Python – How to reliably open a file in the same directory as the currently running script

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I used to open files that were in the same directory as the currently running Python script by simply using a command like:

open("Some file.txt", "r")

However, I discovered that when the script was run in Windows by double-clicking it, it would try to open the file from the wrong directory.

Since then I've used a command of the form

open(os.path.join(sys.path[0], "Some file.txt"), "r")

whenever I wanted to open a file. This works for my particular usage, but I'm not sure if sys.path[0] might fail in some other use case.

So my question is: What is the best and most reliable way to open a file that's in the same directory as the currently running Python script?

Here's what I've been able to figure out so far:

os.getcwd() and os.path.abspath('') return the "current working directory", not the script directory.
os.path.dirname(sys.argv[0]) and os.path.dirname(__file__) return the path used to call the script, which may be relative or even blank (if the script is in the cwd). Also, __file__ does not exist when the script is run in IDLE or PythonWin.
sys.path[0] and os.path.abspath(os.path.dirname(sys.argv[0])) seem to return the script directory. I'm not sure if there's any difference between these two.

Edit:

I just realized that what I want to do would be better described as "open a file in the same directory as the containing module". In other words, if I import a module I wrote that's in another directory, and that module opens a file, I want it to look for the file in the module's directory. I don't think anything I've found is able to do that…

Best Answer

I always use:

__location__ = os.path.realpath(
    os.path.join(os.getcwd(), os.path.dirname(__file__)))

The join() call prepends the current working directory, but the documentation says that if some path is absolute, all other paths left of it are dropped. Therefore, getcwd() is dropped when dirname(__file__) returns an absolute path.

Also, the realpath call resolves symbolic links if any are found. This avoids troubles when deploying with setuptools on Linux systems (scripts are symlinked to /usr/bin/ -- at least on Debian).

You may the use the following to open up files in the same folder:

f = open(os.path.join(__location__, 'bundled-resource.jpg'))
# ...

I use this to bundle resources with several Django application on both Windows and Linux and it works like a charm!

Related Solutions

Bash – How to get the source directory of a Bash script from within the script itself

#!/usr/bin/env bash

SCRIPT_DIR="$( cd -- "$( dirname -- "${BASH_SOURCE[0]}" )" &> /dev/null && pwd )"

is a useful one-liner which will give you the full directory name of the script no matter where it is being called from.

It will work as long as the last component of the path used to find the script is not a symlink (directory links are OK). If you also want to resolve any links to the script itself, you need a multi-line solution:

#!/usr/bin/env bash

SOURCE="${BASH_SOURCE[0]}"
while [ -h "$SOURCE" ]; do # resolve $SOURCE until the file is no longer a symlink
  DIR="$( cd -P "$( dirname "$SOURCE" )" >/dev/null 2>&1 && pwd )"
  SOURCE="$(readlink "$SOURCE")"
  [[ $SOURCE != /* ]] && SOURCE="$DIR/$SOURCE" # if $SOURCE was a relative symlink, we need to resolve it relative to the path where the symlink file was located
done
DIR="$( cd -P "$( dirname "$SOURCE" )" >/dev/null 2>&1 && pwd )"

This last one will work with any combination of aliases, source, bash -c, symlinks, etc.

Beware: if you cd to a different directory before running this snippet, the result may be incorrect!

Also, watch out for $CDPATH gotchas, and stderr output side effects if the user has smartly overridden cd to redirect output to stderr instead (including escape sequences, such as when calling update_terminal_cwd >&2 on Mac). Adding >/dev/null 2>&1 at the end of your cd command will take care of both possibilities.

To understand how it works, try running this more verbose form:

#!/usr/bin/env bash

SOURCE="${BASH_SOURCE[0]}"
while [ -h "$SOURCE" ]; do # resolve $SOURCE until the file is no longer a symlink
  TARGET="$(readlink "$SOURCE")"
  if [[ $TARGET == /* ]]; then
    echo "SOURCE '$SOURCE' is an absolute symlink to '$TARGET'"
    SOURCE="$TARGET"
  else
    DIR="$( dirname "$SOURCE" )"
    echo "SOURCE '$SOURCE' is a relative symlink to '$TARGET' (relative to '$DIR')"
    SOURCE="$DIR/$TARGET" # if $SOURCE was a relative symlink, we need to resolve it relative to the path where the symlink file was located
  fi
done
echo "SOURCE is '$SOURCE'"
RDIR="$( dirname "$SOURCE" )"
DIR="$( cd -P "$( dirname "$SOURCE" )" >/dev/null 2>&1 && pwd )"
if [ "$DIR" != "$RDIR" ]; then
  echo "DIR '$RDIR' resolves to '$DIR'"
fi
echo "DIR is '$DIR'"

And it will print something like:

SOURCE './scriptdir.sh' is a relative symlink to 'sym2/scriptdir.sh' (relative to '.')
SOURCE is './sym2/scriptdir.sh'
DIR './sym2' resolves to '/home/ubuntu/dotfiles/fo fo/real/real1/real2'
DIR is '/home/ubuntu/dotfiles/fo fo/real/real1/real2'

Python – How to find the location of the Python site-packages directory

There are two types of site-packages directories, global and per user.

Global site-packages ("dist-packages") directories are listed in sys.path when you run:
```
python -m site
```
For a more concise list run getsitepackages from the site module in Python code:
```
python -c 'import site; print(site.getsitepackages())'
```
Note: With virtualenvs getsitepackages is not available, sys.path from above will list the virtualenv's site-packages directory correctly, though. In Python 3, you may use the sysconfig module instead:
```
python3 -c 'import sysconfig; print(sysconfig.get_paths()["purelib"])'
```
The per user site-packages directory (PEP 370) is where Python installs your local packages:
```
python -m site --user-site
```
If this points to a non-existing directory check the exit status of Python and see python -m site --help for explanations.

Hint: Running pip list --user or pip freeze --user gives you a list of all installed per user site-packages.

Practical Tips

<package>.__path__ lets you identify the location(s) of a specific package: (details)

$ python -c "import setuptools as _; print(_.__path__)"
['/usr/lib/python2.7/dist-packages/setuptools']

<module>.__file__ lets you identify the location of a specific module: (difference)
```
$ python3 -c "import os as _; print(_.__file__)"
/usr/lib/python3.6/os.py
```

Run pip show <package> to show Debian-style package information:

$ pip show pytest
Name: pytest
Version: 3.8.2
Summary: pytest: simple powerful testing with Python
Home-page: https://docs.pytest.org/en/latest/
Author: Holger Krekel, Bruno Oliveira, Ronny Pfannschmidt, Floris Bruynooghe, Brianna Laugher, Florian Bruhin and others
Author-email: None
License: MIT license
Location: /home/peter/.local/lib/python3.4/site-packages
Requires: more-itertools, atomicwrites, setuptools, attrs, pathlib2, six, py, pluggy

Best Answer

Related Solutions

Bash – How to get the source directory of a Bash script from within the script itself

Python – How to find the location of the Python site-packages directory

Practical Tips

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