How to make cuda unavailable in pytorch

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i'm running some code with cudas, and I need to test the same code on CPU to compare running time. To decide between regular pytorch tensor and cuda float tensor, the library I use calls torch.cuda.is_available(). Is there an easy method to make this function return false? I tried changing the Cuda visible devices with 

os.environ["CUDA_VISIBLE_DEVICES"]=""

but torch.cuda.is_available() still return True. I went into pytorch source code, and in my case, torch.cuda.is_avaible returns 

torch._C._cuda_getDeviceCount() > 0

I assume I should be able to "hide" my GPU at the start of my notebook, so the device count is equal to zero, but i didn't get any success so far. Any help is appreciated :)

pytorch cuda
2 Answers

You can make torch.cuda.is_available() return False by overwriting it. Just run the following code as the first thing in your program:

import torch
torch.cuda.is_available = lambda : False
user_107780550

my code

Instead of trying to trick it, why not rewrite your code? For example,

use_gpu = torch.cuda.is_available() and not os.environ['USE_CPU']

Then you can start your program as python runme.py to run on GPU if available, and USE_CPU=1 python3 runme.py to force CPU execution (or make it semi-permanent by export USE_CPU=1).

I tried changing the Cuda visible devices with

You can also try running your code with CUDA_VISIBLE_DEVICES="" python3 runme.py; if you're setting the environment variable inside your code, it might be set later than PyTorch initialisation, and have no effect.

user_2404430
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