I started using Python few days back and I think I have a very basic question where I am stuck. Maybe I am not doing it correctly in Python so wanted some advice from the experts:
I have a config.cfg & a class test in one package lib as follows:
myProj/lib/pkg1/config.cfg
[api_config]
url = https://someapi.com/v1/
username=sumitk
myProj/lib/pkg1/test.py
class test(object):
def __init__(self, **kwargs):
config = ConfigParser.ConfigParser()
config.read('config.cfg')
print config.get('api_config', 'username')
#just printing here but will be using this as a class variable
def some other foos()..
Now I want to create an object of test in some other module in a different package
myProj/example/useTest.py
from lib.pkg1.test import test
def temp(a, b, c):
var = test()
def main():
temp("","","")
if __name__ == '__main__':
main()
Running useTest.py is giving me error:
...
print config.get('api_config', 'username')
File "C:\Python27\lib\ConfigParser.py", line 607, in get
raise NoSectionError(section)
ConfigParser.NoSectionError: No section: 'api_config'
Now if I place thie useTest.py in the same package it runs perfectly fine:
myProj/lib/pkg1/useTest.py
myProj/lib/pkg1/test.py
myProj/lib/pkg1/config.cfg
I guess there is some very basic package access concept in Python that I am not aware of or is there something I am doing wrong here?
The issue here is that you have a different working directory depending on which module is your main script. You can check the working directory by adding the following lines to the top of each script:
import os
print os.getcwd()
Because you just provide 'config.cfg' as your file name, it will attempt to find that file inside of the working directory.
To fix this, give an absolute path to your config file.
You should be able to figure out the absolute path with the following method since you know that config.cfg and test.py are in the same directory:
# inside of test.py
import os
config_path = os.path.join(os.path.dirname(os.path.abspath(__file__)),
'config.cfg')
Related
I'm using Python 3.9.5.
Based on this post, I'm trying to reuse some functions from the parent directory. Here's my code hierarchy:
github_repository
src
base
string_utilities.py
validation
email_validator.py
I also have __init__.py in all folders. In ALL of them.
Here's the string_utilities.py content:
def isNullOrEmpty(text: str):
return text is not None and len(text) > 0
And here's the email_validator.py content:
from src.base import string_utilities
def is_email(text: str):
if string_utilities.isNullOrEmpty(text):
return False
# logic to check email
return True
Now when I run python email_validator.py, I get this error:
ModuleNotFoundError: No module named 'src'
I have changed that frustrating import statement to all of these different forms, and I still get no results:
from ...src.base import string_utilities
which results in:
ImportError: attempted relative import with no known parent package
import src.base.string_utilities
Which causes compiler to not know the isNullOrEmpty function.
import ...src.base.string_utilities
Which results in:
Relative imports cannot be used with "import .a" form; use "from . import a" instead
I'm stuck at this point on how to reuse that function in this file. Can someone please help?
I am currently having difficulties import some functions which are located in a python file which is in the parent directory of the working directory of the main flask application script. Here's how the structure looks like
project_folder
- public
--app.py
-scripts.py
here's a replica code for app.py:
def some_function():
from scripts import func_one, func_two
func_one()
func_two()
print('done')
if __name__ == "__main__":
some_function()
scripts.py contain the function as such:
def func_one():
print('function one successfully imported')
def func_two():
print('function two successfully imported')
What is the pythonic way of importing those functions in my app.py?
Precede it with a dot so that it searches the current directory (project_folder) instead of your python path:
from .scripts import func_one, func_two
The details of relative imports are described in PEP 328
Edit: I assumed you were working with a package. Consider adding an __init__.py file.
Anyways, you can import anything in python by altering the system path:
import sys
sys.path.append("/path/to/directory")
from x import y
1.
import importlib.util
def loadbasic():
spec = importlib.util.spec_from_file_location("basic", os.path.join(os.path.split(__file__)[0], 'basic.py'))
basic = importlib.util.module_from_spec(spec)
spec.loader.exec_module(basic)
return basic #returns a module
Or create an empty file __init__.py in the directory.
And
do not pollute your path with appends.
File setup:
...\Project_Folder
...\Project_Folder\Project.py
...\Project_folder\Script\TestScript.py
I'm attempting to have Project.py import modules from the folder Script based on user input.
Python Version: 3.4.2
Ideally, the script would look something like
q = str(input("Input: "))
from Script import q
However, python does not recognize q as a variable when using import.
I've tried using importlib, however I cannot figure out how to import from the Script folder mentioned above.
import importlib
q = str(input("Input: "))
module = importlib.import_module(q, package=None)
I'm not certain where I would implement the file path.
