【PyTorch】计算局部相似矩阵

计算局部相似矩阵

代码文档:https://github.com/lartpang/mypython/blob/master/2019-09-25计算局部相关性矩阵/计算局部相关性.ipynb

问题说明

对于给定的数据,其尺寸为N,C,H,W,现在想要计算其局部的相关性,也就是说特定尺寸范围内,例如2*2大小的区域内任意两点之间的点积。

试写出相关的代码。

问题分析

计算局部相关性,而且这里也提到是说使用局部的区域的任意两点之间的点积来计算,所以实际上也就是需要就算对应的2*2范围内的任意两个C维矢量的点积,最终得到一个4*4的关系矩阵。若是在矢量点积的时候,除以各自的模,那么实际上计算的就是两个矢量的余弦距离。

余弦相似度用向量空间中两个向量夹角的余弦值作为衡量两个个体间差异的大小。相比距离度量,余弦相似度更加注重两个向量在方向上的差异,而非距离或长度上。
https://blog.csdn.net/weixin_38659482/article/details/85045537

\[cos<\vec{x}, \vec{y}> = \frac{\vec{x} \cdot \vec{y}}{||\vec{x}|| \cdot ||\vec{y}||} = \frac{\vec{x} }{||\vec{x}||} \cdot \frac{\vec{y}}{||\vec{y}||} \]

所以“除以模”这个归一化操作放在点积之前。实际上也就是除以沿着C维度计算的L2范数。

最直接的思路是简单的遍历计算,但是不实际,太耗时。如何能够利用GPU并行计算的优势,那自然是使用矩阵操作。

对于已有的N,C,H,W的数据,我们需要计算点积,对于三维以上的点积,可以使用torch.matmul,这时,乘法发生在最右侧的几个维度上。

可以构想,我们最终得到的结果应该是N * NumOfRegions * 4 * 4大小的一个张量。而这里的NumOfRegions表示总的计算了的区域的数量。对于pytorch,我所知道的可以收集区域数据,而且没有其他多余操作的方法只有torch.nn.Unfold。所以这里使用它来实现这个过程。

实现过程

对于矩阵乘法,思考的最简单的方式就是维度匹配

import torch
import torch.nn as nn

a = torch.rand(1, 2, 3, 4)
b = torch.rand(1, 2, 3, 4)
print("a=>\n", a)
print("b=>\n", b)

a=>
 tensor([[[[0.4818, 0.9888, 0.8039, 0.7089],
          [0.7667, 0.2273, 0.9956, 0.4739],
          [0.9515, 0.1896, 0.7928, 0.0173]],

         [[0.1723, 0.8767, 0.4832, 0.6515],
          [0.9487, 0.6301, 0.5711, 0.7781],
          [0.2017, 0.9220, 0.2793, 0.2675]]]])
b=>
 tensor([[[[0.1417, 0.3510, 0.1170, 0.1698],
          [0.4311, 0.1535, 0.6087, 0.6646],
          [0.1880, 0.4103, 0.0289, 0.1094]],

         [[0.3398, 0.8751, 0.8299, 0.3514],
          [0.0333, 0.2831, 0.8086, 0.0514],
          [0.3168, 0.2895, 0.5107, 0.4949]]]])

这里先定义两个tensor,二者实际上没有关系,后面的计算也没有关系,只是为了多展示一点。

unfold_func = nn.Unfold(2, 1, 0, 1)

unfold_a = unfold_func(a)
print("unfold_a=>\n", unfold_a)

unfold_b = unfold_func(b)
print("unfold_b=>\n", unfold_b)

unfold_a=>
 tensor([[[0.4818, 0.9888, 0.8039, 0.7667, 0.2273, 0.9956],
         [0.9888, 0.8039, 0.7089, 0.2273, 0.9956, 0.4739],
         [0.7667, 0.2273, 0.9956, 0.9515, 0.1896, 0.7928],
         [0.2273, 0.9956, 0.4739, 0.1896, 0.7928, 0.0173],
         [0.1723, 0.8767, 0.4832, 0.9487, 0.6301, 0.5711],
         [0.8767, 0.4832, 0.6515, 0.6301, 0.5711, 0.7781],
         [0.9487, 0.6301, 0.5711, 0.2017, 0.9220, 0.2793],
         [0.6301, 0.5711, 0.7781, 0.9220, 0.2793, 0.2675]]])
unfold_b=>
 tensor([[[0.1417, 0.3510, 0.1170, 0.4311, 0.1535, 0.6087],
         [0.3510, 0.1170, 0.1698, 0.1535, 0.6087, 0.6646],
         [0.4311, 0.1535, 0.6087, 0.1880, 0.4103, 0.0289],
         [0.1535, 0.6087, 0.6646, 0.4103, 0.0289, 0.1094],
         [0.3398, 0.8751, 0.8299, 0.0333, 0.2831, 0.8086],
         [0.8751, 0.8299, 0.3514, 0.2831, 0.8086, 0.0514],
         [0.0333, 0.2831, 0.8086, 0.3168, 0.2895, 0.5107],
         [0.2831, 0.8086, 0.0514, 0.2895, 0.5107, 0.4949]]])

