为什么自动编码器中的解码器在最后一层使用 sigmoid? [英] Why is the decoder in an autoencoder uses a sigmoid on the last layer?

问题描述

我正在查看这个工作变分自动编码器.

I am looking at this working variational auto encoder.

主类

class VAE(nn.Module):
    def __init__(self):
        super(VAE, self).__init__()

        self.fc1 = nn.Linear(784, 400)
        self.fc21 = nn.Linear(400, 20)
        self.fc22 = nn.Linear(400, 20)
        self.fc3 = nn.Linear(20, 400)
        self.fc4 = nn.Linear(400, 784)

    def encode(self, x):
        h1 = F.relu(self.fc1(x))
        return self.fc21(h1), self.fc22(h1)

    def reparametrize(self, mu, logvar):
        std = logvar.mul(0.5).exp_()
        if torch.cuda.is_available():
            eps = torch.cuda.FloatTensor(std.size()).normal_()
        else:
            eps = torch.FloatTensor(std.size()).normal_()
        eps = Variable(eps)
        return eps.mul(std).add_(mu)

    def decode(self, z):
        h3 = F.relu(self.fc3(z))
        return F.sigmoid(self.fc4(h3))

    def forward(self, x):
        mu, logvar = self.encode(x)
        z = self.reparametrize(mu, logvar)
        return self.decode(z), mu, logvar

有

    def decode(self, z):
        h3 = F.relu(self.fc3(z))
        return F.sigmoid(self.fc4(h3))

我无法向自己解释为什么最后一层在返回之前应该通过一个 sigmoid.

I can't explain to myself why the last layer should be passed through a sigmoid before returning.

请解释.

我只是在没有 sigmoid 的情况下进行了检查.结果还是不错的.现在我不确定是否需要它.

I just checked without the sigmoid. Results are still nice. Now I am not sure if it is needed or not.

推荐答案

正如 Jim J 在回答中提到的，sigmoid 强制输出范围为 [0, 1].在这种情况下，并不是因为我们想将输出解释为概率，而是强制将输出解释为灰度图像的像素强度.

As mentioned in the answer by Jim J, sigmoid forces the output to the range [0, 1]. In this case, it's not because we want to interpret the output as a probability, rather it's done to force the output to be interpreted as pixel intensity of a grey scale image.

如果你移除 sigmoid，神经网络将不得不知道所有的输出都应该在 [0, 1] 的范围内.sigmoid 可能有助于使学习过程更加稳定.

If you remove the sigmoid, the NN will have to learn that all the outputs should be in the range [0, 1]. The sigmoid might help making the learning process more stable.

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