KNN作業要求：
1、掌握KNN算法原理
2、實現具體K值的KNN算法
3、實現對K值的交叉驗證
1、KNN原理見上一小節
2、實現KNN
過程分兩步：
1、計算測試集與訓練集的距離
2、通過比較label出現比例的方式，確定選取的最終label
代碼分析：
cell1 - cell5 對數據的預處理
cell6創建KNN類，初始化類的變量，此處是傳遞測試數據和訓練數據
cell7實現包含兩個循環的KNN算法：
通過計算單一的向量與矩陣之間的距離（在之前的cell中，已經將圖像轉換成列：32*32 的圖像轉換為 1*3072,，
測試集是500張：500*3072，訓練集是5000張：5000*3072）
代碼基礎：使用python 2.7.9 + numpy 1.11.0
技巧：使用help 查看相關函數的用法，或者google
舉例：np.square
q 鍵退出help
可知，np.square() 為了加快運算速度，是用c寫的，在這里查不到具體用法。google查看:
例子為計算數組[-1j,1]里邊各元素的平方，得到的結果為[-1,1]
代碼：實現compute_distances_two_loops(self, X)
1 def compute_distances_two_loops(self, X): 2 """ 3 Compute the distance between each test point in X and each training point 4 in self.X_train using a nested loop over both the training data and the 5 test data. 6 7 Inputs: 8 - X: A numpy array of shape (num_test, D) containing test data. 9 10 Returns: 11 - dists: A numpy array of shape (num_test, num_train) where dists[i, j] 12 is the Euclidean distance between the ith test point and the jth training 13 point. 14 """ 15 num_test = X.shape[0] 16 num_train = self.X_train.shape[0] 17 dists = np.zeros((num_test, num_train)) 18 for i in xrange(num_test): 19 for j in xrange(num_train): 20 ##################################################################### 21 # TODO: # 22 # Compute the l2 distance between the ith test point and the jth # 23 # training point, and store the result in dists[i, j]. You should # 24 # not use a loop over dimension. # 25 ##################################################################### 26 dists[i,j] = np.sqrt(np.sum(np.square(X[i,:]-self.X_train[j,:]))) 27 ##################################################################### 28 # END OF YOUR CODE # 29 ##################################################################### 30 return dists
實現對一張測試圖像對應的矩陣與一張訓練集圖像的矩陣做L2距離。
也可以用numpy.linalg.norm函數實現：
此函數執行的公式：
所以核心代碼可以寫作：
dists[i,j] = np.linalg.norm(self.X_train[j,:]-X[i,:])
cell8 得到的距離可視化，白色表示較大的距離值，黑色是較小距離值
cell9 實現K=1的label預測
代碼：實現 classifier.predict_labels()
1 def predict_labels(self, dists, k=1): 2 """ 3 Given a matrix of distances between test points and training points, 4 predict a label for each test point. 5 6 Inputs: 7 - dists: A numpy array of shape (num_test, num_train) where dists[i, j] 8 gives the distance betwen the ith test point and the jth training point. 9 10 Returns: 11 - y: A numpy array of shape (num_test,) containing predicted labels for the 12 test data, where y[i] is the predicted label for the test point X[i]. 13 """ 14 num_test = dists.shape[0] 15 y_pred = np.zeros(num_test) 16 for i in xrange(num_test): 17 # A list of length k storing the labels of the k nearest neighbors to 18 # the ith test point. 19 closest_y = [] 20 count = [] 21 ######################################################################### 22 # TODO: # 23 # Use the distance matrix to find the k nearest neighbors of the ith # 24 # testing point, and use self.y_train to find the labels of these # 25 # neighbors. Store these labels in closest_y. # 26 # Hint: Look up the function numpy.argsort. # 27 ######################################################################### 28 buf_labels = self.y_train[np.argsort(dists[i,:])] 29 closest_y = buf_labels[0:k] 30 ######################################################################### 31 # TODO: # 32 # Now that you have found the labels of the k nearest neighbors, you # 33 # need to find the most common label in the list closest_y of labels. # 34 # Store this label in y_pred[i]. Break ties by choosing the smaller # 35 # label. # 36 ######################################################################### 37 #for j in closest_y : 38 # count.append(closest_y.count(j)) 39 #m = max(count) 40 #n = count.index(m) 41 #y_pred[i] = closest_y[n] 42 c = Counter(closest_y) 43 y_pred[i] = c.most_common(1)[0][0] 44 ######################################################################### 45 # END OF YOUR CODE # 46 ######################################################################### 47 48 return y_pred
步驟：
1.使用numpy.argsort對所以距離進行排序，得到排序后的索引。
2.通過索引找到對應的label
3.通過collection包的Counter，對label進行統計表示
4.通過counter的Most common方法得到出現最多的label
cell9 在計算完成后，同時實現了准確率的計算
cell10 實現K =5的KNN
cell11 實現compute_distances_one_loop(X_test)
代碼：
1 def compute_distances_one_loop(self, X): 2 """ 3 Compute the distance between each test point in X and each training point 4 in self.X_train using a single loop over the test data. 5 6 Input / Output: Same as compute_distances_two_loops 7 """ 8 num_test = X.shape[0] 9 num_train = self.X_train.shape[0] 10 dists = np.zeros((num_test, num_train)) 11 for i in xrange(num_test): 12 ####################################################################### 13 # TODO: # 14 # Compute the l2 distance between the ith test point and all training # 15 # points, and store the result in dists[i, :]. # 16 ####################################################################### 17 buf = np.square(self.X_train-X[i,:]) 18 dists[i,:] = np.sqrt(np.sum(buf,axis=1)) 19 ####################################################################### 20 # END OF YOUR CODE # 21 ####################################################################### 22 return dists
