各位编码员大家好，
我正在使用完全卷积的自动编码器为黑白图像着色，但是，输出有一个checkerboard pattern，我想摆脱它。到目前为止，我所看到的棋盘状工件都比我的小得多，而摆脱它们的通常方法是用双线性上采样代替所有解池操作（有人告诉我）。
但是我不能简单地替换unpooling操作，因为我使用不同大小的图像，因此需要unpooling操作，否则输出张量的大小可能会与原始张量不同。
TLDR：
如何在不替换分拆操作的情况下摆脱这些棋盘工件？
class AE(nn.Module):
def __init__(self):
super(AE, self).__init__()
self.leaky_reLU = nn.LeakyReLU(0.2)
self.pool = nn.MaxPool2d(kernel_size=2, stride=2, padding=1, return_indices=True)
self.unpool = nn.MaxUnpool2d(kernel_size=2, stride=2, padding=1)
self.softmax = nn.Softmax2d()
self.conv1 = nn.Conv2d(in_channels=3, out_channels=64, kernel_size=3, stride=1, padding=1)
self.conv2 = nn.Conv2d(in_channels=64, out_channels=128, kernel_size=3, stride=1, padding=1)
self.conv3 = nn.Conv2d(in_channels=128, out_channels=256, kernel_size=3, stride=1, padding=1)
self.conv4 = nn.Conv2d(in_channels=256, out_channels=512, kernel_size=3, stride=1, padding=1)
self.conv5 = nn.Conv2d(in_channels=512, out_channels=1024, kernel_size=3, stride=1, padding=1)
self.conv6 = nn.ConvTranspose2d(in_channels=1024, out_channels=512, kernel_size=3, stride=1, padding=1)
self.conv7 = nn.ConvTranspose2d(in_channels=512, out_channels=256, kernel_size=3, stride=1, padding=1)
self.conv8 = nn.ConvTranspose2d(in_channels=256, out_channels=128, kernel_size=3, stride=1, padding=1)
self.conv9 = nn.ConvTranspose2d(in_channels=128, out_channels=64, kernel_size=3, stride=1, padding=1)
self.conv10 = nn.ConvTranspose2d(in_channels=64, out_channels=2, kernel_size=3, stride=1, padding=1)
def forward(self, x):
# encoder
x = self.conv1(x)
x = self.leaky_reLU(x)
size1 = x.size()
x, indices1 = self.pool(x)
x = self.conv2(x)
x = self.leaky_reLU(x)
size2 = x.size()
x, indices2 = self.pool(x)
x = self.conv3(x)
x = self.leaky_reLU(x)
size3 = x.size()
x, indices3 = self.pool(x)
x = self.conv4(x)
x = self.leaky_reLU(x)
size4 = x.size()
x, indices4 = self.pool(x)
######################
x = self.conv5(x)
x = self.leaky_reLU(x)
x = self.conv6(x)
x = self.leaky_reLU(x)
######################
# decoder
x = self.unpool(x, indices4, output_size=size4)
x = self.conv7(x)
x = self.leaky_reLU(x)
x = self.unpool(x, indices3, output_size=size3)
x = self.conv8(x)
x = self.leaky_reLU(x)
x = self.unpool(x, indices2, output_size=size2)
x = self.conv9(x)
x = self.leaky_reLU(x)
x = self.unpool(x, indices1, output_size=size1)
x = self.conv10(x)
x = self.softmax(x)
return x
。
编辑-解决方案：
跳过连接是必经之路！
class AE(nn.Module):
def __init__(self):
super(AE, self).__init__()
self.leaky_reLU = nn.LeakyReLU(0.2)
self.pool = nn.MaxPool2d(kernel_size=2, stride=2, padding=1, return_indices=True)
self.unpool = nn.MaxUnpool2d(kernel_size=2, stride=2, padding=1)
self.softmax = nn.Softmax2d()
self.conv1 = nn.Conv2d(in_channels=3, out_channels=64, kernel_size=3, stride=1, padding=1)
self.conv2 = nn.Conv2d(in_channels=64, out_channels=128, kernel_size=3, stride=1, padding=1)
self.conv3 = nn.Conv2d(in_channels=128, out_channels=256, kernel_size=3, stride=1, padding=1)
self.conv4 = nn.Conv2d(in_channels=256, out_channels=512, kernel_size=3, stride=1, padding=1)
self.conv5 = nn.Conv2d(in_channels=512, out_channels=1024, kernel_size=3, stride=1, padding=1)
self.conv6 = nn.Conv2d(in_channels=1024, out_channels=512, kernel_size=3, stride=1, padding=1)
self.conv7 = nn.Conv2d(in_channels=1024, out_channels=256, kernel_size=3, stride=1, padding=1)
self.conv8 = nn.Conv2d(in_channels=512, out_channels=128, kernel_size=3, stride=1, padding=1)
self.conv9 = nn.Conv2d(in_channels=256, out_channels=64, kernel_size=3, stride=1, padding=1)
self.conv10 = nn.Conv2d(in_channels=128, out_channels=2, kernel_size=3, stride=1, padding=1)
def forward(self, x):
# encoder
x = self.conv1(x)
out1 = self.leaky_reLU(x)
x = out1
size1 = x.size()
x, indices1 = self.pool(x)
x = self.conv2(x)
out2 = self.leaky_reLU(x)
x = out2
size2 = x.size()
x, indices2 = self.pool(x)
x = self.conv3(x)
out3 = self.leaky_reLU(x)
x = out3
size3 = x.size()
x, indices3 = self.pool(x)
x = self.conv4(x)
out4 = self.leaky_reLU(x)
x = out4
size4 = x.size()
x, indices4 = self.pool(x)
######################
x = self.conv5(x)
x = self.leaky_reLU(x)
x = self.conv6(x)
x = self.leaky_reLU(x)
######################
# decoder
x = self.unpool(x, indices4, output_size=size4)
x = self.conv7(torch.cat((x, out4), 1))
x = self.leaky_reLU(x)
x = self.unpool(x, indices3, output_size=size3)
x = self.conv8(torch.cat((x, out3), 1))
x = self.leaky_reLU(x)
x = self.unpool(x, indices2, output_size=size2)
x = self.conv9(torch.cat((x, out2), 1))
x = self.leaky_reLU(x)
x = self.unpool(x, indices1, output_size=size1)
x = self.conv10(torch.cat((x, out1), 1))
x = self.softmax(x)
return x
最佳答案
跳过连接通常在编码器-解码器体系结构中使用，它通过将外观信息从编码器（标识符）的浅层传递到解码器（发生器）的相应深层，来帮助产生准确的结果。 Unet是广泛使用的Encoder-Decoder类型的体系结构。 Linknet也非常流行，它与Unet的不同之处在于将编码器层的外观信息与解码器层融合在一起的方式。对于Unet，传入的特征（来自编码器）在相应的解码器层中串联在一起。另一方面，Linknet执行加法操作，这就是为什么Linknet在一次前向传递中需要较少数量的操作，并且比Unet快得多的原因。
您在Decoder中的每个卷积块可能如下所示：
另外，我在下面附上一个图，描述了Unet和LinkNet的体系结构。希望使用跳过连接会有所帮助。
关于python - 如何摆脱棋盘文物，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/59101333/