I am trying to set up a library in python. I have created a setup.py file and in a folder under that I have a library folder, and then I tried to create an sample and test folder (for sample code and tests that I would include)
Folders:
- setup.py
- cvImageUtils # this is the library
- __init__.py # this includs ColorUtils
- ColorUtils.py # this includes a class called ColorUtils
- examples
- color.py # this is shown below
init.py in ColorUtils folder
from . import ColorUtils
ColorUtils.py
class ColorUtils:
def __...
Color.py
from cvImageUtils import ColorUtils
m1 = cv2.imread(os.path.join(image_folder, "bb.jpeg"), 1) # 1 = load color
cv2.imshow('grayscale', ColorUtils.convert_rbg_to_grayscale(m1))
At first, it said, unable to find the module, so I added to the top of the file the following based on another SO solution:
import sys
sys.path.append('../')
Now that seems broken to me already, but it did get me past the no module found, but now it says ColorUtils has no method convert_rbg_to_grayscale. So Then I had to change it to ColorUtils.ColorUtils.convert_rbg_to_grayscale
cv2.imshow('grayscale', ColorUtils.ColorUtils.convert_rbg_to_grayscale(m1))
How can I setup the folder so that it allows me to include the library without sys, and call it without declaring ColorUtils twice.
change your __init__.py:
from cvImageUtils.ColorUtils import ColorUtils
I don't think you'll need to import sys anymore, and you don't have import ColorUtils twice. but just like you have to instantiate an object, you should create a ColorUtils object.
my personal preference would be not creating a Class for Utils.
you might have done this already, but if you want to use a method straight from a class like you did in python, you might want to declare it static.
class ColorUtils:
#staticmethod
def util_method():
pass
then you can just do:
ColorUtils.util_method()
Update:
you can read more about relative/absolute import from here as well.
to fix your actual problem though, you can do:
color.py
remove your import sys and sys call from color.py
change: import cvImageUtils.ColorUtils as ct
to: from cvImageUtils.ColorUtils import *
remove all your ct reference instead just use the actual functions.
cvImageUtils/__init__.py
change: from . import ColorUtils
to __all__=['ColorUtils']
I was able to run color.py to get all the images printed out on screen.
a image.png was also generated locally as well.
Every directory that you want to expose in module search(we usually hide test.py) in python need a init.py file. That should be rule of thumb, by using sys module you can add the module to your "module search path".
After having init.py in your directories, you need to import packages/modules/funcitons you want to use:-
import cvImageUtils.ColorUtils.convert_rbg_to_grayscale
You can execute following code in python to see, what have included in your sys path(used by python to search for modules/packages)
import sys
sys.path
Look into below links for more detailed explations
https://www.programiz.com/python-programming/package
https://www.programiz.com/python-programming/modules#search
So, I've two python files:
the 1st "m12345.py"
def my():
return 'hello world'
the 2nd "1234.py":
from m12345 import *
a = m12345.my()
print(a)
On ipython I try to exec such cmds:
exec(open("f:\\temp\\m12345.py").read())
exec(open("f:\\temp\\1234.py").read())
the error for the 2nd command is:
ImportError: No module named 'm12345'
Please, help how to add the 1st file as a module for the 2nd?
First off, if you use the universal import (from m12345 import *) then you just call the my() function and not the m12345.my() or else you will get a
NameError: name 'm12345' is not defined
Secondly, you should add the following snippet in every script in which you want to have the ability of directly running it or not (when importing it).
if "__name__" = "__main__":
pass
PS. Add this to the 1st script ("m12345.py").
PS2. Avoid using the universal import method since it has the ability to mess the namespace of your script. (For that reason, it isn't considered best practice).
edit: Is the m12345.py located in the python folder (where it was installed in your hard drive)? If not, then you should add the directory it is located in the sys.path with:
import sys
sys.path.append(directory)
where directory is the string of the location where your m12345.py is located. Note that if you use Windows you should use / and not \.
However it would be much easier to just relocate the script (if it's possible).
You have to create a new module (for example m12345) by calling m12345 = imp.new_module('m12345') and then exec the python script in that module by calling exec(open('path/m12345.py').read(), m12345.__dict__). See the example below:
import imp
pyfile = open('path/m12345.py').read()
m12345 = imp.new_module('m12345')
exec(pyfile, m12345.__dict__)
If you want the module to be in system path, you can add
sys.modules['m12345'] = m12345
After this you can do
import m12345
or
from m12345 import *
Using invoke, how can I change the directory that the run call operates in?
