在我的火车上。py
criteon = nn.CrossEntropyLoss()
loss = criteon(binary_output_c1,labels)
optimizer.zero_grad()
loss.backward()
optimizer.step()
binary_output_c1,标签大小均为[4,224,224],4表示批处理大小,224表示h和w。它有这样的错误
Traceback (most recent call last):
File "D:\Anaconda3\lib\site-packages\IPython\core\interactiveshell.py", line 3296, in run_code
exec(code_obj, self.user_global_ns, self.user_ns)
File "<ipython-input-2-78553e2886de>", line 1, in <module>
runfile('F:/experiment_code/U-net/train.py', wdir='F:/experiment_code/U-net')
File "D:\pycharm\PyCharm Community Edition 2019.1.1\helpers\pydev\_pydev_bundle\pydev_umd.py", line 197, in runfile
pydev_imports.execfile(filename, global_vars, local_vars) # execute the script
File "D:\pycharm\PyCharm Community Edition 2019.1.1\helpers\pydev\_pydev_imps\_pydev_execfile.py", line 18, in execfile
exec(compile(contents+"\n", file, 'exec'), glob, loc)
File "F:/experiment_code/U-net/train.py", line 77, in <module>
loss = criteon(binary_output_c1,labels)
File "D:\Anaconda3\lib\site-packages\torch\nn\modules\module.py", line 493, in __call__
result = self.forward(*input, **kwargs)
File "D:\Anaconda3\lib\site-packages\torch\nn\modules\loss.py", line 942, in forward
ignore_index=self.ignore_index, reduction=self.reduction)
File "D:\Anaconda3\lib\site-packages\torch\nn\functional.py", line 2056, in cross_entropy
return nll_loss(log_softmax(input, 1), target, weight, None, ignore_index, None, reduction)
File "D:\Anaconda3\lib\site-packages\torch\nn\functional.py", line 1881, in nll_loss
out_size, target.size()))
ValueError: Expected target size (4, 224), got torch.Size([4, 224, 224])
我不知道是否可以在交叉熵损失中使用3D张量,该网络用于语义分割
如果使用nn.CrossEntropyLoss
,则您的预测应该有两个通道:一个用于预测nn.CrossEntropyLoss
,另一个用于预测0
。这有点多余,但是损失期望预测具有#channels == #labels。
或者,您可以在传递至亏损之前先进行预测:
1
我将标签的大小设置为[4,256,224,224],其中256是类别数。代码在这里
loss = criteon(torch.cat((-binary_output_c1[:, None, ...], binary_output_c1[:, None,...]), dim=1),labels)
也得到错误
model.train()
outputs = model(imgs) # output B * C * H *W
output_c1 = outputs[:,1,:,:] # 2 channels ,I choose the second channel
Rounding_output_c1 = torch.round(output_c1)
labelss = torch.stack([(labels == i).long() for i in range(256)])
labelss = labelss.permute(1,0,2,3)
Rounding_output_c11 = torch.stack([(Rounding_output_c1 == i).float() for i in range(256)])
Rounding_output_c11 = Rounding_output_c11.permute(1,0,2,3)
loss = criteon(Rounding_output_c11,labelss)
optimizer.zero_grad()
loss.backward()