cs231n assignment1

1.KNN

knn 的三種實現

import numpy as npfrom past.builtins import xrangeclass KNearestNeighbor(object): """ 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. Inputs: - X: A numpy array of shape (num_train, D) containing the training data consisting of num_train samples each of dimension D. - y: A numpy array of shape (N,) containing the training labels, 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. Inputs: - X: A numpy array of shape (num_test, D) containing test data consisting of 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 distances between training points and testing points. 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]. """ 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. 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(X[i,:]-self.X_train[j,:]))) 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 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(X[i] - self.X_train), 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. # # # # 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. # ######################################################################### matrix_1 = (np.sum(np.square(self.X_train), axis=1).reshape(num_train, 1) @ np.ones((1, num_test))).T matrix_2 = (np.ones((num_train, 1)) @ np.sum(np.square(X), axis=1).reshape(-1, num_test)).T merge = -2 * (self.X_train @ X.T).T dists = np.sqrt(matrix_1 + matrix_2 + merge) ######################################################################### # 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. 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. # ######################################################################### # argsort函數返回的是數組值從小到大的索引值 distance = np.argsort(dists[i])[:k] closest_y = self.y_train[distance] ######################################################################### # 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

Two loop version took 22.063287 seconds

One loop version took 80.193520 seconds

No loop version took 0.386157 seconds

這是三種實現的時間對比，可以看見向量化實現的優勢了

交叉驗證的實現

num_folds = 5k_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: accuracies = np.zeros(num_folds)# for num in range(num_folds):# x_train=X_train_folds[num]# y_test=y_train_folds[num] # x_train=[]# y_train=[]# to_train=[x for x in np.arange(num_folds) if x!=num ]# for x in to_train:# x_train.append(X_train_folds[x])# y_train.append(y_train_folds[x])# classifier.train(x_train, y_train) # y_test_pred = classifier.predict_labels(x_test, k=k)# num_correct = np.sum(y_test_pred == y_test)# accuracy = float(num_correct) / num_test# accuracies[k][num]=accuracy for fold in range(num_folds): temp_X = X_train_folds[:] temp_y = y_train_folds[:] X_validate_fold = temp_X.pop(fold) y_validate_fold = temp_y.pop(fold) temp_X = np.array([y for x in temp_X for y in x]) temp_y = np.array([y for x in temp_y for y in x]) classifier.train(temp_X, temp_y) y_test_pred = classifier.predict(X_validate_fold, k=k) num_correct = np.sum(y_test_pred == y_validate_fold) accuracy = float(num_correct) / num_test accuracies[fold] =accuracy k_to_accuracies[k] = accuracies################################################################################# END OF YOUR CODE ################################################################################## Print out the computed accuraciesfor k in sorted(k_to_accuracies): for accuracy in k_to_accuracies[k]: print(k = %d, accuracy = %f % (k, accuracy))我寫的老是報錯，後來參考了別人的，如果有知道的可以指點我一下

參考內容

【實驗小結】cs231n assignment1 knn 部分blog.csdn.net

未完待續