分步示例

Question

我正在研究一个神经网络问题，将数据分类为1或0。我正在使用二进制交叉熵损失进行此操作。损耗很好，但是精度非常低并且没有提高。我假设我在准确性计算中犯了一个错误。在每个时期之后，我在对输出进行阈值处理之后计算正确的预测，然后将该数字除以数据集的总数。我在精度计算中做错了什么吗？为什么它没有改善，却变得更糟？这是我的代码：

net = Model()
criterion = torch.nn.BCELoss(size_average=True)   
optimizer = torch.optim.SGD(net.parameters(), lr=0.1)

num_epochs = 100
for epoch in range(num_epochs):
    for i, (inputs,labels) in enumerate (train_loader):
        inputs = Variable(inputs.float())
        labels = Variable(labels.float())
        output = net(inputs)
        optimizer.zero_grad()
        loss = criterion(output, labels)
        loss.backward()
        optimizer.step()

    #Accuracy
    output = (output>0.5).float()
    correct = (output == labels).float().sum()
    print("Epoch {}/{}, Loss: {:.3f}, Accuracy: {:.3f}".format(epoch+1,num_epochs, loss.data[0], correct/x.shape[0]))

这是我得到的奇怪输出：

Epoch 1/100, Loss: 0.389, Accuracy: 0.035
Epoch 2/100, Loss: 0.370, Accuracy: 0.036
Epoch 3/100, Loss: 0.514, Accuracy: 0.030
Epoch 4/100, Loss: 0.539, Accuracy: 0.030
Epoch 5/100, Loss: 0.583, Accuracy: 0.029
Epoch 6/100, Loss: 0.439, Accuracy: 0.031
Epoch 7/100, Loss: 0.429, Accuracy: 0.034
Epoch 8/100, Loss: 0.408, Accuracy: 0.035
Epoch 9/100, Loss: 0.316, Accuracy: 0.035
Epoch 10/100, Loss: 0.436, Accuracy: 0.035
Epoch 11/100, Loss: 0.365, Accuracy: 0.034
Epoch 12/100, Loss: 0.485, Accuracy: 0.031
Epoch 13/100, Loss: 0.392, Accuracy: 0.033
Epoch 14/100, Loss: 0.494, Accuracy: 0.030
Epoch 15/100, Loss: 0.369, Accuracy: 0.035
Epoch 16/100, Loss: 0.495, Accuracy: 0.029
Epoch 17/100, Loss: 0.415, Accuracy: 0.034
Epoch 18/100, Loss: 0.410, Accuracy: 0.035
Epoch 19/100, Loss: 0.282, Accuracy: 0.038
Epoch 20/100, Loss: 0.499, Accuracy: 0.031
Epoch 21/100, Loss: 0.446, Accuracy: 0.030
Epoch 22/100, Loss: 0.585, Accuracy: 0.026
Epoch 23/100, Loss: 0.419, Accuracy: 0.035
Epoch 24/100, Loss: 0.492, Accuracy: 0.031
Epoch 25/100, Loss: 0.537, Accuracy: 0.031
Epoch 26/100, Loss: 0.439, Accuracy: 0.033
Epoch 27/100, Loss: 0.421, Accuracy: 0.035
Epoch 28/100, Loss: 0.532, Accuracy: 0.034
Epoch 29/100, Loss: 0.234, Accuracy: 0.038
Epoch 30/100, Loss: 0.492, Accuracy: 0.027
Epoch 31/100, Loss: 0.407, Accuracy: 0.035
Epoch 32/100, Loss: 0.305, Accuracy: 0.038
Epoch 33/100, Loss: 0.663, Accuracy: 0.025
Epoch 34/100, Loss: 0.588, Accuracy: 0.031
Epoch 35/100, Loss: 0.329, Accuracy: 0.035
Epoch 36/100, Loss: 0.474, Accuracy: 0.033
Epoch 37/100, Loss: 0.535, Accuracy: 0.031
Epoch 38/100, Loss: 0.406, Accuracy: 0.033
Epoch 39/100, Loss: 0.513, Accuracy: 0.030
Epoch 40/100, Loss: 0.593, Accuracy: 0.030
