Does .item() affect performance?

autograd
1

Hi there, as far I know we can fill our ram by not releasing the pointer to the computational graph, for example-

losses = []
#...training on batch...
losses.append(loss)

The problem above is that loss still has a copy of the graph!!! So its recommended to do as follows in order to release it-

losses.append(loss.item())
# not losses.append(loss)

My question is if we call .item() on loss for let’s say 2 to 3 times at each batch training step, then is it gonna affect the performance? Or are there any other issues that it might have?

2

.item() ensures that you append only the float values to the list rather the tensor itself. You are basically converting a single element tensor value to a python number. This should not affect the performance in any way.

3

as @charan_Vjy mentioned, item() is pretty inexpensive. Are you running into some situation where it is affecting the performance?

4

Thanks @charan_Vjy, @richard. I just read somewhere that .item() has to do something related to the computational graph, so I thought it might affect performance and memory in some way. I ran some tests and found no difference as such (at least with small models like AlexNet).

5

.item() turns a Tensor into a Python number. Autograd can only keep track of computations on Tensors, so using the result of .item() effectively creates a new value that is not a part of the computation graph.