I have two losses: one the usual L1 loss and second one involving torch.rfft()
def dft_amp(img):
fft_im = torch.rfft( img, signal_ndim=2, onesided=False )
fft_amp = fft_im[:,:,:,:,0]**2 + fft_im[:,:,:,:,1]**2
return torch.sqrt(fft_amp)
l1_loss = torch.nn.L1Loss()
loss = l1_loss(pred,gt) + l1_loss(dft_amp(pred),dft(gt_amp))
loss.backward()
This runs for the 1st iteration with both losses not bein being nan but the loss from 2nd iteration onwards becomes nan.
If however only the simple L1 loss is kept and l1_loss(dft_amp(pred),dft(gt_amp)) is omitted, the training proceeds normally.
Does torch.rfft() supports backpropagation? I am using pytorch 1.4.0
Any suggestions would be appreciated
There seems to be some issue with torch.sqrt() and not with the torch.rfft(). torch.sqrt() cannot handle very small values. Thus in the above code replace
return torch.sqrt(fft_amp)
with
return torch.sqrt(fft_amp + 1e-10)
and the NaN values will vanish away.
Though this solution works, it would be nice if someone digs in why square roots are problematic for digital computations.
Related
I am doing a project on multiclass semantic segmentation. I have formulated a model that outputs pretty descent segmented images by decreasing the loss value. However, I cannot evaluate the model performance in metrics, such as meanIoU or Dice coefficient.
In case of binary semantic segmentation it was easy just to set the threshold of 0.5, to classify the outputs as an object or background, but it does not work in the case of multiclass semantic segmentation. Could you please tell me how to obtain model performance on the aforementioned metrics? Any help will be highly appreciated!
By the way, I am using PyTorch framework and CamVid dataset.
If anyone is interested in this answer, please also look at this issue. The author of the issue points out that mIoU can be computed in a different way (and that method is more accepted in literature). So, consider that before using the implementation for any formal publication.
Basically, the other method suggested by the issue-poster is to separately accumulate the intersections and unions over the entire dataset and divide them at the final step. The method in the below original answer computes intersection and union for a batch of images, then divides them to get IoU for the current batch, and then takes a mean of the IoUs over the entire dataset.
However, this below given original method is problematic because the final mean IoU would vary with the batch-size. On the other hand, the mIoU would not vary with the batch size for the method mentioned in the issue as the separate accumulation would ensure that batch size is irrelevant (though higher batch size can definitely help speed up the evaluation).
Original answer:
Given below is an implementation of mean IoU (Intersection over Union) in PyTorch.
def mIOU(label, pred, num_classes=19):
pred = F.softmax(pred, dim=1)
pred = torch.argmax(pred, dim=1).squeeze(1)
iou_list = list()
present_iou_list = list()
pred = pred.view(-1)
label = label.view(-1)
# Note: Following for loop goes from 0 to (num_classes-1)
# and ignore_index is num_classes, thus ignore_index is
# not considered in computation of IoU.
for sem_class in range(num_classes):
pred_inds = (pred == sem_class)
target_inds = (label == sem_class)
if target_inds.long().sum().item() == 0:
iou_now = float('nan')
else:
intersection_now = (pred_inds[target_inds]).long().sum().item()
union_now = pred_inds.long().sum().item() + target_inds.long().sum().item() - intersection_now
iou_now = float(intersection_now) / float(union_now)
present_iou_list.append(iou_now)
iou_list.append(iou_now)
return np.mean(present_iou_list)
Prediction of your model will be in one-hot form, so first take softmax (if your model doesn't already) followed by argmax to get the index with the highest probability at each pixel. Then, we calculate IoU for each class (and take the mean over it at the end).
We can reshape both the prediction and the label as 1-D vectors (I read that it makes the computation faster). For each class, we first identify the indices of that class using pred_inds = (pred == sem_class) and target_inds = (label == sem_class). The resulting pred_inds and target_inds will have 1 at pixels labelled as that particular class while 0 for any other class.
Then, there is a possibility that the target does not contain that particular class at all. This will make that class's IoU calculation invalid as it is not present in the target. So, you assign such classes a NaN IoU (so you can identify them later) and not involve them in the calculation of the mean.
If the particular class is present in the target, then pred_inds[target_inds] will give a vector of 1s and 0s where indices with 1 are those where prediction and target are equal and zero otherwise. Taking the sum of all elements of this will give us the intersection.
If we add all the elements of pred_inds and target_inds, we'll get the union + intersection of pixels of that particular class. So, we subtract the already calculated intersection to get the union. Then, we can divide the intersection and union to get the IoU of that particular class and add it to a list of valid IoUs.
At the end, you take the mean of the entire list to get the mIoU. If you want the Dice Coefficient, you can calculate it in a similar fashion.