Epoch 41/100, Loss: 0.265, Accuracy: 0.036
Epoch 42/100, Loss: 0.576, Accuracy: 0.031
Epoch 43/100, Loss: 0.565, Accuracy: 0.027
Epoch 44/100, Loss: 0.576, Accuracy: 0.030
Epoch 45/100, Loss: 0.396, Accuracy: 0.035
Epoch 46/100, Loss: 0.423, Accuracy: 0.034
Epoch 47/100, Loss: 0.489, Accuracy: 0.033
Epoch 48/100, Loss: 0.591, Accuracy: 0.029
Epoch 49/100, Loss: 0.415, Accuracy: 0.034
Epoch 50/100, Loss: 0.291, Accuracy: 0.039
Epoch 51/100, Loss: 0.395, Accuracy: 0.033
Epoch 52/100, Loss: 0.540, Accuracy: 0.026
Epoch 53/100, Loss: 0.436, Accuracy: 0.033
Epoch 54/100, Loss: 0.346, Accuracy: 0.036
Epoch 55/100, Loss: 0.519, Accuracy: 0.029
Epoch 56/100, Loss: 0.456, Accuracy: 0.031
Epoch 57/100, Loss: 0.425, Accuracy: 0.035
Epoch 58/100, Loss: 0.311, Accuracy: 0.039
Epoch 59/100, Loss: 0.406, Accuracy: 0.034
Epoch 60/100, Loss: 0.360, Accuracy: 0.035
Epoch 61/100, Loss: 0.476, Accuracy: 0.030
Epoch 62/100, Loss: 0.404, Accuracy: 0.034
Epoch 63/100, Loss: 0.382, Accuracy: 0.036
Epoch 64/100, Loss: 0.538, Accuracy: 0.031
Epoch 65/100, Loss: 0.392, Accuracy: 0.034
Epoch 66/100, Loss: 0.434, Accuracy: 0.033
Epoch 67/100, Loss: 0.479, Accuracy: 0.031
Epoch 68/100, Loss: 0.494, Accuracy: 0.031
Epoch 69/100, Loss: 0.415, Accuracy: 0.034
Epoch 70/100, Loss: 0.390, Accuracy: 0.036
Epoch 71/100, Loss: 0.330, Accuracy: 0.038
Epoch 72/100, Loss: 0.449, Accuracy: 0.030
Epoch 73/100, Loss: 0.315, Accuracy: 0.039
Epoch 74/100, Loss: 0.450, Accuracy: 0.031
Epoch 75/100, Loss: 0.562, Accuracy: 0.030
Epoch 76/100, Loss: 0.447, Accuracy: 0.031
Epoch 77/100, Loss: 0.408, Accuracy: 0.038
Epoch 78/100, Loss: 0.359, Accuracy: 0.034
Epoch 79/100, Loss: 0.372, Accuracy: 0.035
Epoch 80/100, Loss: 0.452, Accuracy: 0.034
Epoch 81/100, Loss: 0.360, Accuracy: 0.035
Epoch 82/100, Loss: 0.453, Accuracy: 0.031
Epoch 83/100, Loss: 0.578, Accuracy: 0.030
Epoch 84/100, Loss: 0.537, Accuracy: 0.030
Epoch 85/100, Loss: 0.483, Accuracy: 0.035
Epoch 86/100, Loss: 0.343, Accuracy: 0.036
Epoch 87/100, Loss: 0.439, Accuracy: 0.034
Epoch 88/100, Loss: 0.686, Accuracy: 0.023
Epoch 89/100, Loss: 0.265, Accuracy: 0.039
Epoch 90/100, Loss: 0.369, Accuracy: 0.035
Epoch 91/100, Loss: 0.521, Accuracy: 0.027
Epoch 92/100, Loss: 0.662, Accuracy: 0.027
Epoch 93/100, Loss: 0.581, Accuracy: 0.029
Epoch 94/100, Loss: 0.322, Accuracy: 0.034
Epoch 95/100, Loss: 0.375, Accuracy: 0.035
Epoch 96/100, Loss: 0.575, Accuracy: 0.031
Epoch 97/100, Loss: 0.489, Accuracy: 0.030
Epoch 98/100, Loss: 0.435, Accuracy: 0.033
Epoch 99/100, Loss: 0.440, Accuracy: 0.031
Epoch 100/100, Loss: 0.444, Accuracy: 0.033

