GAN-详解BCELoss和BCEWithLogitsLoss

一、BCELoss()

生成对抗网络的所使用到的loss函数BCELoss和BCEWithLogitsLoss
其中BCELoss的公式为：

其中y是target，x是模型输出的值。

二、例子

import torchfrom torch import autogradfrom torch import nnimport mathinput = autograd.Variable(torch.tensor([[ 1.9072, 1.1079, 1.4906], [-0.6584, -0.0512, 0.7608], [-0.0614, 0.6583, 0.1095]]), requires_grad=True)print(input)print('-'*100)

输出：

tensor([[ 1.9072, 1.1079, 1.4906], [-0.6584, -0.0512, 0.7608], [-0.0614, 0.6583, 0.1095]], requires_grad=True)----------------------------------------------------------------------------------------------------

先用Sigmoid给这些值都搞到0~1之间：

m = nn.Sigmoid()print(m(input))print('-'*100)

输出：

tensor([[0.8707, 0.7517, 0.8162], [0.3411, 0.4872, 0.6815], [0.4847, 0.6589, 0.5273]], grad_fn=)----------------------------------------------------------------------------------------------------

假设Target是：

target = torch.FloatTensor([[0, 1, 1], [1, 1, 1], [0, 0, 0]])print(target)print('-'*100)

输出：

tensor([[0., 1., 1.], [1., 1., 1.], [0., 0., 0.]])----------------------------------------------------------------------------------------------------

计算BCELoss：

r11 = 0 * math.log(0.8707) + (1-0) * math.log((1 - 0.8707))r12 = 1 * math.log(0.7517) + (1-1) * math.log((1 - 0.7517))r13 = 1 * math.log(0.8162) + (1-1) * math.log((1 - 0.8162))r21 = 1 * math.log(0.3411) + (1-1) * math.log((1 - 0.3411))r22 = 1 * math.log(0.4872) + (1-1) * math.log((1 - 0.4872))r23 = 1 * math.log(0.6815) + (1-1) * math.log((1 - 0.6815))r31 = 0 * math.log(0.4847) + (1-0) * math.log((1 - 0.4847))r32 = 0 * math.log(0.6589) + (1-0) * math.log((1 - 0.6589))r33 = 0 * math.log(0.5273) + (1-0) * math.log((1 - 0.5273))r1 = -(r11 + r12 + r13) / 3#0.8447112733378236r2 = -(r21 + r22 + r23) / 3#0.7260397266631787r3 = -(r31 + r32 + r33) / 3#0.8292933181294807bceloss = (r1 + r2 + r3) / 3 print(bceloss)print('-'*100)

输出：

0.8000147727101611----------------------------------------------------------------------------------------------------

下面我们用BCELoss来验证一下Loss：

loss = nn.BCELoss()print(loss(m(input), target))print('-'*100)

输出：

tensor(0.8000, grad_fn=)

和上面计算的结果一样。

BCEWithLogitsLoss()

BCEWithLogitsLoss就是把Sigmoid-BCELoss合成一步。我们直接用刚刚的input验证一下:

loss = nn.BCEWithLogitsLoss()print(loss(input, target))

输出：

tensor(0.8000, grad_fn=)

完整代码

import torchfrom torch import autogradfrom torch import nnimport mathinput = autograd.Variable(torch.tensor([[ 1.9072, 1.1079, 1.4906], [-0.6584, -0.0512, 0.7608], [-0.0614, 0.6583, 0.1095]]), requires_grad=True)print(input)print('-'*100)# from torch import nnm = nn.Sigmoid()print(m(input))print('-'*100)target = torch.FloatTensor([[0, 1, 1], [1, 1, 1], [0, 0, 0]])print(target)print('-'*100)# import mathr11 = 0 * math.log(0.8707) + (1-0) * math.log((1 - 0.8707))r12 = 1 * math.log(0.7517) + (1-1) * math.log((1 - 0.7517))r13 = 1 * math.log(0.8162) + (1-1) * math.log((1 - 0.8162))r21 = 1 * math.log(0.3411) + (1-1) * math.log((1 - 0.3411))r22 = 1 * math.log(0.4872) + (1-1) * math.log((1 - 0.4872))r23 = 1 * math.log(0.6815) + (1-1) * math.log((1 - 0.6815))r31 = 0 * math.log(0.4847) + (1-0) * math.log((1 - 0.4847))r32 = 0 * math.log(0.6589) + (1-0) * math.log((1 - 0.6589))r33 = 0 * math.log(0.5273) + (1-0) * math.log((1 - 0.5273))r1 = -(r11 + r12 + r13) / 3#0.8447112733378236r2 = -(r21 + r22 + r23) / 3#0.7260397266631787r3 = -(r31 + r32 + r33) / 3#0.8292933181294807bceloss = (r1 + r2 + r3) / 3 print(bceloss)print('-'*100)loss = nn.BCELoss()print(loss(m(input), target))print('-'*100)loss = nn.BCEWithLogitsLoss()print(loss(input, target))