In Fabric, one would
from fabric.context_managers import lcd
with lcd('foo'):
local('do')
to run do in the foo directory, but I can't find a similar import in pyinvoke.
as simple as that
import os
os.chdir(path)
Use Context.cd
Take a look at docs
with ctx.cd('/path/targeted'):
# do something in /path/targeted
I am working on project euler and wanted to time all of my code. What I have is directory of files in the form 'problemxxx.py' where xxx is the problem number. Each of these files has a main() function that returns the answer. So I have created a file called run.py, located in the same directory as the problem files. I am able to get the name of the file through command prompt. But when I try to import the problem file, I continue to get ImportError: No module named problem. Below is the code for run.py so far, along with the command prompt used.
# run.py
import sys
problem = sys.argv[1]
import problem # I have also tired 'from problem import main' w/ same result
# will add timeit functions later, but trying to get this to run first
problem.main()
The command prompts that I have tried are the following: (both of which give the ImportError stated above)
python run.py problem001
python run.py problem001.py
How can I import the function main() from the file problem001.py? Does importing not work with the file name stored as a variable? Is there a better solution than trying to get the file name through command prompt? Let me know if I need to add more information, and thank you for any help!
You can do this by using the __import__() function.
# run.py
import sys
problem = __import__(sys.argv[1], fromlist=["main"]) # I have also tired 'from problem import main' w/ same result
problem.main()
Then if you have problem001.py like this:
def main():
print "In sub_main"
Calling python run.py problem001 prints:
In sub_main
A cleaner way to do this (instead of the __import__ way) is to use the importlib module. Your run.py needs to changes:
import importlib
problem = importlib.import_module(sys.argv[1])
Alternatives are mentioned in this question.
For sure! You can use __ import_ built-in function like __import__(problem). However this is not recommended to use, because it is not nice in terms of coding-style. I think if you are using this for testing purposes then you should use unittest module, either way try to avoid these constructions.
Regards
You can use exec() trick:
import sys
problem = sys.argv[1]
exec('import %s' % problem)
exec('%s.main()' % problem)
Is there a universal approach in Python, to find out the path to the file that is currently executing?
Failing approaches
path = os.path.abspath(os.path.dirname(sys.argv[0]))
This does not work if you are running from another Python script in another directory, for example by using execfile in 2.x.
path = os.path.abspath(os.path.dirname(__file__))
I found that this doesn't work in the following cases:
py2exe doesn't have a __file__ attribute, although there is a workaround
When the code is run from IDLE using execute(), in which case there is no __file__ attribute
On Mac OS X v10.6 (Snow Leopard), I get NameError: global name '__file__' is not defined
Test case
Directory tree
C:.
| a.py
\---subdir
b.py
Content of a.py
#! /usr/bin/env python
import os, sys
print "a.py: sys.argv[0]=", sys.argv[0]
print "a.py: __file__=", __file__
print "a.py: os.getcwd()=", os.getcwd()
print
execfile("subdir/b.py")
Content of subdir/b.py
#! /usr/bin/env python
import os, sys
print "b.py: sys.argv[0]=", sys.argv[0]
print "b.py: __file__=", __file__
print "b.py: os.getcwd()=", os.getcwd()
print
Output of python a.py (on Windows)
a.py: __file__= a.py
a.py: os.getcwd()= C:\zzz
b.py: sys.argv[0]= a.py
b.py: __file__= a.py
b.py: os.getcwd()= C:\zzz
Related (but these answers are incomplete)
Find path to currently running file
Path to current file depends on how I execute the program
How can I know the path of the running script in Python?
Change directory to the directory of a Python script
First, you need to import from inspect and os
from inspect import getsourcefile
from os.path import abspath
Next, wherever you want to find the source file from you just use
abspath(getsourcefile(lambda:0))
You can't directly determine the location of the main script being executed. After all, sometimes the script didn't come from a file at all. For example, it could come from the interactive interpreter or dynamically generated code stored only in memory.
However, you can reliably determine the location of a module, since modules are always loaded from a file. If you create a module with the following code and put it in the same directory as your main script, then the main script can import the module and use that to locate itself.
some_path/module_locator.py:
def we_are_frozen():
# All of the modules are built-in to the interpreter, e.g., by py2exe
return hasattr(sys, "frozen")
def module_path():
encoding = sys.getfilesystemencoding()
if we_are_frozen():
return os.path.dirname(unicode(sys.executable, encoding))
return os.path.dirname(unicode(__file__, encoding))
some_path/main.py:
import module_locator
my_path = module_locator.module_path()
If you have several main scripts in different directories, you may need more than one copy of module_locator.
