I see pytorch provides support to write custom loss functions. Consider following hinge loss.
class MarginRankingLossExp(nn.Module):
def __init__(self) -> None:
super(MarginRankingLossExp, self).__init__( )
def forward(self,input1,input2,target):
# loss_without_reduction = max(0, −target * (input1 − input2) + margin)
neg_target = -target
input_diff = input2-input1
mul_target_input = neg_target*input_diff
add_margin = mul_target_input
zeros=torch.zeros_like(add_margin)
loss = torch.max(add_margin, zeros)
return loss.mean()
This has only forward and constructor function defined. How does pytorch calculate gradient for custom functions? Does it differentiate it somehow?
Also, This function is non differentiable at y=margin but it didn't throw any error.
Your function will be differentiable by PyTorch's autograd as long as all the operators used in your function's logic are differentiable. That is, as long as you use torch.Tensor and built-in torch operators that implement a backward function, your custom function will be differentiable out of the box.
In a few words, on inference, a computational graph will be constructed on the fly. That is, for every operation you make, the tensors necessary to compute the gradients will be matched for a later backward pass. Assuming that you use only differentiable operators (i.e. most operators are mathematically differentiable and as such PyTorch provides the backward functionality for them). You will be able to perform backpropagation on the graph: from the end of it from the loss term, up to its leaves on parameters and inputs.
A very easy way to tell if your function is differentiable by Autograd is to infer its output with inputs which require gradient computation. Then check for a grad_fn callback on the output:
>>> x1 = torch.rand(1,10,2,2, requires_grad=True)
>>> x2 = torch.rand(1,10,2,2, requires_grad=True)
>>> y = torch.rand(1,10,2,2)
Here we can check with:
>>> MarginRankingLossExp()(x1, x2, y)
tensor(0.1045, grad_fn=<MeanBackward0>)
Where you notice MeanBackward0 which refers to torch.Tensor.mean, being the very last operator applied by MarginRankingLossExp.forward.
Related
I try to write a cross entropy loss function by myself. My loss function gives the same loss value as the official one, but when i use my loss function in the code instead of official cross entropy loss function, the code does not converge. When i use the official cross entropy loss function, the code converges. Here is my code, please give me some suggestions. Thanks very much
The input 'out' is a tensor (B*C) and 'label' contains class indices (1 * B)
class MylossFunc(nn.Module):
def __init__(self):
super(MylossFunc, self).__init__()
def forward(self, out, label):
out = torch.nn.functional.softmax(out, dim=1)
n = len(label)
loss = torch.FloatTensor([0])
loss = Variable(loss, requires_grad=True)
tmp = torch.log(out)
#print(out)
torch.scalar_tensor(-100)
for i in range(n):
loss = loss - torch.max(tmp[i][label[i]], torch.scalar_tensor(-100) )/n
loss = torch.sum(loss)
return loss
Instead of using torch.softmax and torch.log, you should use torch.log_softmax, otherwise your training will become unstable with nan values everywhere.
This happens because when you take the softmax of your logits using the following line:
out = torch.nn.functional.softmax(out, dim=1)
you might get a zero in one of the components of out, and when you follow that by applying torch.log it will result in nan (since log(0) is undefined). That is why torch (and other common libraries) provide a single stable operation, log_softmax, to avoid the numerical instabilities that occur when you use torch.softmax and torch.log individually.
I am training a PyTorch model to perform binary classification. My minority class makes up about 10% of the data, so I want to use a weighted loss function. The docs for BCELoss and CrossEntropyLoss say that I can use a 'weight' for each sample.
However, when I declare CE_loss = nn.BCELoss() or nn.CrossEntropyLoss() and then do CE_Loss(output, target, weight=batch_weights), where output, target, and batch_weights are Tensors of batch_size, I get the following error message:
forward() got an unexpected keyword argument 'weight'
Another way you could accomplish your goal is to use reduction=none when initializing the loss and then multiply the resulting tensor by your weights before computing the mean.
e.g.
loss = torch.nn.BCELoss(reduction='none')
model = torch.sigmoid
weights = torch.rand(10,1)
inputs = torch.rand(10,1)
targets = torch.rand(10,1)
intermediate_losses = loss(model(inputs), targets)
final_loss = torch.mean(weights*intermediate_losses)
Of course for your scenario you still would need to calculate the weights tensor. But hopefully this helps!
Could it be that you want to apply separate fixed weights to all elements of class 0 and class 1 in your dataset? It is not clear what value you are passing for batch_weights here. If so, then that is not what the weight parameter in BCELoss does. The weight parameter expects you to pass a separate weight for every ELEMENT in the dataset, not for every CLASS. There are several ways around this. You could construct a weight table for every element. Alternatively, you could use a custom loss function that does what you want:
def BCELoss_class_weighted(weights):
def loss(input, target):
input = torch.clamp(input,min=1e-7,max=1-1e-7)
bce = - weights[1] * target * torch.log(input) - (1 - target) * weights[0] * torch.log(1 - input)
return torch.mean(bce)
return loss
Note that it is important to add a clamp to avoid numerical instability.
HTH Jeroen
the issue is wherein your providing the weight parameter. As it is mentioned in the docs, here, the weights parameter should be provided during module instantiation.