这里使用fold和unfold操作之后可以看出来,外侧的括号从原来的四层变为了现在的三层,实际上表示的就是从原来的N,C,H,W变成了现在的N,C*4,H/2*W/2的样子。

而对于H/2*W/2的维度上,在滑窗处理时,也是基于行主序调整成一行的。

unfold_a_reshape = unfold_a.transpose(1, 2).view(1, (3-1)*(4-1), 2, 4)  # N,H'W',C,2*2
print("unfold_a_reshape=>\n", unfold_a_reshape)

unfold_b_reshape = unfold_b.transpose(1, 2).view(1, (3-1)*(4-1), 2, 4)
print("unfold_b_reshape=>\n", unfold_b_reshape)

unfold_a_reshape=>
 tensor([[[[0.4818, 0.9888, 0.7667, 0.2273],
          [0.1723, 0.8767, 0.9487, 0.6301]],

         [[0.9888, 0.8039, 0.2273, 0.9956],
          [0.8767, 0.4832, 0.6301, 0.5711]],

         [[0.8039, 0.7089, 0.9956, 0.4739],
          [0.4832, 0.6515, 0.5711, 0.7781]],

         [[0.7667, 0.2273, 0.9515, 0.1896],
          [0.9487, 0.6301, 0.2017, 0.9220]],

         [[0.2273, 0.9956, 0.1896, 0.7928],
          [0.6301, 0.5711, 0.9220, 0.2793]],

         [[0.9956, 0.4739, 0.7928, 0.0173],
          [0.5711, 0.7781, 0.2793, 0.2675]]]])
unfold_b_reshape=>
 tensor([[[[0.1417, 0.3510, 0.4311, 0.1535],
          [0.3398, 0.8751, 0.0333, 0.2831]],

         [[0.3510, 0.1170, 0.1535, 0.6087],
          [0.8751, 0.8299, 0.2831, 0.8086]],

         [[0.1170, 0.1698, 0.6087, 0.6646],
          [0.8299, 0.3514, 0.8086, 0.0514]],

         [[0.4311, 0.1535, 0.1880, 0.4103],
          [0.0333, 0.2831, 0.3168, 0.2895]],

         [[0.1535, 0.6087, 0.4103, 0.0289],
          [0.2831, 0.8086, 0.2895, 0.5107]],

         [[0.6087, 0.6646, 0.0289, 0.1094],
          [0.8086, 0.0514, 0.5107, 0.4949]]]])

这里调整一下形状,这里可以根据维度匹配的思想进行连接,这里就是为了方便通过后面的矩阵乘法实现对于区域内任意点关系的描述矩阵的构造

mm_unfold_a = torch.matmul(unfold_a_reshape.transpose(2, 3), unfold_a_reshape)  # N,H'W',2*2,2*2
print("mm_unfold_a=>\n", mm_unfold_a)

mm_unfold_b = torch.matmul(unfold_b_reshape.transpose(2, 3), unfold_b_reshape)
print("mm_unfold_b=>\n", mm_unfold_b)
mm_unfold_a=>
 tensor([[[[0.2619, 0.6275, 0.5329, 0.2181],
          [0.6275, 1.7462, 1.5898, 0.7771],
          [0.5329, 1.5898, 1.4878, 0.7720],
          [0.2181, 0.7771, 0.7720, 0.4487]],

         [[1.7462, 1.2184, 0.7771, 1.4851],
          [1.2184, 0.8796, 0.4871, 1.0763],
          [0.7771, 0.4871, 0.4487, 0.5862],
          [1.4851, 1.0763, 0.5862, 1.3174]],

         [[0.8796, 0.8847, 1.0763, 0.7569],
          [0.8847, 0.9270, 1.0779, 0.8429],
          [1.0763, 1.0779, 1.3174, 0.9163],
          [0.7569, 0.8429, 0.9163, 0.8301]],

         [[1.4878, 0.7720, 0.9209, 1.0200],
          [0.7720, 0.4487, 0.3433, 0.6240],
          [0.9209, 0.3433, 0.9459, 0.3664],
          [1.0200, 0.6240, 0.3664, 0.8860]],

         [[0.4487, 0.5862, 0.6240, 0.3562],
          [0.5862, 1.3174, 0.7153, 0.9488],
          [0.6240, 0.7153, 0.8860, 0.4078],
          [0.3562, 0.9488, 0.4078, 0.7065]],

