Python, $PYTHONPATH, sys.path.add()

Question

This code indicates what I wish to do, i.e.

Have a variable to hold the name of a file to import,
Use shutil.which() to look up the FOLDER it resides in (via $PATH),
Automatically get that added to PYTHONPATH to make import find it.

name='numbas'  
d=os.path.dirname(shutil.which(name+'.py'))  
sys.path.append(d)  
from name import *

Now import does not allow a name stored in a variable.
eval() on a string that contains the import statement is also "nonfunctional".

Any ideas on how to get this thing to actually work?

The reason for all this is that I have *.py scripts in a Util/ -folder - which are self-contained scripts AND import modules for other scripts to utilize.

If I move these around or copy them to a friend I wish to have the folder structure remain as is.

To import a module with a specified name, you can use the function importlib.import_module. — Phant
– Phant, Commented Dec 4 at 15:33

Dharman · Accepted Answer · 2025-12-05 11:32:17Z

0

Right, timez.py used as command line script, and by importing and calling main() :

The fact that you get a "<module>" back from importlib.import_module(name) was not clear from the docs I found. a type(...) revealed it though. I would have expected it to be added in the same manner as with a pure import.

$ ./timez.py 10 1000 20
... has rotated 100.0 revs/hour at 2.0 rpm in average, while utilized 50.0% of the power hours.

$ python -c "import importlib;name='timez'; timez=importlib.__import__(name);timez.main(['mama',10, 1000, 20])"
... has rotated 100.0 revs/hour at 2.0 rpm in average, while utilized 50.0% of the power hours.

$ python -c "import importlib;name='timez'; timez=importlib.import_module(name);timez.main(['mama',10, 1000, 20])"
... has rotated 100.0 revs/hour at 2.0 rpm in average, while utilized 50.0% of the power hours.

edited Dec 5 at 11:32

Dharman♦

34k27 gold badges106 silver badges158 bronze badges

answered Dec 4 at 18:28

Hannu

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