For example, something like,
from torch import nn
weights = torch.FloatTensor([2.0, 1.2])
loss = nn.BCELoss(weights=weights)
You can find a more concrete example here or another helpful PT forum discussion here.
you need to pass weights like below:
CE_loss = CrossEntropyLoss(weight=[…])
This is similar to the idea of #Jeroen Vuurens, but the class weights are determined by the target mean:
y_train_mean = y_train.mean()
bi_cls_w2 = 1/(1 - y_train_mean)
bi_cls_w1 = 1/y_train_mean - bi_cls_w2
bce_loss = nn.BCELoss(reduction='none')
loss_fun = lambda pred, target: ((bi_cls_w1*target + bi_cls_w2) * bce_loss(pred, target)).mean()
Using PyTorch, I would like to calculate the Hessian vector product, where the Hessian is the second-derivative matrix of the loss function of some neural net, and the vector will be the vector of gradients of that loss function.
I know how to calculate the Hessian vector product for a regular function thanks to this post. However, I am running into trouble when the function is the loss function of a neural network. This is because the parameters are packaged into a module, accessible via nn.parameters(), and not a torch tensor.
I want to do something like this (doesn't work):
### a simple neural network
linear = nn.Linear(10, 20)
x = torch.randn(1, 10)
y = linear(x).sum()
### compute the gradient and make a copy that is detached from the graph
grad = torch.autograd.grad(y, linear.parameters(),create_graph=True)
v = grad.clone().detach()
### compute the Hessian vector product
z = grad # v
z.backward()
In analogy this this (does work):
x = Variable(torch.Tensor([1, 1]), requires_grad=True)
f = 3*x[0]**2 + 4*x[0]*x[1] + x[1]**2
grad, = torch.autograd.grad(f, x, create_graph=True)
v = grad.clone().detach()
z = grad # v
z.backward()
This post addresses a similar (possibly the same?) issue, but I don't understand the solution.
You are saying it doesn't work but do not show what error you get, this is why you haven't got any answers
torch.autograd.grad(outputs, inputs, grad_outputs=None, retain_graph=None, create_graph=False, only_inputs=True, allow_unused=False)
outputs and inputs are expected to be sequences of tensors. But you
use just a tensor as outputs.
What this is saying is that you should pass a sequence, so pass [y] instead of y
It’s known that sparse_categorical_crossentropy in keras can get the average loss function among each category. But what if only one certain category was I concerned most? Like if I want to define the precision(=TP/(TP+FP)) based on this category as loss function, how can I write it? Thanks！
My codes were like:
from keras import backend as K
def my_loss(y_true,y_pred):
y_true = K.cast(y_true,"float32")
y_pred = K.cast(K.argmax(y_pred),"float32")
nominator = K.sum(K.cast(K.equal(y_true,y_pred) & K.equal(y_true, 0),"float32"))
denominator = K.sum(K.cast(K.equal(y_pred,0),"float32"))
return -(nominator + K.epsilon()) / (denominator + K.epsilon())
And the error is like:
argmax is not differentiable
I don't recommend you to use precision as the loss function.
It is not differentiable that can't be set as a loss function for nn.
you can max it by predicting all the instance as class negative, that makes no sense.
One of the alternative solution is using F1 as the loss function, then tuning the probability cut-off manually for obtaining a desirable level of precision as well as recall is not too low.
You can pass to the fit method a parameter class_weight where you determine which classes are more important.
It should be a dictionary:
{
0: 1, #class 0 has weight 1
1: 0.5, #class 1 has half the importance of class 0
2: 0.7, #....
...
}
Custom loss
If that is not exactly what you need, you can create loss functions like:
import keras.backend as K
def customLoss(yTrue,yPred):
create operations with yTrue and yPred
- yTrue = the true output data (equal to y_train in most examples)
- yPred = the model's calculated output
- yTrue and yPred have exactly the same shape: (batch_size,output_dimensions,....)
- according to the output shape of the last layer
- also according to the shape of y_train
all operations must be like +, -, *, / or operations from K (backend)
return someResultingTensor
You cannot used argmax as it is not differentiable. That means that backprop will not work if loss function can't be differentiated.
Instead of using argmax, do y_true * y_pred.
I have two functions that are suppose to produce equal results: f1(x,theta)=f2(x,theta).
Given input x, I need to find the parameters theta that makes this equality hold as well as possible.
Initially I was thinking of using squared loss and minimizing (f1(x,theta)-f2(x,theta))^2 and solving via SGD.
However I was thinking of making the loss more precise and using huber (or absolute loss) of the difference.
Huber loss is a piecewise function (ie initially it is quadratic and then it changes into a linear function).
How can I take the gradient of my huber loss in theano?
A pretty simple implementation of huber loss in theano can be found here
Here is a code snippet
import theano.tensor as T
delta = 0.1
def huber(target, output):
d = target - output
a = .5 * d**2
b = delta * (abs(d) - delta / 2.)
l = T.switch(abs(d) <= delta, a, b)
return l.sum()
The function huber will return a symbolic representation of the loss which you can then plug in theano.tensor.grad to get the gradient and use it to minimize using SGD