I'm training a model in pytorch. Every 10 epochs, I'm evaluating the train and test error on the entire train and test dataset. For some reason the evaluation function is causing out-of-memory on my GPU. This is strange because I have the same batchsize for training and evaluation. I believe it's due to the net.forward() method being called repeated and having all the hidden values stored in memory but I'm not sure how to get around this?
def evaluate(self, data):
correct = 0
total = 0
loader = self.train_loader if data == "train" else self.test_loader
for step, (story, question, answer) in enumerate(loader):
story = Variable(story)
question = Variable(question)
answer = Variable(answer)
_, answer = torch.max(answer, 1)
if self.config.cuda:
story = story.cuda()
question = question.cuda()
answer = answer.cuda()
pred_prob = self.mem_n2n(story, question)[0]
_, output_max_index = torch.max(pred_prob, 1)
toadd = (answer == output_max_index).float().sum().data[0]
correct = correct + toadd
total = total + captions.size(0)
acc = correct / total
return acc
I think it fails during Validation because you don't use optimizer.zero_grad(). The zero_grad executes detach, making the tensor a leaf. It is commonly used every epoch in the training part.
The use of volatile flag in Variable from PyTorch 0.4.0 has been removed.
Ref - migration_guide_to_0.4.0
Starting from 0.4.0, to avoid the gradient being computed during validation, use torch.no_grad()
Code example from the migration guide.
# evaluate
with torch.no_grad(): # operations inside don't track history
for input, target in test_loader:
...
For 0.3.X, using volatile should work.
I would suggest to use volatile flag set to True for all variables used during the evaluation,
story = Variable(story, volatile=True)
question = Variable(question, volatile=True)
answer = Variable(answer, volatile=True)
Thus, the gradients and operation history is not stored and you will save a lot of memory.
Also, you could delete references to those variables at the end of the batch processing:
del story, question, answer, pred_prob
Don't forget to set the model to the evaluation mode (and back to the train mode after you finished the evaluation). For instance, like this
model.eval()
This release of PyTorch seems provide the PackedSequence for variable lengths of input for recurrent neural network. However, I found it's a bit hard to use it correctly.
Using pad_packed_sequence to recover an output of a RNN layer which were fed by pack_padded_sequence, we got a T x B x N tensor outputs where T is the max time steps, B is the batch size and N is the hidden size. I found that for short sequences in the batch, the subsequent output will be all zeros.
Here are my questions.
For a single output task where the one would need the last output of all the sequences, simple outputs[-1] will give a wrong result since this tensor contains lots of zeros for short sequences. One will need to construct indices by sequence lengths to fetch the individual last output for all the sequences. Is there more simple way to do that?
For a multiple output task (e.g. seq2seq), usually one will add a linear layer N x O and reshape the batch outputs T x B x O into TB x O and compute the cross entropy loss with the true targets TB (usually integers in language model). In this situation, do these zeros in batch output matters?
Question 1 - Last Timestep
This is the code that i use to get the output of the last timestep. I don't know if there is a simpler solution. If it is, i'd like to know it. I followed this discussion and grabbed the relative code snippet for my last_timestep method. This is my forward.
class BaselineRNN(nn.Module):
def __init__(self, **kwargs):
...
def last_timestep(self, unpacked, lengths):
# Index of the last output for each sequence.
idx = (lengths - 1).view(-1, 1).expand(unpacked.size(0),
unpacked.size(2)).unsqueeze(1)
return unpacked.gather(1, idx).squeeze()
def forward(self, x, lengths):
embs = self.embedding(x)
# pack the batch
packed = pack_padded_sequence(embs, list(lengths.data),
batch_first=True)
out_packed, (h, c) = self.rnn(packed)
out_unpacked, _ = pad_packed_sequence(out_packed, batch_first=True)
# get the outputs from the last *non-masked* timestep for each sentence
last_outputs = self.last_timestep(out_unpacked, lengths)
# project to the classes using a linear layer
logits = self.linear(last_outputs)
return logits
Question 2 - Masked Cross Entropy Loss
Yes, by default the zero padded timesteps (targets) matter. However, it is very easy to mask them. You have two options, depending on the version of PyTorch that you use.
PyTorch 0.2.0: Now pytorch supports masking directly in the CrossEntropyLoss, with the ignore_index argument. For example, in language modeling or seq2seq, where i add zero padding, i mask the zero padded words (target) simply like this:
loss_function = nn.CrossEntropyLoss(ignore_index=0)
PyTorch 0.1.12 and older: In the older versions of PyTorch, masking was not supported, so you had to implement your own workaround. I solution that i used, was masked_cross_entropy.py, by jihunchoi. You may be also interested in this discussion.
A few days ago, I found this method which uses indexing to accomplish the same task with a one-liner.
