在我们使用k-NN模型时,需要计算测试集中每一点到训练集中每一点的欧氏距离,即需要求得两矩阵之间的欧氏距离。在实现k-NN算法时通常有三种方案,分别是使用两层循环,使用一层循环和不使用循环。
分别对训练集和测试集中的数据进行循环遍历,计算每两个点之间的欧式距离,然后赋值给dist矩阵。此算法没有经过任何优化。
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. # ##################################################################### # pass dists[i][j] = np.sqrt(np.sum(np.square(X[i] - self.X_train[j]))) ##################################################################### # END OF YOUR CODE # ##################################################################### return dists
使用矩阵表示训练集的数据,计算测试集中每一点到训练集矩阵的距离,可以对算法优化为只使用一层循环。
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, :]. # ####################################################################### # pass dists[i] = np.sqrt(np.sum(np.square(self.X_train - X[i]), axis = 1)) ####################################################################### # END OF YOUR CODE # ####################################################################### return dists
运算效率最高的算法是将训练集和测试集都使用矩阵表示,然后使用矩阵运算的方法替代之前的循环操作。但此操作需要我们对矩阵的运算规则非常熟悉。接下来着重记录如何计算两个矩阵之间的欧式距离。
记录测试集矩阵P的大小为M*D,训练集矩阵C的大小为N*D(测试集中共有M个点,每个点为D维特征向量。训练集中共有N个点,每个点为D维特征向量)
记是P的第i行,记是C的第j行
首先计算和之间的距离dist(i,j)
我们可以推广到距离矩阵的第i行的计算公式
继续将公式推广为整个距离矩阵
表示为python代码:
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. # ######################################################################### # pass dists = np.sqrt(-2*np.dot(X, self.X_train.T) + np.sum(np.square(self.X_train), axis = 1) + np.transpose([np.sum(np.square(X), axis = 1)])) ######################################################################### # END OF YOUR CODE # ######################################################################### return dists
联系客服