.concatenate.html
14. https://docs.scipy.org/doc/numpy/reference/generated/numpy.ones.html
15. https://docs.scipy.org/doc/numpy/reference/generated/numpy.count_nonzero.html#numpy.count_nonzero
16. https://docs.scipy.org/doc/numpy/reference/generated/numpy.argsort.html
17. https://docs.scipy.org/doc/numpy/reference/generated/numpy.bincount.html
18. https://docs.scipy.org/doc/numpy/reference/generated/numpy.multiply.html
19. https://docs.scipy.org/doc/numpy/reference/generated/numpy.dot.html
20. https://docs.scipy.org/doc/numpy/reference/generated/numpy.array_split.html