1 问题描述
矩阵P的大小为[m, d] 用行向量表示为P1, P2,...,Pm
矩阵C的大小为[n, d] 用行向量表示为C1, C2,...,Cn
求矩阵P的每个行向量与矩阵C的每个行向量的欧氏距离
典型的例子是KNN算法应用于二维的点的聚类时，求取点与点之间的欧式距离时的情况。
2 解决办法1——两层循环
使用两层循环， 计算矩阵P的第i个行向量与矩阵C的第j个行向量的欧式距离
def l2distanceForMatrix_2loop(a, b):time1 = time.time()# 两层循环计算两个矩阵中每个样本之间的距离num_a = a.shape[0]num_b = b.shape[0]print(f"矩阵a数据的条数:{num_a}, 矩阵b数据的条数:{num_b}")distance = torch.zeros((num_a, num_b))for i in range(num_a):for j in range(num_b):# 首先做减法, 对应元素相减# 然后求平方# 再做加法# 最后开方# (x1, y1) 与 (x2, y2)的欧式距离:# sqrt( (x1-x2)^2 + (y1-y2)^2 )distance[i][j] = torch.sqrt(torch.sum(torch.square(a[i] - b[j])))time2 = time.time()print(f"花费时间:{time2 - time1}")print(distance)return distance3 解决办法2——一层循环
计算矩阵P的第i个行向量与矩阵C的欧式距离
def l2distanceForMatrix_1loop(a, b):time1 = time.time()# 1层循环计算两个矩阵中每个样本之间的距离num_a = a.shape[0]num_b = b.shape[0]distance = torch.zeros((num_a, num_b))for i in range(num_a):# 矩阵a中第i个样本与矩阵b中的样本的欧式距离# temp = torch.square(a[i] - b)# print(temp.shape)# print(temp)# temp = torch.sum(temp, dim=1)# print(temp)# temp = torch.sqrt(temp)# print(temp)# distance[i] = tempdistance[i] = torch.sqrt(torch.sum(torch.square(a[i] - b),dim=1))time2 = time.time()print(f"花费时间:{time2 - time1}")print(distance)return distance4 解决办法3——没有循环，使用矩阵运算
,
4.1 写法1
def l2distanceForMatrix(a, b):time1 = time.time()# 使用矩阵运算的方式求取两个矩阵中各个样本的欧式距离m = a.shape[0]n = b.shape[0]# 对矩阵的每个元素求平方aa = torch.pow(a, 2) # [m, d]# 按行求和, 并且保留维度数量不变aa = torch.sum(aa, dim=1, keepdim=True) # [m, 1]# 将矩阵aa从[m, 1]的形状扩展为[m, n]aa = aa.expand(m, n) # [m, n]# 处理矩阵bbb = torch.pow(b, 2).sum(dim=1, keepdim=True).expand(n ,m) # [n, m]bb = torch.transpose(bb, 0, 1) # [m, n]# 计算第三项 [m, d] * [d, n] = [m, n]tail = 2 * torch.matmul(a, torch.transpose(b, 0, 1))# 计算最后的结果distance = torch.sqrt(aa + bb - tail)time2 = time.time()print(f"花费时间:{time2 - time1}")print(distance)return distance4.2 写法2
def l2distanceForMatrix2(a, b):print("方法l2distanceForMatrix2")time1 = time.time()m = a.shape[0]n = b.shape[0]# 计算a*a^Tmatrix_a = torch.matmul(a, torch.transpose(a, 0, 1))matrix_b = torch.matmul(b, torch.transpose(b, 0, 1))# 取出matrix_a矩阵中的主对角线元素, 取出来的结果就是行向量的2范数diag_a = torch.diag(matrix_a) # [m]# print(diag_a.shape)# 扩展维度aa = diag_a.unsqueeze(1) # [m, 1]aa = aa.expand(m, n) # [m, n]# print(aa.shape)# print(aa)# 同样的方法处理矩阵bdiag_b = torch.diag(matrix_b) # [n]diag_b = diag_b.unsqueeze(1) # [n, 1]# 扩展维度bb = diag_b.expand(n, m) # [n, m]# 转置bb = torch.transpose(bb, 0, 1) # [m, n]# print(bb.shape)# print(bb)# 计算第三项 [m, d] * [d, n] = [m, n]tail = 2 * torch.matmul(a, torch.transpose(b, 0, 1))# 计算最后的结果distance = torch.sqrt(aa + bb - tail)time2 = time.time()print(f"花费时间:{time2 - time1}")print(distance)return distance5 测试代码
import os
import torch
import torch.nn as nn