Answer 1

我认为最简单的答案是来自the cifar10 tutorial的答案：

total = 0
with torch.no_grad():
    net.eval()
    for data in testloader:
        images, labels = data
        outputs = net(images)
        _, predicted = torch.max(outputs.data, 1)
        total += labels.size(0)
        correct += (predicted == labels).sum().item()

print('Accuracy of the network on the 10000 test images: %d %%' % (
    100 * correct / total))

如此：

acc = (true == pred).sum().item()

如果您有计数器，别忘了最终要除以数据集或类似值的大小。

我用过：

N = data.size(0) # since usually it's size (batch_size, D1, D2, ...)
correct += (1/N) * correct

自包含代码：

# testing accuracy function
# https://discuss.pytorch.org/t/calculating-accuracy-of-the-current-minibatch/4308/11
# https://stackoverflow.com/questions/51503851/calculate-the-accuracy-every-epoch-in-pytorch

import torch
import torch.nn as nn

D = 1
true = torch.tensor([0,1,0,1,1]).reshape(5,1)
print(f'true.size() = {true.size()}')

batch_size = true.size(0)
print(f'batch_size = {batch_size}')
x = torch.randn(batch_size,D)
print(f'x = {x}')
print(f'x.size() = {x.size()}')

mdl = nn.Linear(D,1)
logit = mdl(x)
_, pred = torch.max(logit.data, 1)

print(f'logit = {logit}')

print(f'pred = {pred}')
print(f'true = {true}')

acc = (true == pred).sum().item()
print(f'acc = {acc}')

此外，我发现这段代码是很好的参考：

def calc_accuracy(mdl, X, Y):
    # reduce/collapse the classification dimension according to max op
    # resulting in most likely label
    max_vals, max_indices = mdl(X).max(1)
    # assumes the first dimension is batch size
    n = max_indices.size(0)  # index 0 for extracting the # of elements
    # calulate acc (note .item() to do float division)
    acc = (max_indices == Y).sum().item() / n
    return acc

解释pred = mdl(x).max(1)见此https://discuss.pytorch.org/t/how-does-one-get-the-predicted-classification-label-from-a-pytorch-model/91649

最主要的是，您必须缩小/折叠分类原始值/ logit具有最大值的维，然后使用.indices选择它。通常这是尺寸1，因为dim 0具有批量大小，例如[batch_size,D_classification]，其中原始数据的大小可能为[batch_size,C,H,W]

具有一维原始数据的合成示例，如下所示：

import torch
import torch.nn as nn

# data dimension [batch-size, D]
D, Dout = 1, 5
batch_size = 16
x = torch.randn(batch_size, D)
y = torch.randint(low=0,high=Dout,size=(batch_size,))

mdl = nn.Linear(D, Dout)
logits = mdl(x)
print(f'y.size() = {y.size()}')
# removes the 1th dimension with a max, which is the classification layer
# which means it returns the most likely label. Also, note you need to choose .indices since you want to return the
# position of where the most likely label is (not it's raw logit value)
pred = logits.max(1).indices
print(pred)

print('--- preds vs truth ---')
print(f'predictions = {pred}')
print(f'y = {y}')

acc = (pred == y).sum().item() / pred.size(0)
print(acc)

