Reputation: 63
For my model I'm using a roberta transformer model and the Trainer from the Huggingface transformer library.
I calculate two losses:
lloss
is a Cross Entropy Loss and dloss
calculates the loss inbetween hierarchy layers.
The total loss is the sum of lloss and dloss. (Based on this)
When calling total_loss.backwards()
however, I get the error:
RuntimeError: Trying to backward through the graph a second time, but the buffers have already been freed
Any idea why that happens? Can I force it to only call backwards once? Here is the loss calculation part:
dloss = calculate_dloss(prediction, labels, 3)
lloss = calculate_lloss(predeiction, labels, 3)
total_loss = lloss + dloss
total_loss.backward()
def calculate_lloss(predictions, true_labels, total_level):
'''Calculates the layer loss.
'''
loss_fct = nn.CrossEntropyLoss()
lloss = 0
for l in range(total_level):
lloss += loss_fct(predictions[l], true_labels[l])
return self.alpha * lloss
def calculate_dloss(predictions, true_labels, total_level):
'''Calculate the dependence loss.
'''
dloss = 0
for l in range(1, total_level):
current_lvl_pred = torch.argmax(nn.Softmax(dim=1)(predictions[l]), dim=1)
prev_lvl_pred = torch.argmax(nn.Softmax(dim=1)(predictions[l-1]), dim=1)
D_l = self.check_hierarchy(current_lvl_pred, prev_lvl_pred, l) #just a boolean tensor
l_prev = torch.where(prev_lvl_pred == true_labels[l-1], torch.FloatTensor([0]).to(self.device), torch.FloatTensor([1]).to(self.device))
l_curr = torch.where(current_lvl_pred == true_labels[l], torch.FloatTensor([0]).to(self.device), torch.FloatTensor([1]).to(self.device))
dloss += torch.sum(torch.pow(self.p_loss, D_l*l_prev)*torch.pow(self.p_loss, D_l*l_curr) - 1)
return self.beta * dloss
Upvotes: 0
Views: 608
Reputation: 330
After backward() your comp. graph is freed so for the second backward you need to create a new graph by providing inputs again. If you want to reiterate the same graph after backward (for some reason) you need to specify retain_graph flag in backward as True. see retain_graph here.
P.S. As the summation of Tensors is automatically differentiable, summing the losses would not cause any issue in the backward.
Upvotes: 0
Reputation: 20482
There is nothing wrong with having a loss that is the sum of two individual losses, here is a small proof of principle adapted from the docs:
import torch
import numpy
from sklearn.datasets import make_blobs
class Feedforward(torch.nn.Module):
def __init__(self, input_size, hidden_size):
super(Feedforward, self).__init__()
self.input_size = input_size
self.hidden_size = hidden_size
self.fc1 = torch.nn.Linear(self.input_size, self.hidden_size)
self.relu = torch.nn.ReLU()
self.fc2 = torch.nn.Linear(self.hidden_size, 1)
self.sigmoid = torch.nn.Sigmoid()
def forward(self, x):
hidden = self.fc1(x)
relu = self.relu(hidden)
output = self.fc2(relu)
output = self.sigmoid(output)
return output
def blob_label(y, label, loc): # assign labels
target = numpy.copy(y)
for l in loc:
target[y == l] = label
return target
x_train, y_train = make_blobs(n_samples=40, n_features=2, cluster_std=1.5, shuffle=True)
x_train = torch.FloatTensor(x_train)
y_train = torch.FloatTensor(blob_label(y_train, 0, [0]))
y_train = torch.FloatTensor(blob_label(y_train, 1, [1,2,3]))
x_test, y_test = make_blobs(n_samples=10, n_features=2, cluster_std=1.5, shuffle=True)
x_test = torch.FloatTensor(x_test)
y_test = torch.FloatTensor(blob_label(y_test, 0, [0]))
y_test = torch.FloatTensor(blob_label(y_test, 1, [1,2,3]))
model = Feedforward(2, 10)
criterion = torch.nn.BCELoss()
optimizer = torch.optim.SGD(model.parameters(), lr = 0.01)
model.eval()
y_pred = model(x_test)
before_train = criterion(y_pred.squeeze(), y_test)
print('Test loss before training' , before_train.item())
model.train()
epoch = 20
for epoch in range(epoch):
optimizer.zero_grad() # Forward pass
y_pred = model(x_train) # Compute Loss
lossCE= criterion(y_pred.squeeze(), y_train)
lossSQD = (y_pred.squeeze()-y_train).pow(2).mean()
loss=lossCE+lossSQD
print('Epoch {}: train loss: {}'.format(epoch, loss.item())) # Backward pass
loss.backward()
optimizer.step()
There must be a real second time that you call directly or indirectly backward
on some varaible that then traverses through your graph. It is a bit too much to ask for the complete code here, only you can check this or at least reduce it to a minimal example (while doing so, you might already find the issue). Apart from that, I would start checking:
detach
?backward
on your losses individually lloss.backward()
followed by dloss.backward()
(this has the same effect as adding them together first as gradients are accumulated): what happens? This will let you track down for which of the two losses the error occurs.Upvotes: 1