import timedef l2distanceForMatrix_2loop(a, b):time1 = time.time()# 两层循环计算两个矩阵中每个样本之间的距离num_a = a.shape[0]num_b = b.shape[0]print(f"矩阵a数据的条数:{num_a}, 矩阵b数据的条数:{num_b}")distance = torch.zeros((num_a, num_b))for i in range(num_a):for j in range(num_b):# 首先做减法, 对应元素相减# 然后求平方# 再做加法# 最后开方# (x1, y1) 与 (x2, y2)的欧式距离:# sqrt( (x1-x2)^2 + (y1-y2)^2 )distance[i][j] = torch.sqrt(torch.sum(torch.square(a[i] - b[j])))time2 = time.time()print(f"花费时间:{time2 - time1}")print(distance)return distancedef l2distanceForMatrix_1loop(a, b):time1 = time.time()# 1层循环计算两个矩阵中每个样本之间的距离num_a = a.shape[0]num_b = b.shape[0]distance = torch.zeros((num_a, num_b))for i in range(num_a):# 矩阵a中第i个样本与矩阵b中的样本的欧式距离# temp = torch.square(a[i] - b)# print(temp.shape)# print(temp)# temp = torch.sum(temp, dim=1)# print(temp)# temp = torch.sqrt(temp)# print(temp)# distance[i] = tempdistance[i] = torch.sqrt(torch.sum(torch.square(a[i] - b),dim=1))time2 = time.time()print(f"花费时间:{time2 - time1}")print(distance)return distancedef l2distanceForMatrix(a, b):print("方法l2distanceForMatrix")time1 = time.time()# 使用矩阵运算的方式求取两个矩阵中各个样本的欧式距离m = a.shape[0]n = b.shape[0]# 对矩阵的每个元素求平方aa = torch.pow(a, 2) # [m, d]# 按行求和, 并且保留维度数量不变aa = torch.sum(aa, dim=1, keepdim=True) # [m, 1]# 将矩阵aa从[m, 1]的形状扩展为[m, n]aa = aa.expand(m, n) # [m, n]# print(aa.shape)# print(aa)# 处理矩阵bbb = torch.pow(b, 2).sum(dim=1, keepdim=True).expand(n ,m) # [n, m]bb = torch.transpose(bb, 0, 1) # [m, n]# print(bb.shape)# print(bb)# 计算第三项 [m, d] * [d, n] = [m, n]tail = 2 * torch.matmul(a, torch.transpose(b, 0, 1))# 计算最后的结果distance = torch.sqrt(aa + bb - tail)time2 = time.time()print(f"花费时间:{time2 - time1}")print(distance)return distancedef l2distanceForMatrix2(a, b):print("方法l2distanceForMatrix2")time1 = time.time()m = a.shape[0]n = b.shape[0]# 计算a*a^Tmatrix_a = torch.matmul(a, torch.transpose(a, 0, 1))matrix_b = torch.matmul(b, torch.transpose(b, 0, 1))# 取出matrix_a矩阵中的主对角线元素, 取出来的结果就是行向量的2范数diag_a = torch.diag(matrix_a) # [m]# print(diag_a.shape)# 扩展维度aa = diag_a.unsqueeze(1) # [m, 1]aa = aa.expand(m, n) # [m, n]# print(aa.shape)# print(aa)# 同样的方法处理矩阵bdiag_b = torch.diag(matrix_b) # [n]diag_b = diag_b.unsqueeze(1) # [n, 1]# 扩展维度bb = diag_b.expand(n, m) # [n, m]# 转置bb = torch.transpose(bb, 0, 1) # [m, n]# print(bb.shape)# print(bb)# 计算第三项 [m, d] * [d, n] = [m, n]tail = 2 * torch.matmul(a, torch.transpose(b, 0, 1))# 计算最后的结果distance = torch.sqrt(aa + bb - tail)time2 = time.time()print(f"花费时间:{time2 - time1}")print(distance)return distancedef main():# a = torch.randn((600, 2))# b = torch.randn((600, 2))a = torch.tensor([[1, 2],[3, 4],[5, 6],[7, 8],[9, 10]]).float()b = torch.tensor([[3, 4],[5, 6],[7, 8],[9, 10]]).float()l2distanceForMatrix_2loop(a, b)l2distanceForMatrix_1loop(a, b)l2distanceForMatrix(a, b)l2distanceForMatrix2(a, b)if __name__ == "__main__":main()6 参考
计算两个矩阵之间的欧式距离_frankzd的博客-CSDN博客_两个矩阵的欧式距离
快速计算矩阵之间的欧式距离_SP FA的博客-CSDN博客_计算两个矩阵的欧式距离
![[机器学习-概念] 什么是欧式距离、标准化欧式距离、马氏距离、余弦距离](https://img-blog.csdnimg.cn/20200703235519658.png)
