I've found that I'm using this pattern a lot :
os.path.join(os.path.dirname(__file__), file_path)
so I've decided to put in a function in a file that has many such small utilities:
def filepath_in_cwd(file_path):
return os.path.join(os.path.dirname(__file__), file_path)
The thing is, __file__ returns the current file and therefore the current folder, and I've missed the whole point. I could do this ugly hack (or just keep writing the pattern as is):
def filepath_in_cwd(py_file_name, file_path):
return os.path.join(os.path.dirname(py_file_name), file_path)
and then the call to it will look like this:
filepath_in_cwd(__file__, "my_file.txt")
but I'd prefer it if I had a way of getting the __file__ of the function that's one level up in the stack. Is there any way of doing this?