输出：


y.size() = torch.Size([16])
tensor([3, 1, 1, 3, 4, 1, 4, 3, 1, 1, 4, 4, 4, 4, 3, 1])
--- preds vs truth ---
predictions = tensor([3, 1, 1, 3, 4, 1, 4, 3, 1, 1, 4, 4, 4, 4, 3, 1])
y = tensor([3, 3, 3, 0, 3, 4, 0, 1, 1, 2, 1, 4, 4, 2, 0, 0])
0.25

参考：

Answer 2

一种更好的方法是在优化步骤之后立即计算正确

for epoch in range(num_epochs):

    correct = 0
    for i, (inputs,labels) in enumerate (train_loader):
        ...
        output = net(inputs)
        ...
        optimizer.step()

        correct += (output == labels).float().sum()

    accuracy = 100 * correct / len(trainset)
    # trainset, not train_loader
    # probably x in your case

    print("Accuracy = {}".format(accuracy))

Answer 3

这是我的解决方案：

def evaluate(model, validation_loader, use_cuda=True):
    model.eval()        
    with torch.no_grad():
        acc = .0
        for i, data in enumerate(validation_loader):
            X = data[0]
            y = data[1]
            if use_cuda:
                X = X.cuda()
                y = y.cuda()
            predicted = model(X)
            acc+=(predicted.round() == y).sum()/float(predicted.shape[0])       
    model.train()
    return (acc/(i+1)).detach().item()

注意1：：在验证时将模型设置为eval模式，然后返回到train模式。

注意2：：我不确定是否需要禁用autograd。这是thread on it

对于单项结果，可以使用torch.max。 Example：

correct = 0
total = 0
with torch.no_grad():
    for data in testloader:
        images, labels = data
        outputs = net(images)
        _, predicted = torch.max(outputs.data, 1)
        total += labels.size(0)
        correct += (predicted == labels).sum().item()

print('Accuracy of the network on the 10000 test images: %d %%' % (
    100 * correct / total))

Answer 4

让我们看一下基础知识：

  Accuracy = Total Correct Observations / Total Observations

在代码中，当您计算准确度时，您将一个时期内的总正确观测值除以不正确的总观测值

correct/x.shape[0]

相反，您应该将其除以每个时期的观察次数，即批量大小。假设您的批量大小= batch_size

Solution 1. Accuracy = correct/batch_size
Solution 2. Accuracy = correct/len(labels)
Solution 3. Accuracy = correct/len(input)

理想情况下，在每个时期，您的批处理大小，输入长度（行数）和标签长度应相同。

Answer 5

一个衬里以获得准确性

acc == (true == mdl(x).max(1).item() / true.size(0)

假设第0维是批次大小，第1维是分类标签的对数/原始值。

更多详细信息：

def calc_accuracy(mdl, X, Y):
    # reduce/collapse the classification dimension according to max op
    # resulting in most likely label
    max_vals, max_indices = mdl(X).max(1)
    # assumes the first dimension is batch size
    n = max_indices.size(0)  # index 0 for extracting the # of elements
    # calulate acc (note .item() to do float division)
    acc = (max_indices == Y).sum().item() / n
    return acc

Answer 6

在这里检查这些定义：

def train(model, train_loader):
model.train()
    train_acc, correct_train, train_loss, target_count = 0, 0, 0, 0
    for i, (input, target) in enumerate(train_loader):
        target = target.cuda()
        input_var = Variable(input)
        target_var = Variable(target)

        optimizer.zero_grad()
        output = model(input_var)
        loss = criterion(output, target_var)
        loss.backward()
        optimizer.step()

        # accuracy
        _, predicted = torch.max(output.data, 1)
        target_count += target_var.size(0)
        correct_train += (target_var == predicted).sum().item()
        train_acc = (100 * correct_train) / target_count
    return train_acc, train_loss / target_count


def validate(model, val_loader):
    model.eval()
    val_acc, correct_val, val_loss, target_count = 0, 0, 0, 0
    for i, (input, target) in enumerate(val_loader):
        target = target.cuda()
        input_var = Variable(input, volatile=True)
        target_var = Variable(target, volatile=True)
        output = model(input_var)
        loss = criterion(output, target_var)
        val_loss += loss.item()

