PyTorch gradient accumulation training loop

gradient_accumulation.py

model.zero_grad()                                   # Reset gradients tensors
for i, (inputs, labels) in enumerate(training_set):
    predictions = model(inputs)                     # Forward pass
    loss = loss_function(predictions, labels)       # Compute loss function
    loss = loss / accumulation_steps                # Normalize our loss (if averaged)
    loss.backward()                                 # Backward pass
    if (i+1) % accumulation_steps == 0:             # Wait for several backward steps
        optimizer.step()                            # Now we can do an optimizer step
        model.zero_grad()                           # Reset gradients tensors
        if (i+1) % evaluation_steps == 0:           # Evaluate the model when we...
            evaluate_model()                        # ...have no gradients accumulated

hi thom..
if we have to simulate the case where we to process more number of images in a batch say due to GPU limitation i can keep batch size to be only 64 . But suppose i want to process 128 images . In that case shouldnt we do this way

total_loss=0
for i, (inputs, labels) in enumerate(training_set):
    predictions = model(inputs)                     # Forward pass
    loss = loss_function(predictions, labels)       # Compute loss function
  
    if (i+1) % accumulation_steps == 0:             # Wait for several backward steps
        total_loss = (total_loss+loss )/ accumulation_steps                # Normalize our loss (if averaged)
        loss.backward()                                 # Backward pass
        optimizer.step()                            # Now we can do an optimizer step
        model.zero_grad()                           # Reset gradients tensors
        total_loss=0
        if (i+1) % evaluation_steps == 0:           # Evaluate the model when we...
            evaluate_model()    
   else :
        total_loss=loss+total_loss

Hi I'm using the similar idea. Do you have any further suggestions? Thanks

@jaideep11061982 , @vaneshieh , this code is incorrect and has multiple mistakes.
Explanation (a bit rewritten for readability):

total_loss = 0
for i, (inputs, labels) in enumerate(training_set, 1):
    # Given absense of torch.no_grad(),
    # `loss` will contain all immediate states of forward pass, needed for backward computation
    predictions = model(inputs)
    loss = loss_function(predictions, labels)
  
    # Summation here aggregates loss from N accumulation steps
    # Thus total_loss has immediate states of NUM_ACCUMULATES forwards
    # So no memory saving here, but leak
    total_loss += loss
    if i % NUM_ACCUMULATES == 0:
        total_loss = total_loss / NUM_ACCUMULATES 
        
        # Here we do backward on loss from last step, so grads in model are influenced only by each Nth batch
        loss.backward()
        optimizer.step()
        optimizer.zero_grad()
        total_loss = 0
        
        if i % (NUM_ACCUMULATES * NUM_UPDATES)== 0:
            evaluate_model()    

So this code updates weights using only 1/N of training set, eating N times more memory than training with single accumulation step.
Thus it's equal to those (to make mistakes more obvious):

total_loss = 0
batches = []
for i, (inputs, labels) in enumerate(training_set, 1):
    batches.append((inputs, labels))
    
    if i % NUM_ACCUMULATES == 0:
        # Store immediate buffers from NUM_ACCUMULATES forward passes
        losses = [loss_function(model(inputs), labels) for inputs in batches]
        total_loss = sum(losses) / NUM_ACCUMULATES
        
        # Do backward only on last loss
        losses[-1].backward()
    
        optimizer.step()
        optimizer.zero_grad()
        batches = []
        
        if i % (NUM_ACCUMULATES * NUM_UPDATES) == 0:
            evaluate_model()

Original code of @thomwolf is also not without mistakes, though they are less striking.
I have rewriten it to be more error-prone:

# Zero saved gradients
optimizer.zero_grad()

total_loss = 0
for i, (inputs, labels) in enumerate(training_set, 1):
    # Do forward pass and store immediate buffers
    predictions = model(inputs)
    loss = loss_function(predictions, labels)

    # Add gradients from this batch to saved ones, divide loss by NUM_ACCUMULATES if it's averaged over samples
    (loss / NUM_ACCUMULATES).backward()
    
    # Drop immediate buffers, mandatory
    # This guaranties that total loss doesn't bring some immediate buffers with it
    predictions.detach_()
    loss.detach_()
    total_loss += loss

    if i % NUM_ACCUMULATES == 0:
        # Update parameters using saved gradients
        optimizer.step()

        # Zero saved gradients
        optimizer.zero_grad()

        if i % (NUM_ACCUMULATES * NUM_UPDATES) == 0:
            print(f'train loss: {total_loss / NUM_ACCUMULATES / NUM_UPDATES}')
            total_loss = 0
            evaluate_model()

But seriously, use some more high level wrapper for Pytorch, like catalyst, or pytorch-lightning, or even ignite,
as they will prevent you from those mistakes.

@arquolo - I was hoping to understand the changes you have suggested to @thomwolf's code. Are these changes optimization changes (drop immediate buffers) or do these changes actually affect the value of the gradient ? I apologise in advance if this a rudimentary question.

@Alex-Mathai-98 W.r.t. @thomwolf's code my changes are optimization to not keep gradient-linked data when it's not needed anymore and they don't affect the value of the gradient.
The main difference is detach() calls to not keep gradient used by last backward after end of the loop or in-between of iterations.

@arquolo - okay thankyou for the clarification.

@thomwolf Thanks for the code. Just a little fix in the condition at line 7:

if (i+1) % accumulation_steps == 0:

This assumes that the number of batches is perfectly divisble by the accumulation steps. However, if there are, say, 10, batches, and the accumulation steps are 4, the last two batches would not make to the optimizer.step().

And even if you add an extra condition, you will still need to adjust the normalization denominator because there will be only 2 not 4 accumulation steps.

So the updated code would be:

n = len(training_set)
remainder_batches = n % accumulation_steps # calculate number of remainder batches

for i, (inputs, labels) in enumerate(training_set):
    predictions = model(inputs)
    loss = loss_function(predictions, labels)
    remaining = n - i

    # update the denominator if the remaining batches are leq number of remainder batches 
    denominator = remainder_batches if remaining <= remainder_batches else accumulation_steps

    loss = loss / denominator
    loss.backward()
    if (i+1) % accumulation_steps == 0 or i == n - 1: # add condition for last iteration
        optimizer.step()
        model.zero_grad()

You can emulate the logic in a standalone script as follows:

def get_value():
    return 5


n = 10
steps = 4
values = []
val = 0
remainder = n % steps
for i in range(n):
    a = get_value()
    remaining = n - i
    if remaining <= remainder:
        denom = n % steps
    else:
        denom = steps
    val += a / denom
    print(i, denom)
    if (i + 1) % steps == 0 or i == n - 1:
        values.append(val)
        val = 0
        print("update")