如何在pytorch中为Fashion_MNIST使用MSELoss函数？

Question

我想通过 Fashion_Mnist 数据，我想看看输出梯度可能是第一层和第二层之间的均方和

下面是我的代码

#import the nescessary libs
import numpy as np
import torch
import time

# 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")
torch.cuda.get_device_name(0)


# Define a transform to normalize the data
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 = 128, shuffle = True, num_workers=4)
testloader = torch.utils.data.DataLoader(testset, batch_size = 128, 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 = [], []
squared_sum = []
# start = time.time()
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)
        


        optimizer.zero_grad()
    
        output = model[0].forward(images)
        loss = criterion(output[0], labels.float())
        
        loss.backward()
        
        
             
        
        optimizer.step()
        running_loss += loss.item()
    
    else:

        print(running_loss)
        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[0])
                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)))

我想得到什么，

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)


        optimizer.zero_grad()
    
        output = model[0].forward(images)
        loss = criterion(output[0], labels.float())
        
        loss.backward()
                
        optimizer.step()
        running_loss += loss.item()

在这里，model[0]（这可能是第一层 nn.Linear(784, 128)），我希望只获得第一层和第二层的均方误差，

如果我运行此代码，我会在下面收到此错误

RuntimeError: The size of tensor a (128) must match the size of tensor b (96) at non-singleton dimension 0

如果我想正确运行这段代码来获得 MSELoss，我需要做什么？

Answer 1

错误是由数据集中的样本数量和批量大小引起的。

更详细地说，训练 MNIST 数据集包括 60,000 个样本，您当前的 batch_size 为 128，您将需要 60000/128=468.75 循环来完成一个 epoch 的训练。所以问题就出在这里，对于 468 个循环，您的数据将有 128 个样本，但最后一个循环仅包含 60000 - 468*128 = 96 样本。

要解决这个问题，我认为你需要找到合适的batch_size 以及模型中的神经元数量。

我认为它应该适用于计算损失

trainloader = torch.utils.data.DataLoader(trainset, batch_size = 96, shuffle = True, num_workers=0)
testloader = torch.utils.data.DataLoader(testset, batch_size = 96, shuffle = True, num_workers=0)
model = nn.Sequential(nn.Linear(784, 96),
                      nn.ReLU(),
                      nn.Linear(96, 10),
                      nn.LogSoftmax(dim = 1)
                     )