卷积自动编码器的一个构造解码器如何?假设我有这个
(input -> conv2d -> maxpool2d -> maxunpool2d -> convTranspose2d -> output)
:
# CIFAR images shape = 3 x 32 x 32
class ConvDAE(nn.Module):
def __init__(self):
super().__init__()
# input: batch x 3 x 32 x 32 -> output: batch x 16 x 16 x 16
self.encoder = nn.Sequential(
nn.Conv2d(3, 16, 3, stride=1, padding=1), # batch x 16 x 32 x 32
nn.ReLU(),
nn.BatchNorm2d(16),
nn.MaxPool2d(2, stride=2) # batch x 16 x 16 x 16
)
# input: batch x 16 x 16 x 16 -> output: batch x 3 x 32 x 32
self.decoder = nn.Sequential(
# this line does not work
# nn.MaxUnpool2d(2, stride=2, padding=0), # batch x 16 x 32 x 32
nn.ConvTranspose2d(16, 16, 3, stride=2, padding=1, output_padding=1), # batch x 16 x 32 x 32
nn.ReLU(),
nn.BatchNorm2d(16),
nn.ConvTranspose2d(16, 3, 3, stride=1, padding=1, output_padding=0), # batch x 3 x 32 x 32
nn.ReLU()
)
def forward(self, x):
print(x.size())
out = self.encoder(x)
print(out.size())
out = self.decoder(out)
print(out.size())
return out
pytorch特定问题:为什么我不能在解码器部分中使用maxunpool2d。这给了我以下错误:
TypeError: forward() missing 1 required positional argument: 'indices'
和概念问题:我们不应该在解码器中进行倒数,而不是我们在编码器中所做的一切?我看到了一些实现,看来他们只关心解码器的输入和输出的尺寸。这里和这里有一些示例。
对于问题的火炬部分,UndOl模块作为所需的位置参数,从池化模块中返回的索引将以return_indices=True
返回。所以你可以做
class ConvDAE(nn.Module):
def __init__(self):
super().__init__()
# input: batch x 3 x 32 x 32 -> output: batch x 16 x 16 x 16
self.encoder = nn.Sequential(
nn.Conv2d(3, 16, 3, stride=1, padding=1), # batch x 16 x 32 x 32
nn.ReLU(),
nn.BatchNorm2d(16),
nn.MaxPool2d(2, stride=2, return_indices=True)
)
self.unpool = nn.MaxUnpool2d(2, stride=2, padding=0)
self.decoder = nn.Sequential(
nn.ConvTranspose2d(16, 16, 3, stride=2, padding=1, output_padding=1),
nn.ReLU(),
nn.BatchNorm2d(16),
nn.ConvTranspose2d(16, 3, 3, stride=1, padding=1, output_padding=0),
nn.ReLU()
)
def forward(self, x):
print(x.size())
out, indices = self.encoder(x)
out = self.unpool(out, indices)
out = self.decoder(out)
print(out.size())
return out
至于问题的一般部分,我认为最新的状态不是使用对称解码器部分,因为已经表明,Devonvolution/transped卷积会产生棋盘板效应,许多方法倾向于使用UPSMPLING模块反而。您将通过Pytorch频道更快地找到更多信息。