         [[1.3174, 0.9163, 0.9488, 0.1700],
          [0.9163, 0.8301, 0.5930, 0.2164],
          [0.9488, 0.5930, 0.7065, 0.0884],
          [0.1700, 0.2164, 0.0884, 0.0719]]]])
mm_unfold_b=>
 tensor([[[[0.1355, 0.3471, 0.0724, 0.1180],
          [0.3471, 0.8891, 0.1805, 0.3017],
          [0.0724, 0.1805, 0.1869, 0.0756],
          [0.1180, 0.3017, 0.0756, 0.1037]],

         [[0.8891, 0.7674, 0.3017, 0.9213],
          [0.7674, 0.7025, 0.2530, 0.7424],
          [0.3017, 0.2530, 0.1037, 0.3224],
          [0.9213, 0.7424, 0.3224, 1.0244]],

         [[0.7025, 0.3115, 0.7424, 0.1204],
          [0.3115, 0.1523, 0.3875, 0.1309],
          [0.7424, 0.3875, 1.0244, 0.4461],
          [0.1204, 0.1309, 0.4461, 0.4443]],

         [[0.1869, 0.0756, 0.0916, 0.1865],
          [0.0756, 0.1037, 0.1186, 0.1450],
          [0.0916, 0.1186, 0.1357, 0.1689],
          [0.1865, 0.1450, 0.1689, 0.2522]],

         [[0.1037, 0.3224, 0.1450, 0.1490],
          [0.3224, 1.0244, 0.4839, 0.4306],
          [0.1450, 0.4839, 0.2522, 0.1597],
          [0.1490, 0.4306, 0.1597, 0.2616]],

         [[1.0244, 0.4461, 0.4306, 0.4668],
          [0.4461, 0.4443, 0.0455, 0.0982],
          [0.4306, 0.0455, 0.2616, 0.2559],
          [0.4668, 0.0982, 0.2559, 0.2569]]]])

这里计算了乘法,实际上结果计算出来的就是对应的关系矩阵。这里结果的尺寸为N, NumOfRegion, 2*2, 2*2。(这里没有计算范数,实际上应该除以范数)

a_ = a[0, :2, :2, :2]
b_ = b[0, :2, :2, :2]

print(a_.shape, b_.shape)

a_ = a_.reshape(1, 2, 2*2)  # N,C,2*2
b_ = b_.reshape(1, 2, 2*2)

print("torch.matmul(a_.t, a_)=>\n", torch.matmul(a_.transpose(1, 2), a_))
print("torch.matmul(b_.t, b_)=>\n", torch.matmul(b_.transpose(1, 2), b_))

print(torch.matmul(a_.transpose(1, 2), a_)[0] == mm_unfold_a[0, 0])
print(torch.matmul(b_.transpose(1, 2), b_)[0] == mm_unfold_b[0, 0])
torch.Size([2, 2, 2]) torch.Size([2, 2, 2])
torch.matmul(a_.t, a_)=>
 tensor([[[0.2619, 0.6275, 0.5329, 0.2181],
         [0.6275, 1.7462, 1.5898, 0.7771],
         [0.5329, 1.5898, 1.4878, 0.7720],
         [0.2181, 0.7771, 0.7720, 0.4487]]])
torch.matmul(b_.t, b_)=>
 tensor([[[0.1355, 0.3471, 0.0724, 0.1180],
         [0.3471, 0.8891, 0.1805, 0.3017],
         [0.0724, 0.1805, 0.1869, 0.0756],
         [0.1180, 0.3017, 0.0756, 0.1037]]])
tensor([[True, True, True, True],
        [True, True, True, True],
        [True, True, True, True],
        [True, True, True, True]])
tensor([[True, True, True, True],
        [True, True, True, True],
        [True, True, True, True],
        [True, True, True, True]])

从这里可以看出来,通过fold、reshape(view)、matmul实现了对于N,C,H,W形状的数据的局部(这里对应为滑窗操作的kernel_size)关联矩阵的计算,而且速度又快(相较于最原始朴素的“滑窗式”计算方法)。

对于运算过程代码的书写,这里验证了一个想法,简单的按照矩阵的维度匹配的原则,是可以直接写出来这个局部关系矩阵的

N,C,H,W --(Ws*Ws)--> 
N,C*Ws*Ws,H/Ws*W/Ws --> 
N,H/Ws*W/Ws,C*Ws*Ws --> 
N,H/Ws*W/Ws,C*Ws*Ws --> 
N,H/Ws*W/Ws,C,Ws*Ws --> 
N,H/Ws*W/Ws,Ws*Ws,Ws*Ws

这里的H/Ws*W/Ws实际上反映出来的是分块的数量,这里直接使用除法对应的是滑窗大小正好可以被数据长宽整除,同时步长等于滑窗大小,没有padding的情况。

前面给出的代码中可以看出来,这里的值对于步长为1的时候,是需要进行调整的。

unfold_func = nn.Unfold(2, 1, 0, 1)
...
unfold_a_reshape = unfold_a.transpose(1, 2).view(1, (3-1)*(4-1), 2, 4)
posted @ 2019-09-25 12:17  lart  阅读(2100)  评论(0编辑  收藏  举报