这代码ok吗

[5464jmy]

x3 = torch.cat((x1, x2), dim=1)
b, n, h, w = x3 .size()
b_n = b * n // 2
y = x3 .reshape(b_n, 2, h * w)
y = y.permute(1, 0, 2)
y = y.reshape(2, -1, n // 2, h, w)
等于
y = torch.cat([x1[:, ::2], x2[:, ::2], x1[:, 1::2], x2[:, 1::2]], dim=1)
这与两个卷积的对应通道数挨在一起不符合

[alanli1997]

@5464jmy 请注意，这行代码“y = torch.cat([x1[:, ::2], x2[:, ::2], x1[:, 1::2], x2[:, 1::2]], dim=1)”并不等价于shuffle操作，“x1[:, ::2], x2[:, ::2], x1[:, 1::2], x2[:, 1::2]”是四个tensor的整体的串接，并不是各tensor内channel的交替串接。
欢迎继续讨论。

[5464jmy]

各tensor内channel的交替串接是怎样的呢？
假如是（16,32,40,40）大小的x1, x2， 第一个batch x1,x2大小分别为（32,40,40），假定其中channel的排列方式如图中一样，
shuffle表达的是图中的序号一的操作吗，

[alanli1997]

@5464jmy “shuffle”操作的结果是图中②的样子；“y = torch.cat([x1[:, ::2], x2[:, ::2], x1[:, 1::2], x2[:, 1::2]], dim=1)”的结果则是图中①的样子。随着channel的增加，①的shuffle强度会越来越弱，②的shuffle强度始终保持一致而不受channel数量变化的影响。

[5464jmy]

class GSConv(nn.Module):
# GSConv https://github.com/AlanLi1997/slim-neck-by-gsconv
def init(self, c1, c2, k=1, s=1, p=None, g=1, d=1, act=True):
super().init()
self.c_ = c2 // 2
self.cv1 = Conv(c1, self.c_, k, s, p, g, d, Conv.default_act)
self.cv2 = Conv(self.c_, self.c_, 5, 1, p, self.c_, d, Conv.default_act)

def forward(self, x):
    x1 = self.cv1(x)
    x2 = self.cv2(x1)
    x3 = torch.cat((x1, x2), dim=1)
    b, n, h, w = x3.size()
    b_n = b * n // 2
    y = x3.reshape(b_n, 2, h * w)
    y = y.permute(1, 0, 2)
    y = y.reshape(2, -1, n // 2, h, w)

    return (torch.cat((y[0], y[1]), 1), 
            torch.cat([x1[:, ::2], x2[:, ::2], x1[:, 1::2], x2[:, 1::2]], dim=1), 
            torch.cat([x1.unsqueeze(2), x2.unsqueeze(2)], dim=2).flatten(1, 2)

device = torch.device("cuda:0" if torch.cuda.is_available() else "cpu")
a = GSConv(16, 64, k=3, s=2).to(device)
b = torch.rand((16, 16, 40, 40), device=device)
c1, c2, c3 = a(b)
print(c1.size())
print((c1 == c2).sum())
print((c1 == c3).sum())

torch.Size([16, 64, 20, 20])
tensor(409600, device='cuda:0')
tensor(25600, device='cuda:0')

我无法解释

[alanli1997]

@5464jmy 我注意到了，你使用的shuffle操作和仓库中不一致，根据你使用shuffle操作(可能是最早期的GSConv的shuffle版本)来说确实和“torch.cat([x1[:, ::2], x2[:, ::2], x1[:, 1::2], x2[:, 1::2]], dim=1)”相同。
这是一段演示代码，运行后可以直观对比两种方法的差别，供你参考：

import torch
x1 = torch.randn(1,4,2,2)
x2 = torch.randn(1,4,2,2)

x3 = torch.cat((x1, x2), dim=1)
# method 1 (last version)
y1 = x3.reshape(x3.shape[0], 2, x3.shape[1] // 2, x3.shape[2], x3.shape[3])
y1 = y1.permute(0, 2, 1, 3, 4)
y1 = y1.reshape(y1.shape[0], -1, y1.shape[3], y1.shape[4])
# method 2， equation yours shuffle version
y2 = torch.cat((x1[:, ::2], x2[:, ::2], x1[:, 1::2], x2[:, 1::2]), dim=1)

print(x1)
print(x2)
print(y1)
print(y2)

print(y1.size())
print(y2.size())
print(y1==y2)

[5464jmy]

谢谢你的指正