        # accuracy
        _, predicted = torch.max(output.data, 1)
        target_count += target_var.size(0)
        correct_val += (target_var == predicted).sum().item()
        val_acc = 100 * correct_val / target_count
    return (val_acc * 100) / target_count, val_loss / target_count                            

for epoch in range(0, n_epoch):
    train_acc, train_loss = train(model, train_loader)
    val_loss = validate(model, val_loader)
    print("Epoch {0}: train_acc {1} \t train_loss {2} \t val_acc {3} \t val_loss {4}".format(epoch, train_acc, train_loss, val_acc, val_loss))

Answer 7

x是整个输入数据集吗？如果是这样，您可能要除以correct/x.shape[0]中整个输入数据集的大小（而不是迷你批处理的大小）。尝试将其更改为correct/output.shape[0]

Answer 8

只需阅读此答案：

https://stackoverflow.com/a/63271002/1601580

分步示例

这里以自包含代码为例一步步解释：

#%%

# refs:
# https://stackoverflow.com/questions/51503851/calculate-the-accuracy-every-epoch-in-pytorch
# https://discuss.pytorch.org/t/how-to-calculate-accuracy-in-pytorch/80476/5
# https://discuss.pytorch.org/t/how-does-one-get-the-predicted-classification-label-from-a-pytorch-model/91649

# how to get the class prediction

batch_size = 4
n_classes = 2
y_logits = torch.randn(batch_size, n_classes)  # usually the scores
print('scores (logits) for each class for each example in batch (how likely a class is unnormalized)')
print(y_logits)
print('the max over entire tensor (not usually what we want)')
print(y_logits.max())
print('the max over the n_classes dim. For each example in batch returns: '
      '1) the highest score for each class (most likely class)\n, and '
      '2) the idx (=class) with that highest score')
print(y_logits.max(1))

print('-- calculate accuracy --')

# computing accuracy in pytorch
"""
random.choice(a, size=None, replace=True, p=None)
Generates a random sample from a given 1-D array

for pytorch random choice https://stackoverflow.com/questions/59461811/random-choice-with-pytorch
"""

import torch
import torch.nn as nn

in_features = 1
n_classes = 10
batch_size = n_classes

mdl = nn.Linear(in_features=in_features, out_features=n_classes)

x = torch.randn(batch_size, in_features)
y_logits = mdl(x)  # scores/logits for each example in batch [B, n_classes]
# get for each example in batch the label/idx most likely according to score
# y_max_idx[b] = y_pred[b] = argmax_{idx \in [n_classes]} y_logit[idx]
y_max_scores, y_max_idx = y_logits.max(dim=1)
y_pred = y_max_idx  # predictions are really the inx \in [n_classes] with the highest scores
y = torch.randint(high=n_classes, size=(batch_size,))
# accuracy for 1 batch
assert (y.size(0) == batch_size)
acc = (y == y_pred).sum() / y.size(0)
acc = acc.item()

print(y)
print(y_pred)
print(acc)

输出：

scores (logits) for each class for each example in batch (how likely a class is unnormalized)
tensor([[ 0.4912,  1.5143],
        [ 1.2378,  0.3172],
        [-1.0164, -1.2786],
        [-1.6685, -0.6693]])
the max over entire tensor (not usually what we want)
tensor(1.5143)
the max over the n_classes dim. For each example in batch returns: 1) the highest score for each class (most likely class)
, and 2) the idx (=class) with that highest score
torch.return_types.max(
values=tensor([ 1.5143,  1.2378, -1.0164, -0.6693]),
indices=tensor([1, 0, 0, 1]))
-- calculate accuracy --
tensor([6, 1, 3, 5, 3, 9, 6, 5, 6, 6])
tensor([5, 5, 5, 5, 5, 7, 7, 5, 5, 7])
0.20000000298023224

计算PyTorch中每个时期的准确性

8 个答案:

分步示例