python: Change the scripts working directory to the script's own directory
Question:
I run a python shell from crontab every minute:
* * * * * /home/udi/foo/bar.py
/home/udi/foo has some necessary subdirectories, like /home/udi/foo/log and /home/udi/foo/config, which /home/udi/foo/bar.py refers to.
The problem is that crontab runs the script from a different working directory, so trying to open ./log/bar.log fails.
Is there a nice way to tell the script to change the working directory to the script’s own directory? I would fancy a solution that would work for any script location, rather than explicitly telling the script where it is.
EDIT:
os.chdir(os.path.dirname(sys.argv[0]))
Was the most compact elegant solution. Thanks for your answers and explanations!
Answers:
This will change your current working directory to so that opening relative paths will work:
import os
os.chdir("/home/udi/foo")
However, you asked how to change into whatever directory your Python script is located, even if you don’t know what directory that will be when you’re writing your script. To do this, you can use the os.path functions:
import os
abspath = os.path.abspath(__file__)
dname = os.path.dirname(abspath)
os.chdir(dname)
This takes the filename of your script, converts it to an absolute path, then extracts the directory of that path, then changes into that directory.
Don’t do this.
Your scripts and your data should not be mashed into one big directory. Put your code in some known location (site-packages or /var/opt/udi or something) separate from your data. Use good version control on your code to be sure that you have current and previous versions separated from each other so you can fall back to previous versions and test future versions.
Bottom line: Do not mingle code and data.
Data is precious. Code comes and goes.
Provide the working directory as a command-line argument value. You can provide a default as an environment variable. Don’t deduce it (or guess at it)
Make it a required argument value and do this.
import sys
import os
working= os.environ.get("WORKING_DIRECTORY","/some/default")
if len(sys.argv) > 1: working = sys.argv[1]
os.chdir( working )
Do not “assume” a directory based on the location of your software. It will not work out well in the long run.
You can get a shorter version by using sys.path[0].
os.chdir(sys.path[0])
From http://docs.python.org/library/sys.html#sys.path
As initialized upon program startup, the first item of this list,
path[0], is the directory containing the script that was used to
invoke the Python interpreter
Change your crontab command to
* * * * * (cd /home/udi/foo/ || exit 1; ./bar.py)
The (...) starts a sub-shell that your crond executes as a single command. The || exit 1 causes your cronjob to fail in case that the directory is unavailable.
Though the other solutions may be more elegant in the long run for your specific scripts, my example could still be useful in cases where you can’t modify the program or command that you want to execute.
Just in case, you could also use pathlib
import os
from pathlib import Path
file = Path(__file__)
parent = file.parent
os.chdir(parent)
# print(file, parent, os.getcwd())
Or in single line:
os.chdir(Path(__file__).parent)
I run a python shell from crontab every minute:
* * * * * /home/udi/foo/bar.py
/home/udi/foo has some necessary subdirectories, like /home/udi/foo/log and /home/udi/foo/config, which /home/udi/foo/bar.py refers to.
The problem is that crontab runs the script from a different working directory, so trying to open ./log/bar.log fails.
Is there a nice way to tell the script to change the working directory to the script’s own directory? I would fancy a solution that would work for any script location, rather than explicitly telling the script where it is.
EDIT:
os.chdir(os.path.dirname(sys.argv[0]))
Was the most compact elegant solution. Thanks for your answers and explanations!
This will change your current working directory to so that opening relative paths will work:
import os
os.chdir("/home/udi/foo")
However, you asked how to change into whatever directory your Python script is located, even if you don’t know what directory that will be when you’re writing your script. To do this, you can use the os.path functions:
import os
abspath = os.path.abspath(__file__)
dname = os.path.dirname(abspath)
os.chdir(dname)
This takes the filename of your script, converts it to an absolute path, then extracts the directory of that path, then changes into that directory.
Don’t do this.
Your scripts and your data should not be mashed into one big directory. Put your code in some known location (site-packages or /var/opt/udi or something) separate from your data. Use good version control on your code to be sure that you have current and previous versions separated from each other so you can fall back to previous versions and test future versions.
Bottom line: Do not mingle code and data.
Data is precious. Code comes and goes.
Provide the working directory as a command-line argument value. You can provide a default as an environment variable. Don’t deduce it (or guess at it)
Make it a required argument value and do this.
import sys
import os
working= os.environ.get("WORKING_DIRECTORY","/some/default")
if len(sys.argv) > 1: working = sys.argv[1]
os.chdir( working )
Do not “assume” a directory based on the location of your software. It will not work out well in the long run.
You can get a shorter version by using sys.path[0].
os.chdir(sys.path[0])
From http://docs.python.org/library/sys.html#sys.path
As initialized upon program startup, the first item of this list,
path[0], is the directory containing the script that was used to
invoke the Python interpreter
Change your crontab command to
* * * * * (cd /home/udi/foo/ || exit 1; ./bar.py)
The (...) starts a sub-shell that your crond executes as a single command. The || exit 1 causes your cronjob to fail in case that the directory is unavailable.
Though the other solutions may be more elegant in the long run for your specific scripts, my example could still be useful in cases where you can’t modify the program or command that you want to execute.
Just in case, you could also use pathlib
import os
from pathlib import Path
file = Path(__file__)
parent = file.parent
os.chdir(parent)
# print(file, parent, os.getcwd())
Or in single line:
os.chdir(Path(__file__).parent)