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Import python package from local directory into interpreter

I'm developing/testing a package in my local directory. I want to import it in the interpreter (v2.5), but sys.path does not include the current directory. Right now I type in sys.path.insert(0,'.') . Is there a better way?

Also,

from . import mypackage

fails with this error:

ValueError: Attempted relative import in non-package

You can use relative imports only from in a module that was in turn imported as part of a package -- your script or interactive interpreter wasn't, so of course from . import from . import (which means "import from the same package I got imported from") doesn't work. import mypackage will be fine once you ensure the parent directory of mypackage is in sys.path (how you managed to get your current directory away from sys.path I don't know -- do you have something strange in site.py, or...?)

To get your current directory back into sys.path there is in fact no better way than putting it there.

See the documentation for sys.path:

http://docs.python.org/library/sys.html#sys.path

To quote:

If the script directory is not available (eg if the interpreter is invoked interactively or if the script is read from standard input), path[0] is the empty string, which directs Python to search modules in the current directory first.

So, there's no need to monkey with sys.path if you're starting the python interpreter from the directory containing your module.

Also, to import your package, just do:

import mypackage

Since the directory containing the package is already in sys.path, it should work fine.

Keep it simple:

 try:
     from . import mymodule     # "myapp" case
 except:
     import mymodule            # "__main__" case

If you want to run an unmodified python script so it imports libraries from a specific local directory you can set the PYTHONPATH environment variable - eg in bash:

export PYTHONPATH=/home/user/my_libs
python myscript.py

If you just want it to import from the current working directory use the . notation:

export PYTHONPATH=.
python myscript.py

如果在一个包里面有setup.py ,那么最好安装它

pip install -e .

A simple way to make it work is to run your script from the parent directory using python's -m flag, eg python -m packagename.scriptname . Obviously in this situation you need an __init__.py file to turn your directory into a package.

Using sys.path should include current directory already.

Try:

import .

or:

from . import sth

however it may be not a good practice, so why not just use:

import mypackage

A bit late to the party, but this is what worked for me:

>>> import sys
>>> sys.path.insert(0, '')

Apparently, if there is an empty string, Python knows that it should look in the current directory. I did not have the empty string in sys.path , which caused this error.

I used pathlib to add my module directory to my system path as I wanted to avoid installing the module as a package and @maninthecomputer response didn't work for me

import sys
from pathlib import Path

cwd = str(Path(__file__).parent)
sys.path.insert(0, cwd)
from my_module import my_function

Speaking for python3.. I wanted to use an improved version of a library that's installed in my environment. There are some extra print statements it makes to show that it and not the original lib is being used.

I placed the lib's folder next to the python script. Ran the script.. it ran with the local lib with the modifications.

Removed the folder and ran it again - this time it ran with the installed lib.

So, solution was simple : place the lib's folder (with same name as in your import statement) in your project folder. That does the job, at least at my end.

This is on a standard Linux Mint 20.04 system, with a python 3.8 virutal environment activated (so "(py3.8)" appears in my terminal when I'm in the virtual env)

You can import package_name if the package is a module: this needs you have init .py under the package and things that you want to use are needed to import in the init .py

Or if you want to import class under the package, you can use from package_name import class_name

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