Jupyter import Python function

Question

I have the following folder structure

main/ jupyter/ nb.ipynb helper/ text.txt foo/ foo.py

The file foo.py contains

def foo(): open("../text.txt", "r")

In the jupyter notebook I have

import sys sys.path.append("../helper/foo") from foo import foo foo()

which gives a file not found error. What's the cleanest way of fixing that? (If possible, I'd like to keep foo.py unchanged.)

Martin Ueding · Accepted Answer · 2021-04-16 14:16:41Z

2

You get this error because the path is relative to your working directory. You will need to change it using os.chdir('../helper/foo').

It would be a bit better to change foo.py and use os.path.join(os.path.dirname(os.path.dirname(__file__)), 'text.txt') as path.

edited Apr 16, 2021 at 14:16

answered Apr 16, 2021 at 13:10

Martin Ueding

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2 Comments

julian Over a year ago

Thanks a lot for the answer. I also think that the second part is actually better. Only a small typo: it should be __file__.

Martin Ueding Over a year ago

In the grander scheme you could also think about making your helper code an actual package and then load the file with pkgresources. — And thanks for the __FILE__ hint, I was still in C++ mode in my mind.

Wuhuuu2 · Accepted Answer · 2021-04-16 13:19:49Z

It is sufficient to just go one level up and search there for your folder & module

import sys sys.path.append("../") # just one level higher from folder.module import function function()

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