How to import other file using relative path to executed file not python command path?

Question

My folder trees:

./
├── README.MD
├── basic
│   └── thresh.py
├── images
│   └── figure.jpg
└── utils
    ├── util.py
    └── util.pyc

I want to import util.py in thresh.py:

import sys
sys.path.append('../utils')
import util

When I run command $ python thresh.py in the basic folder, it's allright. But run $ python ./basic/thresh.py in the topmost folder, I will get the error:

ImportError: No module named util

So how to make $ python ./basic/thresh.py and $ python thresh.py both work to import file by given the file's relative path to executed file regardless of python command path?

sancho.s ReinstateMonicaCellio · Accepted Answer · 2017-12-28 07:34:47Z

1

You can get the absolute path of the script you are executing with (there are other variants also using __file__, but this should work)

import os
wk_dir = os.path.dirname(os.path.realpath('__file__'))
print( wk_dir )

and then get your dir with it, e.g.

import sys
sys.path.append(wk_dir+'/../utils')

PS: You might need to use __file__ instead of '__file__'.

edited Dec 28, 2017 at 7:34

answered Dec 27, 2017 at 4:46

sancho.s ReinstateMonicaCellio

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3 Comments

tomfriwel Over a year ago

This work. But you need change '__file__' to __file__ in your code.

sancho.s ReinstateMonicaCellio Over a year ago

@tomfriwel - The code posted was tested and it works. I found it somehwere else too. The change you suggest does not work for me, although I have seen other posts using it similarly. I did not explore this in detail, but the important points are: 1) you solved your problem, 2) others are aware that they might need to try both.

tomfriwel Over a year ago

Maybe we use different version of Python.

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