Get a consistent path for the file pool (or another way to isolate a subpart of a file name)

JKellogg

In my experiment I want to isolate a part of various sound file names. For instance, if I have the file name Dutch_112.wav , I want to isolate the string 112. I want a way to automize this so that it can apply for various sound files that are called in a sequence. I've been struggling, though, to find a way to call a sound file name without the file pool path. I could simply trim the file path from the name, like the following code shows:

(full sound file path is /var/folders/sg/8vk99wn53bckkgs0j95ydyh0000gn/T/tmph0k3emlj.opensesamepool/Dutch_112.wav ).

sound_file = pool["Dutch_112.wav"]

sound_name = sound_file.replace("/var/folders/sg/8vk99wn53bckkgs0j95ydyh0000gn/T/tmph0k3emlj.opensesamepool/", "")

seq = sound_name.replace("Dutch_", " ")

seq = seq.replace(".wav", " ")

Which gets me just 112.

A problem I'm running into, however, is that every time I reopen my experiment, the file pool path has changed. I have to re-copy-and-paste the file pool path into my replace() code. This is doable, but is there a more efficient way to achieve this? One idea I had was to "set" a consistent file pool directory so the path to trim is always the same, but I can't figure out how to do that (if it's possible). Alternatively there may be a simpler way to get the name of the sound file without the extraneous path. Any ideas?

eduard

Hi Jordan,

You don't need to reinvent the wheel. The os package has many functions to operate with paths. You could do:

filename = '/var/folders/sg/8vk99wn53bckkgs0j95ydyh0000gn/T/tmph0k3emlj.opensesamepool/Dutch_112.wav'
import os
sound_file = os.path.splitext(os.path.basename(filename))[0]
seq = sound_file.split('_')[1]

Eduard

JKellogg

This worked, thank you!