How do I open a file in python with a relative path using with open()?

Question

I am trying to not use an absolute path for my configuration file because I need this deployed in multiple environments, what is my best option here

The Below code is what I have tried and it is not able to find the path, however I am able to cat the file in the same location. I am using Python3.6 on a Redhat server.

with open("~/scripts/config.yml", 'r') as ymlfile:
    cfg = yaml.load(ymlfile)

I am getting the below error:

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
FileNotFoundError: [Errno 2] No such file or directory: '~/scripts/config.yml'

What you are providing is not a relative path. ``~`` is a shell expansion (https://www.gnu.org/software/bash/manual/html_node/Tilde-Expansion.html), not unlike ``$HOME``. You have to do this expansion in Python explicitly. — MisterMiyagi, Jul 02 '19 at 11:19
Possible duplicate of [How to get the home directory in Python?](https://stackoverflow.com/questions/4028904/how-to-get-the-home-directory-in-python) — MisterMiyagi, Jul 02 '19 at 11:21

score 7 · Answer 1 · answered Jul 02 '19 at 11:15

7

First of all, ~/path/to/file is always an absolute path (~ expands to $HOME). To make this substitution in Python, you need to use os.path.expanduser such as:

with open(os.path.expanduser("~/scripts/config.yml"), 'r') as ymlfile:
    cfg = yaml.load(ymlfile)

answered Jul 02 '19 at 11:15

ig0774

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An even neater response, thank you, this worked as well – EngineJanwaar Jul 03 '19 at 08:57

score 1 · Accepted Answer · answered Jul 02 '19 at 11:13

1

You can do it with:

import os
path = os.getenv('HOME') + '/scripts/config.yaml'

~ only works in shell, not in a Python string

answered Jul 02 '19 at 11:13

jacalvo

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Thank you, this method worked for me, thanks for the quick response. – EngineJanwaar Jul 03 '19 at 08:56

How do I open a file in python with a relative path using with open()?

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