Repeat of my answer originally posted at How to import a module given the full path?
as this is a Python 3.4 specific question:
This area of Python 3.4 seems to be extremely tortuous to understand, mainly because the documentation doesn't give good examples! This was my attempt using non-deprecated modules. It will import a module given the path to the .py file. I'm using it to load "plugins" at runtime.
def import_module_from_file(full_path_to_module):
"""
Import a module given the full path/filename of the .py file
Python 3.4
"""
module = None
try:
# Get module name and path from full path
module_dir, module_file = os.path.split(full_path_to_module)
module_name, module_ext = os.path.splitext(module_file)
# Get module "spec" from filename
spec = importlib.util.spec_from_file_location(module_name,full_path_to_module)
module = spec.loader.load_module()
except Exception as ec:
# Simple error printing
# Insert "sophisticated" stuff here
print(ec)
finally:
return module
# load module dynamically
path = "<enter your path here>"
module = import_module_from_file(path)
# Now use the module
# e.g. module.myFunction()
I did this by defining the entire import line as a string, formatting the string with q and then using the exec command:
imp = 'from Script import %s' %q
exec imp
I am trying to calling a function from a module which is located in different directory than the current.
This is code I am using
c:\commands contains a file nlog.py which contains function parselog . i would like to importing this function
into a python file in other directory c:\applications
def parselog(inFileDir = )
### This is in directory c:\commands which need to imported ####
c:\applications\pscan.py is the script which is calling / importing from the above file / directory
if __name__ == '__main__':
#sys.path.append("C:/commands")
sys.path.insert(0,'C:/commands')
from commands import nlog
def pscan (infileDir = )
parselog(inFileDir=r'os.getcwd()') # I am calling the function here
We are getting error
NameError: global name 'parselog' is not defined
I am not sure if I am importing in a wrong way. The C:\commands folder contains _init_.py file. Please let me know what I am missing.
Make and empty __init__.py file in the folder of the module you want to import.
Below code should work.
import sys
sys.path.append(r'c:\commands')
import nlog
parselog(inFileDir=r'os.getcwd()')
I have a module called spellnum. It can be used as a command-line utility (it has the if __name__ == '__main__': block) or it can be imported like a standard Python module.
The module defines a class named Speller which looks like this:
class Speller(object):
def __init__(self, lang="en"):
module = __import__("spelling_" + lang)
# use module's contents...
As you can see, the class constructor loads other modules at runtime. Those modules (spelling_en.py, spelling_es.py, etc.) are located in the same directory as the spellnum.py itself.
Besides spellnum.py, there are other files with utility functions and classes. I'd like to hide those files since I don't want to expose them to the user and since it's a bad idea to pollute the Python's lib directory with random files. The only way to achieve this that I know of is to create a package.
I've come up with this layout for the project (inspired by this great tutorial):
spellnum/ # project root
spellnum/ # package root
__init__.py
spellnum.py
spelling_en.py
spelling_es.py
squash.py
# ... some other private files
test/
test_spellnum.py
example.py
The file __init__.py contains a single line:
from spellnum import Speller
Given this new layout, the code for dynamic module loading had to be changed:
class Speller(object):
def __init__(self, lang="en"):
spelling_mod = "spelling_" + lang
package = __import__("spellnum", fromlist=[spelling_mod])
module = getattr(package, spelling_mod)
# use module as usual
So, with this project layout a can do the following:
Successfully import spellnum inside example.py and use it like a simple module:
# an excerpt from the example.py file
import spellnum
speller = spellnum.Speller(es)
# ...
import spellnum in the tests and run those tests from the project root like this:
$ PYTHONPATH="`pwd`:$PYTHONPATH" python test/test_spellnum.py
The problem
I cannot execute spellnum.py directly with the new layout. When I try to, it shows the following error:
Traceback (most recent call last):
...
File "spellnum/spellnum.py", line 23, in __init__
module = getattr(package, spelling_mod)
AttributeError: 'module' object has no attribute 'spelling_en'
The question
What's the best way to organize all of the files required by my module to work so that users are able to use the module both from command line and from their Python code?
Thanks!
How about keeping spellnum.py?
spellnum.py
spelling/
__init__.py
en.py
es.py
Your problem is, that the package is called the same as the python-file you want to execute, thus importing
from spellnum import spellnum_en
will try to import from the file instead of the package. You could fiddle around with relative imports, but I don't know how to make them work with __import__, so I'd suggest the following:
def __init__(self, lang="en"):
mod = "spellnum_" + lang
module = None
if __name__ == '__main__':
module = __import__(mod)
else:
package = getattr(__import__("spellnum", fromlist=[mod]), mod)