I have my dataset batch first ([batch size, sequence length, features]), so for me:
unpacked_out = unpacked_out[np.arange(unpacked_out.shape[0]), lengths - 1, :]
where unpacked_out is the output of torch.nn.utils.rnn.pad_packed_sequence.
I have compared it with the method described here, which looks similar to the last_timestep() method Christos Baziotis is using above (also recommended here), and the results are the same in my case.
For diagnostic purposes, I am grabbing the gradients of the network periodically. One way to do this is to return the gradients as output of the theano function. However, copying the gradients from the GPU to CPU memory every time may be costly so I would prefer to do it only periodically. At the moment, I am achieving this by creating two function objects, one which returns the gradient and one which doesn't.
However, I do not know whether this is optimal and am looking for a more elegant way to achieve the same thing.
Your first function obviously executes a training step and updates all your parameters.
The second function must return the gradients of your parameters.
The fastest way to do what you are asking is to add the updates for the training step to the second function and when logging the gradients, don't call the first function, but only the second.
gradients = [ ... ]
train_f = theano.function([x, y], [], updates=updates)
train_grad_f = theano.function([x, y], gradients, updates=updates)
num_iters = 1000
grad_array = []
for i in range(num_iters):
# every 10 training steps keep log of gradients
if i % 10 == 0:
grad_array.append(train_grad_f(...))
else:
train_f(...)
Update
if you wish to have a single function to do this, you can do the following
from theano.ifelse import ifelse
no_grad = T.iscalar('no_grad')
example_gradient = T.grad(example_cost, example_variable)
# if no_grad is > 0 then return the gradient, otherwise return zeros array
out_grad = ifelse(T.gt(no_grad,0), example_gradient, T.zeros_like(example_variable))
train_f = theano.function([x, y, no_grad], [out_grad], updates=updates)
So when you want to retrieve the gradients you call
train_f(x_data, y_data, 1)
otherwise
train_f(x_data, y_data, 0)
I want to make use of Theano's logistic regression classifier, but I would like to make an apples-to-apples comparison with previous studies I've done to see how deep learning stacks up. I recognize this is probably a fairly simple task if I was more proficient in Theano, but this is what I have so far. From the tutorials on the website, I have the following code:
def errors(self, y):
# check if y has same dimension of y_pred
if y.ndim != self.y_pred.ndim:
raise TypeError(
'y should have the same shape as self.y_pred',
('y', y.type, 'y_pred', self.y_pred.type)
)
# check if y is of the correct datatype
if y.dtype.startswith('int'):
# the T.neq operator returns a vector of 0s and 1s, where 1
# represents a mistake in prediction
return T.mean(T.neq(self.y_pred, y))
I'm pretty sure this is where I need to add the functionality, but I'm not certain how to go about it. What I need is either access to y_pred and y for each and every run (to update my confusion matrix in python) or to have the C++ code handle the confusion matrix and return it at some point along the way. I don't think I can do the former, and I'm unsure how to do the latter. I've done some messing around with an update function along the lines of:
def confuMat(self, y):
x=T.vector('x')
classes = T.scalar('n_classes')
onehot = T.eq(x.dimshuffle(0,'x'),T.arange(classes).dimshuffle('x',0))
oneHot = theano.function([x,classes],onehot)
yMat = T.matrix('y')
yPredMat = T.matrix('y_pred')
confMat = T.dot(yMat.T,yPredMat)
confusionMatrix = theano.function(inputs=[yMat,yPredMat],outputs=confMat)
def confusion_matrix(x,y,n_class):
return confusionMatrix(oneHot(x,n_class),oneHot(y,n_class))
t = np.asarray(confusion_matrix(y,self.y_pred,self.n_out))
print (t)
But I'm not completely clear on how to get this to interface with the function in question and give me a numpy array I can work with.
I'm quite new to Theano, so hopefully this is an easy fix for one of you. I'd like to use this classifer as my output layer in a number of configurations, so I could use the confusion matrix with other architectures.
I suggest using a brute force sort of a way. You need an output for a prediction first. Create a function for it.
prediction = theano.function(
inputs = [index],
outputs = MLPlayers.predicts,
givens={
x: test_set_x[index * batch_size: (index + 1) * batch_size]})
In your test loop, gather the predictions...
labels = labels + test_set_y.eval().tolist()
for mini_batch in xrange(n_test_batches):
wrong = wrong + int(test_model(mini_batch))
predictions = predictions + prediction(mini_batch).tolist()
Now create confusion matrix this way:
correct = 0
confusion = numpy.zeros((outs,outs), dtype = int)
for index in xrange(len(predictions)):
if labels[index] is predictions[index]:
correct = correct + 1
confusion[int(predictions[index]),int(labels[index])] = confusion[int(predictions[index]),int(labels[index])] + 1
You can find this kind of an implementation in this repository.