Neural network 计算Pytork中每个历元的精度

我正在研究一个神经网络问题，将数据分类为1或0。我用二进制交叉熵损失来做这个。损失很小，但是，准确度很低，没有改善。我假设我在准确度计算中犯了一个错误。在每个历元之后，我在对输出进行阈值化并将该数字除以数据集的总数之后，计算正确的预测。我在精度计算中有没有做错什么？为什么情况没有改善，但越来越糟？ 这是我的代码：

net=Model（）
标准=torch.nn.BCELoss（尺寸平均值=True）
optimizer=torch.optim.SGD（net.parameters（），lr=0.1）
num_epochs=100
对于范围内的历元（num_历元）：
对于枚举（列车装载机）中的i（输入、标签）：
输入=变量（inputs.float（））
labels=变量（labels.float（））
输出=净（输入）
optimizer.zero_grad（）
损失=标准（输出、标签）
loss.backward（）
optimizer.step（）
#准确度
输出=（输出>0.5）.float（）
correct=（output==labels）.float（）.sum（）
打印（“历元{}/{}，丢失：{.3f}，精度：{.3f}”。格式（历元+1，历元数，丢失.data[0]，正确/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

---

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（与小批量的大小相反）。尝试将其更改为

correct/output.shape[0]

更好的方法是在优化步骤之后立即计算correct

用于范围内的历元（num_历元）：
正确=0
对于枚举（列车装载机）中的i（输入、标签）：
...
输出=净（输入）
...
optimizer.step（）
正确+=（输出==标签）.float（）.sum（）
精度=100*正确/长度（列车组）
#列车组，非列车装载机
#你的情况可能是x
打印（“精度={}”。格式（精度））

以下是我的解决方案：

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1:验证时将模型设置为模式，然后返回模式

注意2:我不确定是否需要禁用

自动加载。这是一个

例如，可以使用热结果：

请阅读以下答案：



古老的
我认为最简单的答案是：

因此：

如果您有一个计数器，不要忘记最终除以数据集的大小或类似值

我用过：

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


独立代码：

# 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）
参见此

最重要的是，您必须使用max减少/折叠分类原始值/logit所在的维度，然后使用
.index
选择它。通常这是维度
1
，因为dim 0具有批量大小，例如
[批量大小，D\U分类]
，其中原始数据的大小可能为
[批量大小，C，H，W]

1D中原始数据的合成示例如下：

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

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


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




因此：
让我们看一下基本知识：

  Accuracy = Total Correct Observations / Total Observations

在您的代码中，当您计算精度时，您将一个历元中的所有正确观测值除以所有不正确的观测值

correct/x.shape[0]

相反，您应该将其除以每个历元的观察数，即批量大小。假设您的批量大小=批量大小

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

理想情况下，在每个历元中，批大小、输入长度（行数）和标签长度都应该相同。
一行以获得准确度

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

假设第0维度是批次大小，第1维度保留分类标签的logits/raw值


更多详情：

def calc_error(mdl: torch.nn.Module, X: torch.Tensor, Y):
    # acc == (true != mdl(x).max(1).item() / true.size(0)
    train_acc = calc_accuracy(mdl, X, Y)
    train_err = 1.0 - train_acc
    return train_err

def calc_accuracy(mdl: torch.nn.Module, X: torch.Tensor, Y: torch.Tensor) -> float:
    """
    Get the accuracy with respect to the most likely label

    :param mdl:
    :param X:
    :param Y:
    :return:
    """
    # get the scores for each class (or logits)
    y_logits = mdl(X)  # unnormalized probs
    # return the values & indices with the largest value in the dimension where the scores for each class is
    # get the scores with largest values & their corresponding idx (so the class that is most likely)
    max_scores, max_idx_class = mdl(X).max(dim=1)  # [B, n_classes] -> [B], # get values & indices with the max vals in the dim with scores for each class/label
    # usually 0th coordinate is batch size
    n = X.size(0)
    assert( n == max_idx_class.size(0))
    # calulate acc (note .item() to do float division)
    acc = (max_idx_class == Y).sum().item() / n
    return acc

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))

请阅读以下答案：





循序渐进的例子
以下是以自包含代码为例的逐步说明：

#%%

# 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：


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

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

你能发布更多的代码来更好地理解吗？你的准确度公式对我来说很正确，请提供更多的代码你能发布
x.shape[0]
中的
x
是什么吗？你能澄清这一行中的
i
是什么吗：
对于枚举（火车加载器）中的i，（输入，标签）：
？这个指数是训练次数吗？一行获得准确度
acc==（true==mdl（x）.max（1）.item（）/true.size（0）
假设第0维是批次大小，第1维包含分类标签的logits/raw值。我将其除以数据集的总数，因为我已经完成了一个历元。您可以看到print语句在历元循环中，而不是批次循环中。是的，我看到了。但是，“正确”仍然只有一个小批量那么大是的。我想你是对的。本例中的输出是最后一个小批量输出，我们将在其中对每个历元进行验证。因此，我们应该划分历元最后一次迭代的小批量大小。感谢你的回答也许使用
.item（）
比
.float（）更好
您作为伪代码/注释编写的代码是其中最棘手的部分，也是我寻求解释的部分：
max\u vals，max\u index=torch.max（mdl（X），1）
@CharlieParker.item（）在张量中正好有1个值时工作。否则，它将给出一个错误。（输出==标签）是一个具有许多值的布尔张量，通过将其转换为浮点，False被转换为0，True被转换为1。然后我们对True的数量求和（.sum（）本身可能就足够了，因为它应该进行转换）我通常更喜欢在我的实验脚本顶部调用它
device=torch.device（“cuda”如果torch.cuda.is\u可用（）否则“cpu”）
然后执行
mdl.to（device）
或
tensor.to（device）
以使其更短（并且模型也不可知），我认为您不需要
.detac