loss.sum().backward()中对于sum()的理解 - 知乎 - 知乎专栏
网页对其中的的 l.sum().backward() 并不是很理解，对损失函数求导即可，为什么要先求和再求导。 其实在2.5节，沐神已经明确的讲解了这个问题，只不过当时我并没有get到这个点。
对模型训练代码中l.sum().backward()的粗浅理解 - 知乎
网页l.sum()其实就对应着公式中的： z \leftarrow \sum_{i \in \mathcal{B}} l^{(i)}(\mathbf{w},b). 接下来用z表示l.sum()，写成这样方便看懂如何求导： z \leftarrow l^{(1)}(\mathbf{w},b) +l^{(2)}(\mathbf{w},b) + \dots + l^{(\mathcal{B})}(\mathbf{w},b). 这时对z求关于权重w和偏置b的偏导数，也就是l.sum ...
loss.sum().backward()中对于sum()的理解 - CSDN博客
网页2023年9月16日 · loss.sum ().backward ()中对于sum ()的理解. 李沐 书中提到的：. # 输出： tensor([ 0., 4., 8., 12.]) # 输出： tensor([1., 1., 1., 1.]) #输出： tensor([0., 2., 4., 6.]) 我们通过了解可以确定的 梯度只能为标量（即一个数）输出隐式地创建. 所以 backward 前面都是标 …
Loss reduction sum vs mean: when to use each? - PyTorch Forums
网页2021年3月23日 · While experimenting with my model I see that the various Loss classes for pytorch will accept a reduction parameter (none | sum | mean) for example. The differences are rather obvious regarding what will be returned, but I’m curious when it would be useful to use sum as opposed to mean?
关于梯度累加pytorch实现的一个细节——loss = loss / …
网页loss = loss / accumulation_steps这一步是额外的操作。 但如果loss的定义不是 \sum_{i=1}^{100}{(ypred-ytrue)^2} ，而是 \frac{1}{batchsize}\sum_{i=1}^{100}{(ypred-ytrue)^2}
MSELoss — PyTorch 2.4 documentation
网页classtorch.nn.MSELoss(size_average=None, reduce=None, reduction='mean')[source] ¶. Creates a criterion that measures the mean squared error (squared L2 norm) between each element in the input x x x and target y y y.
About the difference between `none` and `sum` in `reduction` …
网页2019年8月21日 · According to my understanding, the following nloss.sum() and sloss should be equal, but the actual is different. Why is this? import torch.nn as nn. mean_loss = nn.BCELoss(reduction='mean') sum_loss = nn.BCELoss(reduction='sum') none_loss = nn.BCELoss(reduction='none') prediction = torch.rand((4, 1, 256, 256))
Should i do loss.backward () or loss.mean ().backward ()
网页2019年1月25日 · If your loss is not a scalar value, then you should certainly use either loss.mean() or loss.sum() to convert it to a scalar before calling the backward. Otherwise, it will cause an error as in the following example:
How can i process multi loss in pytorch? - Stack Overflow
网页2019年1月1日 · Two different loss functions. If you have two different loss functions, finish the forwards for both of them separately, and then finally you can do (loss1 + loss2).backward(). It’s a bit more efficient, skips quite some computation. Extra tip: Sum the loss. In your code you want to do: loss_sum += loss.item()
loss.sum().backward()中对于sum()的理解 - 腾讯云开发者社区-腾 …
网页.sum() 函数主要有两个作用，一个是用来求和，一个是用来降维。 在深度学习中，损失函数都是标量，所以一般情况下可以直接调用backward()就可以了。 发表于: 2024-04-17 2024-04-17 08:00:00