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I have a script named test1.py which is not in a module. It just has code that should execute when the script itself is run. There are no functions, classes, methods, etc. I have another script which runs as a service. I want to call test1.py from the script running as a service.
For example:
File test1.py:
print "I am a test"
print "see! I do nothing productive."
File service.py:
# Lots of stuff here
test1.py # do whatever is in test1.py
I'm aware of one method which is opening the file, reading the contents, and basically evaluating it. I'm assuming there's a better way of doing this. Or at least I hope so.
The usual way to do this is something like the following.
test1.py
def some_func():
print 'in test 1, unproductive'
if __name__ == '__main__':
# test1.py executed as script
# do something
some_func()
service.py
import test1
def service_func():
print 'service func'
if __name__ == '__main__':
# service.py executed as script
# do something
service_func()
test1.some_func()
This is possible in Python 2 using
execfile("test2.py")
See the documentation for the handling of namespaces, if important in your case.
In Python 3, this is possible using (thanks to #fantastory)
exec(open("test2.py").read())
However, you should consider using a different approach; your idea (from what I can see) doesn't look very clean.
Another way:
File test1.py:
print "test1.py"
File service.py:
import subprocess
subprocess.call("test1.py", shell=True)
The advantage to this method is that you don't have to edit an existing Python script to put all its code into a subroutine.
Documentation: Python 2, Python 3
import os
os.system("python myOtherScript.py arg1 arg2 arg3")
Using os you can make calls directly to your terminal. If you want to be even more specific you can concatenate your input string with local variables, ie.
command = 'python myOtherScript.py ' + sys.argv[1] + ' ' + sys.argv[2]
os.system(command)
If you want test1.py to remain executable with the same functionality as when it's called inside service.py, then do something like:
test1.py
def main():
print "I am a test"
print "see! I do nothing productive."
if __name__ == "__main__":
main()
service.py
import test1
# lots of stuff here
test1.main() # do whatever is in test1.py
I prefer runpy:
#!/usr/bin/env python
# coding: utf-8
import runpy
runpy.run_path(path_name='script-01.py')
runpy.run_path(path_name='script-02.py')
runpy.run_path(path_name='script-03.py')
You should not be doing this. Instead, do:
test1.py:
def print_test():
print "I am a test"
print "see! I do nothing productive."
service.py
#near the top
from test1 import print_test
#lots of stuff here
print_test()
Use import test1 for the 1st use - it will execute the script. For later invocations, treat the script as an imported module, and call the reload(test1) method.
When reload(module) is executed:
Python modules’ code is recompiled and the module-level code reexecuted, defining a new set of objects which are bound to names in the module’s dictionary. The init function of extension modules is not called
A simple check of sys.modules can be used to invoke the appropriate action. To keep referring to the script name as a string ('test1'), use the 'import()' builtin.
import sys
if sys.modules.has_key['test1']:
reload(sys.modules['test1'])
else:
__import__('test1')
As it's already mentioned, runpy is a nice way to run other scripts or modules from current script.
By the way, it's quite common for a tracer or debugger to do this, and under such circumstances methods like importing the file directly or running the file in a subprocess usually do not work.
It also needs attention to use exec to run the code. You have to provide proper run_globals to avoid import error or some other issues. Refer to runpy._run_code for details.
Why not just import test1? Every python script is a module. A better way would be to have a function e.g. main/run in test1.py, import test1 and run test1.main(). Or you can execute test1.py as a subprocess.
I found runpy standard library most convenient. Why? You have to consider case when error raised in test1.py script, and with runpy you are able to handle this in service.py code. Both traceback text (to write error in log file for future investigation) and error object (to handle error depends on its type): when with subprocess library I wasn't able to promote error object from test1.py to service.py, only traceback output.
Also, comparing to "import test1.py as a module" solution, runpy is better cause you have no need to wrap code of test1.py into def main(): function.
Piece of code as example, with traceback module to catch last error text:
import traceback
import runpy #https://www.tutorialspoint.com/locating-and-executing-python-modules-runpy
from datetime import datetime
try:
runpy.run_path("./E4P_PPP_2.py")
except Exception as e:
print("Error occurred during execution at " + str(datetime.now().date()) + " {}".format(datetime.now().time()))
print(traceback.format_exc())
print(e)
This process is somewhat un-orthodox, but would work across all python versions,
Suppose you want to execute a script named 'recommend.py' inside an 'if' condition, then use,
if condition:
import recommend
The technique is different, but works!
Add this to your python script.
import os
os.system("exec /path/to/another/script")
This executes that command as if it were typed into the shell.
An example to do it using subprocess.
from subprocess import run
import sys
run([sys.executable, 'fullpathofyourfile.py'])
This is an example with subprocess library:
import subprocess
python_version = '3'
path_to_run = './'
py_name = '__main__.py'
# args = [f"python{python_version}", f"{path_to_run}{py_name}"] # works in python3
args = ["python{}".format(python_version), "{}{}".format(path_to_run, py_name)]
res = subprocess.Popen(args, stdout=subprocess.PIPE)
output, error_ = res.communicate()
if not error_:
print(output)
else:
print(error_)
According to the given example, this is the best way:
# test1.py
def foo():
print("hellow")
# test2.py
from test1 import foo # might be different if in different folder.
foo()
But according to the title, using os.startfile("path") is the best way as its small and it works. This would execute the file specified. My python version is 3.x +.
I have a file called hotel_helper.py from which I want to import a function called demo1, but I am unable to import it.
