问题定义:
我不得不用MSELoss函数来定义损失分类问题。因此,它一直在说关于张量形状的错误信息。
完整错误消息:
torch.Size(32,10)跟踪(最近一次调用)在53输出=模型中的torch.Size(32).forward(图像) 54打印(output.shape,labels.shape) -> 55损失=标准(输出,标签) 56 loss.backward() 57 optimizer.step()
/opt/conda/lib/python3.7/site-packages/torch/nn/modules/module.py in call(self,*input,**kwargs) 530结果=self._slow_forward(*输入,**kwargs) 531其他:-> 532 =self.forward(*输入,**kwargs) 533用于self._forward_hooks.values()中的钩子: 534 hook_result =钩子(self,input,result)
/opt/conda/lib/python3.7/site-packages/torch/nn/modules/loss.py在前进(自我,输入,目标)429430前进(自我,输入,目标):-> 431返回F.mse_loss(输入,目标,reduction=self.reduction) 432 433
/opt/conda/lib/python3.7/site-packages/torch/nn/functional.py in mse_loss(输入、目标、size_average、缩减、缩减) 2213
ret = torch.mean(ret)如果还原==‘意味着’其他torch.sum(ret)
其它:-> 2215 expanded_input,expanded_target =torch.broadcast_tensors(输入,目标) 2216 ret = torch._C._nn.mse_loss(expanded_input,expanded_target,_Reduction.get_enum(还原))
/opt/conda/lib/python3.7/site-packages/torch/functional.py in broadcast_tensors(*张量) 50 0,1,2]) 51“”--> 52返回torch._C._VariableFunctions.broadcast_tensors(tensors) 53 54
> RuntimeError:张量a (10)的大小必须与非单例维的张量b (32)的大小相匹配。
我如何重塑张量,我应该改变哪一个张量(输出或标签)来计算损失?
整个代码附在下面。
import numpy as np
import torch
# Loading the Fashion-MNIST dataset
from torchvision import datasets, transforms
# Get GPU Device
device = torch.device("cuda:0" if torch.cuda.is_available() else "cpu")
transform = transforms.Compose([transforms.ToTensor(),
transforms.Normalize((0.5,), (0.5,))])
# Download and load the training data
trainset = datasets.FashionMNIST('MNIST_data/', download = True, train = True, transform = transform)
testset = datasets.FashionMNIST('MNIST_data/', download = True, train = False, transform = transform)
trainloader = torch.utils.data.DataLoader(trainset, batch_size = 32, shuffle = True, num_workers=4)
testloader = torch.utils.data.DataLoader(testset, batch_size = 32, shuffle = True, num_workers=4)
# Examine a sample
dataiter = iter(trainloader)
images, labels = dataiter.next()
# Define the network architecture
from torch import nn, optim
import torch.nn.functional as F
model = nn.Sequential(nn.Linear(784, 128),
nn.ReLU(),
nn.Linear(128, 10),
nn.LogSoftmax(dim = 1))
model.to(device)
# Define the loss
criterion = nn.MSELoss()
# Define the optimizer
optimizer = optim.Adam(model.parameters(), lr = 0.001)
# Define the epochs
epochs = 5
train_losses, test_losses = [], []
for e in range(epochs):
running_loss = 0
for images, labels in trainloader:
# Flatten Fashion-MNIST images into a 784 long vector
images = images.to(device)
labels = labels.to(device)
images = images.view(images.shape[0], -1)
# Training pass
optimizer.zero_grad()
output = model.forward(images)
print(output.shape, labels.shape)
loss = criterion(output, labels)
loss.backward()
optimizer.step()
running_loss += loss.item()
else:
test_loss = 0
accuracy = 0
# Turn off gradients for validation, saves memory and computation
with torch.no_grad():
# Set the model to evaluation mode
model.eval()
# Validation pass
for images, labels in testloader:
images = images.to(device)
labels = labels.to(device)
images = images.view(images.shape[0], -1)
ps = model(images)
test_loss += criterion(ps, labels)
top_p, top_class = ps.topk(1, dim = 1)
equals = top_class == labels.view(*top_class.shape)
accuracy += torch.mean(equals.type(torch.FloatTensor))
model.train()
print("Epoch: {}/{}..".format(e+1, epochs),
"Training loss: {:.3f}..".format(running_loss/len(trainloader)),
"Test loss: {:.3f}..".format(test_loss/len(testloader)),
"Test Accuracy: {:.3f}".format(accuracy/len(testloader)))发布于 2020-06-15 08:00:36
从您在其错误之前打印的输出,torch.Size([32, 10]) torch.Size([32])。
左边的是模型给你的,右边的是trainloader的,通常你用它来做nn.CrossEntropyLoss之类的事情。
从完整的错误日志中,错误来自以下一行
loss = criterion(output, labels)做这个工作的方法叫做“一热编码”,如果是因为我的懒惰,我会写成这样。
ones = torch.sparse.torch.eye(10).to(device) # number of class class
labels = ones.index_select(0, labels)发布于 2021-09-22 22:34:09
或者,您可以将丢失函数从nn.MSELoss()更改为nn.CrossEntropyLoss()。交叉熵损失通常比MSE更适合于这类分类任务,而且在PyTorch的实现中,这个丢失函数负责遮罩下的许多形状转换,因此您可以为它提供一个类概率向量和一个类标签。
从根本上说,您的模型试图通过计算每个可能的类的分数(您可以称之为“信任分数”)来预测输入属于哪个类。因此,如果有10个类,模型的输出将是一个10维列表(在PyTorch中是张量形状的[10]),预测将是得分最高的指标。通常情况下,人们会应用softmax (https://en.wikipedia.org/wiki/Softmax_function)函数将这些分数转换为概率分布,所以所有分数都在0到1之间,所有的元素都等于1。
然后,交叉熵是此任务损失函数的一个常见选择:它将预测列表与单一热编码标签进行比较。例如,如果您有3个类,标签将类似于[1, 0, 0]来表示第一个类。这也被称为“一热编码”。同时,一个预测可能看起来像[0.7, 0.1, 0.2]。在PyTorch中,nn.CrossEntropyLoss()期望标签作为单个值张量出现,其值表示类标签,因为不需要在内存中移动长而稀疏的向量。因此,这个丢失函数完成了您想要做的比较,我猜它的实现比实际创建一个热编码更有效。
https://stackoverflow.com/questions/62383595
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