cs231n-Assignment-1 python.

import os import numpy as np import pickle import matplotlib.pyplot as plt from CIFAR import load_CIFAR10 class NearestNeighbor(object) : def __init__(self) : pass def train(self, X, y) : """X is N X D where each row is an example.Y is 1-dimension of size N""" # just remember all the data self.Xtr = X self.ytr = y def predict(self, X) : num_test = X.shape[0] # make sure the output type matches the input type Ypred = np.zeros(num_test, dtype=self.ytr.type) # loop over all test rows for i in range(num_test) : # find the nearest training image to the i'th test image(using L1distance) distances = np.sum(np.abs(self.Xtr - X[i, :]), axis=1) min_index = np.argmin(distances) # get the index with smallest distance Ypred[i] = self.ytr[min_index] # predict the label of the nearest example return Ypred # "其中有几个语句要注意一下: # "X = X.reshape(10000, 3, 32, 32).transpose(0, 2, 3, 1).astype("float") # "起初，X的size为(10000, 3072(3*32*32))。首先reshape很好理解，最后astype的格式转换也很好理解。 # "可是为什么要调用transpose，转置轴呢？就我认为只需要把一幅图像转成行向量就可以了。是为了方便检索吗？ # "xs.append(X)将5个batch整合起来；np.concatenate(xs)使得最终Xtr的尺寸为(50000,32,32,3) # "当然还需要一步Xtr_rows = Xtr.reshape(Xtr.shape[0], 32 * 32 * 3)使得每一副图像称为一个行向量，最终就有了50000个行向量（Xtr_rows的尺寸为（50000,3072）） # "综上，为了方便，难道不应该直接从最开始就不要调用reshape(10000, 3, 32, 32).transpose(0, 2, 3, 1).astype("float")，直接append再concatenate不就能导出Xtr_rows了吗？" class KNearestNeighbor(object) : """a Knn classifier with L2 distance""" def __init__(self) : pass def train(self, X, y) : """ Train the classifier. 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_train,D) containing the training data consisting of num_train samples each of dimension D. - k: The number of nearest neighbors that vote for the predicted labels. - num_Loops:O Determings 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 points in X and each training points 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. return: - 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 range(num_test) : for j in range(num_train) : dists[i, j] = np.sqrt(np.dot(X[i] - self.X_train[j], X[i] - self.X_train[j])) return dists def compute_distances_one_loop(self, X) : num_test = X.shape[0] num_train = self.X_train.shape[0] dists = np.zeros((num_test, num_train)) for i in range(num_test) : dists[i, :] = np.sqrt(np.sum(np.square(X[i] - self.X_train), axis=1)) return dists def compute_distances_no_loops(self, X) : """ Compute the distance between each test point in X and training point in self.X_train using no explicit loops. """ num_test = X.shape[0] num_train = self.X_train.shape[0] dists = np.zeros((num_test, num_train)) dists = np.sqrt(self.getNormMatrix(X, num_train).T + self.getNormMatrix(self.X_train, num_test) - 2 * np.dot(X,self.X_train.T)) 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) 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 between 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 range(num_test) : closest_y = [] # 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]) # 找到距离最小的那个索引 closest_y = self.y_train[kids[:k]] # 找到距离最小的那个索引对应的训练数据的标签 # 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. # 找到贡献最大的那一个标签 # count = 0 label = 0 for j in closest_y : tmp = 0 for kk in closest_y : tmp += (kk == j) if tmp > count : count = tmp label = j y_pred[i] = label return y_pred plt.rcParams['figure.figsize'] = (10.0, 8.0) # set default size of plots plt.rcParams['image.interpolation'] = 'nearest' plt.rcParams['image.cmap'] = 'gray' # laod cifar10 data cifar10_dir = 'E:/research/CS231n/cifar-10-python/cifar-10-batches-py' X_train, y_train, X_test, y_test = load_CIFAR10(cifar10_dir) # 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) #随机显示一些样本图片 # classes = ['plane', 'car', '