# Backward() for costume loss function

hi,
i created a loss function, and it returns a float
i get the error ‘float’ object has no attribute ‘backward’
which is makes sense…

if i return tensor(some_float)
i get the error:
element 0 of tensors does not require grad and does not have a grad_fn
how can i handle it?

U have to return a FloatTensor that is the result of some operation of ur loss function, so ur loss function won’t break the gradients graph. To be more specific, can u provide an example code of ur loss function?

the function:

def loss_function(p, y):
losses = 0
for i in range(len§):
temp_p = torch.tensor(p[i])
temp_y = torch.tensor(y[i])
k = len(temp_y)
loss1 = -tensot_mul(vec_minus(one_Vector(k),y[i]), list(np.log(vec_minus(one_Vector(k),p[i]))))
l = y[i]*p[i].log()
prod = vec_minus(vec_minus(one_Vector(k),y[i]), - l)
loss2 = loss1 + sum(softmax(vec_minus(one_Vector(k),prod)))
losses = losses + loss2
returnfloat(losses/(len§))

p - prediction
y - target

As I said, I think in this code u r returning a Python float object, and that doesn’t have a `.backward()` method.
I’m unclear about what the `tensor_mul`, `vec_minus` are, but the operations u apply to the outputs should all the Pytorch operations, involving some numpy operations would probably break the graph(I’m not sure).
If u return a `FloatTensor` like u described. The tensor would be a “Leaf variable”(I’m not sure what exactly it should be called), and these kind of variable does not have gradients because u created them out of a `float`, which holds no gradients information.

1 Like

im sorry, it didnt work…

i changed the function, and i also tried to add Variable()

def loss_function(p, y, b):
losses = 0
for i in range(b):
temp_p = torch.tensor(p[i])
temp_y = torch.tensor(y[i])
k = len(temp_y)
loss1 = -(torch.ones(k).cuda()-temp_y)@((torch.ones(k).cuda()-temp_p).log())
l = temp_y*(temp_p.log())
prod = (torch.ones(k).cuda()-temp_y) - l
loss2 = loss1 + torch.max(torch.ones(k).cuda()-prod)
losses = loss1 + loss2
f = (losses/(b))
return Variable(torch.FloatTensor([f]))

i still get:

element 0 of tensors does not require grad and does not have a grad_fn

what is the type of “p” and “y”?

p,y - tensor
b - int (batch size)

can u reformat the code? I’m having trouble distinguishing what is inside the loop.

Are you sure you have set `tensor.requires_grad()=True` ?

def loss_function(p, y, b):
—losses = 0
—for i in range(b):
------temp_p = torch.tensor(p[i])
------temp_y = torch.tensor(y[i])
------k = len(temp_y)
------loss1 = -(torch.ones(k).cuda()- temp_y)@((torch.ones(k).cuda()-temp_p).log())
------l = temp_y*(temp_p.log())
------prod = (torch.ones(k).cuda()-temp_y) - l
------loss2 = loss1 + torch.max(torch.ones(k).cuda()-
prod)
------losses = loss1 + loss2
—f = (losses/(b))
—return Variable(torch.FloatTensor([f]))

Where should i define it?

I would say there :

``````temp_p = torch.tensor(p[i], requires_grad=True)
temp_y = torch.tensor(y[i], requires_grad=True)
``````

By default, the requires_grad are set to False. You need to activate them yourself. More on this here : https://pytorch.org/docs/stable/notes/autograd.html

May be you should do it even before … At the beginning of your function I would try:

``````p = torch.tensor(p, requires_grad=True)
y = torch.tensor(y, requires_grad=True)
``````

i get
Only Tensors of floating point dtype can require gradients

1. `Variable` is now deprecated.
2. When applying the tensor to some linear transformation with scalers, u can just use Python value. For example `a: tc.Tensor = 1 - b`
3. Using torch.tensor(OLD_TENSOR) will not pass the gradients of the OLD_TENSOR to the new tensor.

Note: u should read more about pytorch docs and the autograd mechanism.

1 Like

ok
tnx for your note, i will do it

right now, i have this

``````def loss_function(p, y, b):
losses = 0
for i in range(b):
k = len(y[i])
loss1 = -(torch.ones(k).cuda()-y[i])@((torch.ones(k).cuda()-p[i]).log())
l = y[i]*(p[i].log())
prod = (torch.ones(k).cuda()-y[i]) - l
loss2 = loss1 + torch.max(torch.ones(k).cuda()-prod)
losses = loss1 + loss2
return losses/(b)
``````

so im not doing this: torch.tensor(OLD_TENSOR) anymore

Just to note: U can change `torch.ones(k).cuda() - TEN` to `1 - TEN`. Also, ur loss function looks like cross entropy loss to me, and Pytorch has a implementation for that.

it is not cross entropy,
it is a loss for a multilabel problem, that i want to minimize when the model correct in one lable of all the labels of an object in my data