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For single-file scripts, I can always execute them directly, at least on linux, by setting executable permissions on the script and adding a shebang (#! /bin/python) at the top of the file.

Is an equivalent possible for packages, so that I don't need to prefix the package name with python -m some_package on the command line?

Update: I'm aware there are workarounds, such as wrapping it in a script. Not only is this not always practical, but also my question isn't about finding any way to do it, rather to understand what the "official" analog to running "some_script.py" directly, but for packages, assuming that such an analog even exists at all or is simply not intended or not defined anywhere.

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  • Typo: The 1st line should be "#!/bin/python" (no space). Read man python and try #!/bin/python -m some_package. Commented Dec 26, 2023 at 23:26
  • shebang is correct and more legible and always works. if not, please provide references, since "man python" doesn't seem to mention it. Moreover, your second form is a more complicated form of what I'm already doing. I don't think you even read the question, let alone "man python". Commented Dec 27, 2023 at 0:13
  • I do not have python anymore, but only have python3. So all my shebangs are #!/usr/bin/python3 Or #!/usr/bin/env python3 Seen discussions on which is better. Commented Dec 27, 2023 at 14:47

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Simply wrap it in a bash script:

#!/bin/bash
python -m some_package your_script

Store the above on your $PATH (~/bin), call it foo, make it executable (chmod u+x ~/bin/foo), then call it as foo.

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