计算两个矩阵之间的欧式距离

xiaoxiao2021-02-28  21

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。

使用两层循环

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

使用一层循环

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

不使用循环

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 Pi P i 是P的第i行,记 Cj C j 是C的第j行 Pi=[Pi1Pi2PiD] P i = [ P i 1 P i 2 ⋯ P i D ] Cj=[Cj1Cj2CjD] C j = [ C j 1 C j 2 ⋯ C j D ]


首先计算 Pi P i Cj C j 之间的距离dist(i,j) d(Pi,Cj)=(Pi1Cj1)2+(Pi2Cj2)2++(PiDCjD)2=(P2i1+P2i2++P2iD)+(C2j1+C2j2++C2jD)2×(Pi1Cj1+Pi2Cj2++PiDCiD)=Pi2+Cj22×PiCTj d ( P i , C j ) = ( P i 1 − C j 1 ) 2 + ( P i 2 − C j 2 ) 2 + ⋯ + ( P i D − C j D ) 2 = ( P i 1 2 + P i 2 2 + ⋯ + P i D 2 ) + ( C j 1 2 + C j 2 2 + ⋯ + C j D 2 ) − 2 × ( P i 1 C j 1 + P i 2 C j 2 + ⋯ + P i D C i D ) = ‖ P i ‖ 2 + ‖ C j ‖ 2 − 2 × P i C j T


我们可以推广到距离矩阵的第i行的计算公式 dist[i]=(Pi2Pi2Pi2)+(C12C22CN2)2×Pi(CT1CT2CTN)=(Pi2Pi2Pi2)+(C12C22CN2)2×PiCT d i s t [ i ] = ( ‖ P i ‖ 2 ‖ P i ‖ 2 ⋯ ‖ P i ‖ 2 ) + ( ‖ C 1 ‖ 2 ‖ C 2 ‖ 2 ⋯ ‖ C N ‖ 2 ) − 2 × P i ( C 1 T C 2 T ⋯ C N T ) = ( ‖ P i ‖ 2 ‖ P i ‖ 2 ⋯ ‖ P i ‖ 2 ) + ( ‖ C 1 ‖ 2 ‖ C 2 ‖ 2 ⋯ ‖ C N ‖ 2 ) − 2 × P i C T


继续将公式推广为整个距离矩阵 dist=P12P22PM2P12P22PM2P12P22PM2+C12C12C12C22C22C22CN2CN2CN22×PCT d i s t = ( ‖ P 1 ‖ 2 ‖ P 1 ‖ 2 ⋯ ‖ P 1 ‖ 2 ‖ P 2 ‖ 2 ‖ P 2 ‖ 2 ⋯ ‖ P 2 ‖ 2 ⋮ ⋮ ⋱ ⋮ ‖ P M ‖ 2 ‖ P M ‖ 2 ⋯ ‖ P M ‖ 2 ) + ( ‖ C 1 ‖ 2 ‖ C 2 ‖ 2 ⋯ ‖ C N ‖ 2 ‖ C 1 ‖ 2 ‖ C 2 ‖ 2 ⋯ ‖ C N ‖ 2 ⋮ ⋮ ⋱ ⋮ ‖ C 1 ‖ 2 ‖ C 2 ‖ 2 ⋯ ‖ C N ‖ 2 ) − 2 × P C T

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
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