My hotel_helper.py file:
def demo1():
print('\n\n trying to import this function ')
My other file:
from hotel.helpers.hotel_helper import demo1
demo1()
but I get:
ImportError: cannot import name 'demo1' from 'hotel.helpers.hotel_helper'
When I import using from hotel.helpers.hotel_helper import * instead of from hotel.helpers.hotel_helper import demo1 it works and the function gets called. I tried importing the whole file with from hotel.helpers import hotel_helper and then call the function with hotel_helper.demo1() and it works fine. I don't understand what's wrong in first method. I want to directly import function rather using * or importing the whole file.
If you filename is hotel_helper.py you have to options how to import demo1:
You can import the whole module hotel_helper as and then call your func:
import hotel_helper as hh
hh.demo1()
You can import only function demo1 from module as:
from hote_helpers import demo1
demo1()
From your fileName import your function
from hotel.helpers import demo1
demo1()
You can import a py file with the following statement:
# Other import
import os
import sys
if './hotel' not in sys.path:
sys.path.insert(0, './hotel')
from hotel import *
NOTE:
For IDE like PyCharm, you can specify the import path using the Project Structure setting tab (CTRL+ALT+S)
Helpful stack overflow questions [maybe off topic]:
What is the right way to create project structure in pycharm?
Manage import with PyCharm documentation:
https://www.jetbrains.com/help/pycharm/configuring-project-structure.html
This is probably a duplicate of: https://stackoverflow.com/posts/57944151/edit
I created two files (defdemo.py and rundefdemo.py) from your posted 2 files and substituted 'defdemo' for 'hotel.helpers.hotel_helper' in the code. My 2 files are in my script directory for Python 3.7 on windows 10 and my script directory is in the python path file python37._pth. It worked.
defdemo.py
def demo1():
print('\n\n trying to import this function ')
rundefdemo.py
from defdemo import demo1
demo1()
output
trying to import this function
I was able to solve the issue, it was related to some imports I was making in my file, when I removed all the import statement in my hotel_helper.py ,the code started working as expected , Still I don't understand reason why the issue was occurring. anyway it works.
This ImportError can also arise when the function being imported is already defined somewhere else in the main script (i.e. calling script) or notebook, or when it is defined in a separate dependency (i.e. another module). This happens most often during development, when the developer forgets to comment out or delete the function definition in the body of a main file or nb after moving it to a module.
Make sure there are no other versions of the function in your development environment and dependencies.
I want to make my own programming language based on python which will provide additional features that python wasn't provide, for example to make multiline anonymous function with custom syntax. I want my programming language is so simple to be used, just import my script, then I read the script file which is imported my script, then process it's code and stop anymore execution of the script which called my script to prevent error on syntax...
Let say there are 2 py file, main.py and MyLanguage.py
The main.py imported MyLanguage.py
Then how to get the main.py file from MyLanguage.py if main.py can be another name(Dynamic Name)?
Additional information:
I using python 3.4.4 on Windows 7
Like Colonder, I believe the project you have in mind is far more difficult than you imagine.
But, to get you started, here is how to get the main.py file from inside MyLanguage.py. If your importing module looks like this
# main.py
import MyLanguage
if __name__ == "__main__":
print("Hello world from main.py")
and the module it is importing looks like this, in Python 3:
#MyLanguage.py
import inspect
def caller_discoverer():
print('Importing file is', inspect.stack()[-1].filename)
caller_discoverer()
or (edit) like this, in Python 2:
#MyLanguage.py
import inspect
def caller_discoverer():
print 'Importing file is', inspect.stack()[-1][1]
caller_discoverer()
then the output you will get when you run main.py is
Importing file is E:/..blahblahblah../StackOverflow-3.6/48034902/main.py
Hello world from main.py
I believe this answers the question you asked, though I don't think it goes very far towards achieving what you want. The reason for my scepticism is simple: the import statement expects a file containing valid Python, and if you want to import a file with your own non-Python syntax, then you are going to have to do some very clever stuff with import hooks. Without that, your program will simply fail at the import statement with a syntax error.
Best of luck.
I am using Python 2.7 on Windows 7 Professional.
I am trying to call a function saved in another file to run in this file's code.
Function called dosomething is found in anotherfile.py
anotherfile.py is in the same directory as current code.
My call in this file is simple:
import anotherfile
print anotherfile.dosomething
I am getting an error: No module named anotherfile
The problem is the same as I found in this post
I don't understand the solution but I'd like any insight?
Thank you.
EDIT: The other question/answers discuss resetting CLASSPATH and setting PYTHONPATH. I explored this but was not sure how to do this. Perhaps relevant?
Let us have two files in the same directory. Files are called main.py and another.py.
First write a method in another.py:
def do_something():
return "This is the do something method"
Then call the method from main.py. Here is the main.py:
import another
print another.do_something()
Run main.py and you will get output like this:
This is the do something method
N.B.: The above code is being executed using Python 2.7 in Windows 10.
Specify the module then the file then the import like so:
from this_module.anotherfile import dosomething
or if you want all functions from "anotherfile.py"
from this_module.anotherfile import *
and then you can call the "dosomething" command without the "anotherfile" prefix.
I ran into same problem. After ample of trials, I ended up solving it with the below mentioned solution:
Make sure your current file and anotherfile.py lies in same location of system path.
Say your another.py and current file lies at location : "C:/Users/ABC"
In case, one is not aware of system path. Use below code in current file:
import sys
print(sys.path)
import sys
sys.path.append('/C:/Users/ABC/')
Then you do below code in same current code:
from another import dosomething
I found the issue. Python was looking in another directory for the files. I explicitly set my working directory as the path to where thisfile.py and anotherfile.py reside and it works. Thank you for all the quick replies.
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 *