Of course, if your main script is loaded by some other tool that doesn't let you import modules that are co-located with your script, then you're out of luck. In cases like that, the information you're after simply doesn't exist anywhere in your program. Your best bet would be to file a bug with the authors of the tool.
This solution is robust even in executables:
import inspect, os.path
filename = inspect.getframeinfo(inspect.currentframe()).filename
path = os.path.dirname(os.path.abspath(filename))
I was running into a similar problem, and I think this might solve the problem:
def module_path(local_function):
''' returns the module path without the use of __file__. Requires a function defined
locally in the module.
from http://stackoverflow.com/questions/729583/getting-file-path-of-imported-module'''
return os.path.abspath(inspect.getsourcefile(local_function))
It works for regular scripts and in IDLE. All I can say is try it out for others!
My typical usage:
from toolbox import module_path
def main():
pass # Do stuff
global __modpath__
__modpath__ = module_path(main)
Now I use _modpath_ instead of _file_.
You have simply called:
path = os.path.abspath(os.path.dirname(sys.argv[0]))
instead of:
path = os.path.dirname(os.path.abspath(sys.argv[0]))
abspath() gives you the absolute path of sys.argv[0] (the filename your code is in) and dirname() returns the directory path without the filename.
The short answer is that there is no guaranteed way to get the information you want, however there are heuristics that work almost always in practice. You might look at How do I find the location of the executable in C?. It discusses the problem from a C point of view, but the proposed solutions are easily transcribed into Python.
See my answer to the question Importing modules from parent folder for related information, including why my answer doesn't use the unreliable __file__ variable. This simple solution should be cross-compatible with different operating systems as the modules os and inspect come as part of Python.
First, you need to import parts of the inspect and os modules.
from inspect import getsourcefile
from os.path import abspath
Next, use the following line anywhere else it's needed in your Python code:
abspath(getsourcefile(lambda:0))
How it works:
From the built-in module os (description below), the abspath tool is imported.
OS routines for Mac, NT, or Posix depending on what system we're on.
Then getsourcefile (description below) is imported from the built-in module inspect.
Get useful information from live Python objects.
abspath(path) returns the absolute/full version of a file path
getsourcefile(lambda:0) somehow gets the internal source file of the lambda function object, so returns '<pyshell#nn>' in the Python shell or returns the file path of the Python code currently being executed.
Using abspath on the result of getsourcefile(lambda:0) should make sure that the file path generated is the full file path of the Python file.
This explained solution was originally based on code from the answer at How do I get the path of the current executed file in Python?.
This should do the trick in a cross-platform way (so long as you're not using the interpreter or something):
import os, sys
non_symbolic=os.path.realpath(sys.argv[0])
program_filepath=os.path.join(sys.path[0], os.path.basename(non_symbolic))
sys.path[0] is the directory that your calling script is in (the first place it looks for modules to be used by that script). We can take the name of the file itself off the end of sys.argv[0] (which is what I did with os.path.basename). os.path.join just sticks them together in a cross-platform way. os.path.realpath just makes sure if we get any symbolic links with different names than the script itself that we still get the real name of the script.
I don't have a Mac; so, I haven't tested this on one. Please let me know if it works, as it seems it should. I tested this in Linux (Xubuntu) with Python 3.4. Note that many solutions for this problem don't work on Macs (since I've heard that __file__ is not present on Macs).
Note that if your script is a symbolic link, it will give you the path of the file it links to (and not the path of the symbolic link).
You can use Path from the pathlib module:
from pathlib import Path
# ...
Path(__file__)
You can use call to parent to go further in the path:
Path(__file__).parent
Simply add the following:
from sys import *
path_to_current_file = sys.argv[0]
print(path_to_current_file)
Or:
from sys import *
print(sys.argv[0])
If the code is coming from a file, you can get its full name
sys._getframe().f_code.co_filename
You can also retrieve the function name as f_code.co_name
The main idea is, somebody will run your python code, but you need to get the folder nearest the python file.
My solution is:
import os
print(os.path.dirname(os.path.abspath(__file__)))
With
os.path.dirname(os.path.abspath(__file__))
You can use it with to save photos, output files, ...etc
import os
current_file_path=os.path.dirname(os.path.realpath('__file__'))