[solved] get path of the experiment file

embodiment

Hi. I'm wondering if it is possible to automatically obtain the path of the current opensesame file. The reason is that I want opensesame to search for the stimuli inside a folder placed in the same path of the experiment file. I don't want to use the file pool because I have about 2000 images, so opensesame is slow in loading and saving the experiment. On the other hand, of course I can provide the absolute paths to the sketchpad, but then I will have to modify them when I move the experiment on the laboratory PC.
Thanks.

lvanderlinden

Hi,

The get_file() function first looks in the file pool and then looks in the folder of the current experiment. So path = exp.get_file("my_image.png") will return the full path of the image, even if it's not stored in the file pool but merely in the same folder as your experiment. And you're right, storing your stimuli this way will make loading and saving much faster!

Fore more info, see:

• http://osdoc.cogsci.nl/python-inline-code/experiment-functions/#experiment.get_file

Does this help?

Best,

Lotje

embodiment

Yes, but in my case the stimuli are in a subfolder of the folder of the current experiment. Have I necessarily to move the opensesame file inside the stimuli folder? I would prefer to leave a folder with only the pictures because in this way everything looks neater.

boris

Hi,

Imagine that your image are in the "images" folder of the current experiment.
Then I guess that

path = exp.get_file("\\images\\my_image.png")

should work...

embodiment

Hi boris. Unfortunately it does not work.

lvanderlinden

Hi Andrea,

You could determine the experiment folder by using the built-in Python module os:

# Some important paths. On Windows, the working directory
# is always the OpenSesame program folder, on other platforms
# it may not be.  

import os

print 'The experiment path is %s' % exp.experiment_path
print 'The working directory is %s' % os.getcwd()
print 'The path to the file pool is %s' % exp.pool_folder

Alternatively, you could use exp.get_file() like so:
path = exp.get_file('subfolder/image.png')

Best wishes,

Lotje

embodiment

Ok, solved! I was doing and incorrect use of the exp.get_file() function.

Many many thanks.

lvanderlinden

Great to hear!