並通過計算一個循環與兩個循環分別得到的結果的差值平方，來衡量准確性。
cell12 實現完全的數組操作，不使用循環。
1 def compute_distances_no_loops(self, X): 2 """ 3 Compute the distance between each test point in X and each training point 4 in self.X_train using no explicit loops. 5 6 Input / Output: Same as compute_distances_two_loops 7 """ 8 num_test = X.shape[0] 9 num_train = self.X_train.shape[0] 10 dists = np.zeros((num_test, num_train)) 11 ######################################################################### 12 # TODO: # 13 # Compute the l2 distance between all test points and all training # 14 # points without using any explicit loops, and store the result in # 15 # dists. # 16 # # 17 # You should implement this function using only basic array operations; # 18 # in particular you should not use functions from scipy. # 19 # # 20 # HINT: Try to formulate the l2 distance using matrix multiplication # 21 # and two broadcast sums. # 22 ######################################################################### 23 #buf = np.tile(X,(1,num_train)) 24 buf = np.dot(X, self.X_train.T) 25 buf_test = np.square(X).sum(axis = 1) 26 buf_train = np.square(self.X_train).sum(axis = 1) 27 dists = np.sqrt(-2*buf+buf_train+np.matrix(buf_test).T) 28 ######################################################################### 29 # END OF YOUR CODE # 30 ######################################################################### 31 return dists
使用(a-b)2=a2+b2-2ab 的公式
27行： 此時buf_test 為500*1數組 buf_train為5000*1的數組 需要得到500*5000的數組 此處通過構造矩陣的方式進行broadcast
cell13 比較3種方案的執行效率
cell14 交叉驗證
交叉驗證的思想在上一節有解釋過了
代碼：（其中帶有注釋）
1 num_folds = 5 2 k_choices = [1, 3, 5, 8, 10, 12, 15, 20, 50, 100] 3 4 X_train_folds = [] 5 y_train_folds = [] 6 ################################################################################ 7 # TODO: # 8 # Split up the training data into folds. After splitting, X_train_folds and # 9 # y_train_folds should each be lists of length num_folds, where # 10 # y_train_folds[i] is the label vector for the points in X_train_folds[i]. # 11 # Hint: Look up the numpy array_split function. # 12 ################################################################################ 13 X_train_folds = np.array_split(X_train, num_folds) 14 y_train_folds = np.array_split(y_train, num_folds) 15 ################################################################################ 16 # END OF YOUR CODE # 17 ################################################################################ 18 19 # A dictionary holding the accuracies for different values of k that we find 20 # when running cross-validation. After running cross-validation, 21 # k_to_accuracies[k] should be a list of length num_folds giving the different 22 # accuracy values that we found when using that value of k. 23 k_to_accuracies = {} 24 25 26 ################################################################################ 27 # TODO: # 28 # Perform k-fold cross validation to find the best value of k. For each # 29 # possible value of k, run the k-nearest-neighbor algorithm num_folds times, # 30 # where in each case you use all but one of the folds as training data and the # 31 # last fold as a validation set. Store the accuracies for all fold and all # 32 # values of k in the k_to_accuracies dictionary. # 33 ################################################################################ 34 for k in k_choices: 35 k_to_accuracies[k] = [] 36 37 for k in k_choices: 38 print 'evaluating k=%d' % k 39 for j in range(num_folds): 40 #get validation 41 X_train_cv = np.vstack(X_train_folds[0:j]+X_train_folds[j+1:]) 42 X_test_cv = X_train_folds[j] 43 y_train_cv = np.hstack(y_train_folds[0:j]+y_train_folds[j+1:]) 44 y_test_cv = y_train_folds[j] 45 #train 46 classifier.train(X_train_cv, y_train_cv) 47 dists_cv = classifier.compute_distances_no_loops(X_test_cv) 48 #get accuracy 49 y_test_pred = classifier.predict_labels(dists_cv, k) 50 num_correct = np.sum(y_test_pred == y_test_cv) 51 accuracy = float(num_correct) / num_test 52 #add j th accuracy of k to array 53 k_to_accuracies[k].append(accuracy) 54 ################################################################################ 55 # END OF YOUR CODE # 56 ################################################################################ 57 58 # Print out the computed accuracies 59 for k in sorted(k_to_accuracies): 60 for accuracy in k_to_accuracies[k]: 61 print 'k = %d, accuracy = %f' % (k, accuracy)
cell15 顯示k值對應的准確率
上述包含了均值和標准差
cell16 使用最優值，得到較好的准確率
總結：
整體來說，第一個作業難度較大，主要難度不是在算法部分，而是在熟悉python相關函數與相應的用法方面。對於python大神而言，難度低。
但是經過扎實的第一次作業后，后邊的作業相對簡單了。之后的作業細節方面講解的少些。
附：通關CS231n企鵝群：578975100 validation：DL-CS231n