I need to execute the class name with functions inside it and execute the functions inside the class individually on run time by passing arguments.
Python code saved as test.py
import sys
import threading
from threading import Thread
class test():
def func1(self):
print('hi im func1')
def func2(self):
print('hi im func2')
def func3(self):
print('hi im func3')
def runall(self):
if __name__ == '__main__':
Thread(target=self.func1).start()
Thread(target=self.func2).start()
Thread(target=self.func3).start()
if __name__ == '__main__':
try:
run = test()
run.runall()
globals()[sys.argv[1]]()
except KeyError:
raise KeyError('Invalid Function Name Passed in Argument! refer the code for valid Name.')
Trying to execute all functions inside class:
Runtime Execution: c:\ > python test.py test
Passed but gave error on
File "test.py", line 180, in
globals()sys.argv[2]
TypeError: 'test' object is not callable
Trying to execute only particular functions inside the class
Runtime Execution: c:\ > python test.py func1
Keyerror is getting thrown.
Can someone guide me on how to execute complete class and individual functions inside the class at runtime?
The first step works by me (on python 3.7.2)
> python3 test.py test
hi im func1
hi im func2
hi im func3
However this is triggered by the run.runall() statement. What version of Python do you run? don't you have another test variable in your work-space ?
For the second point, solution inspired by here, you could get the individual methods of the class running like this:
if __name__ == '__main__':
try:
run = test()
run.runall()
#globals()[sys.argv[1]]()
getattr(test(), sys.argv[1])()
except KeyError:
raise KeyError('Invalid Function Name Passed in Argument! refer the code for valid Name.')
result:
> python3 test.py func1
hi im func1
hi im func2
hi im func3
hi im func1
It worked for me Now I can Execute all Functions and separate functions individually. Thanks All For Your support and guide!!
if __name__ == '__main__':
try:
if sys.argv[2]=='test':
run = test()
run.runall()
else:
getattr(test(), sys.argv[2])()
except KeyError:
raise KeyError('Invalid Function Name Passed in Argument! refer the code for valid Name.')
I have 2 files a.py and b.py as follows:
a.py
import b.py
Test="abc"
def main(args):
global Test
if args.target=="this":
Test="klm"
b.fun()
#rest of the body which I intend to execute only once
#hence I cannot call main() again
if __name__ == "__main__":
#some arguments are parsed
args = parser.parse_args()
main(args)
b.py
import a
print a.Test
EDIT: Output:
python a.py
abc
So basically my question is why is the Test variable not getting updated in b.py and how can I make this work? Thanks.
import a
a.main()
print a.Test
a.Test = "new Value"
print a.Text
You never invoke the main function. When you import a module, __name__ is not "__main__", so your main() never runs. When you run a.py directly it will run main()
Added due to question edit:
You need to consider the ordering of the imports execution. Consider these working files.
a.py
print("Entering a")
import argparse
import b
Test="abc"
print("Id of Test: ", id(Test))
def main(args):
global Test
if args.target=="this":
Test="klm"
b.fun()
#rest of the body which I intend to execute only once
#hence I cannot call main() again
if __name__ == "__main__":
#some arguments are parsed
print('Entering main')
parser = argparse.ArgumentParser()
parser.add_argument('--target', dest='target', type=str)
args = parser.parse_args()
main(args)
b.py
print("Entering b")
import a
print a.Test
def fun():
pass
The console produces the following:
$ python a.py
Entering a
Entering b
Entering a
('Id of Test: ', 40012016L)
abc
('Id of Test: ', 40012016L)
Entering main
The problem is, when you import a python module/file, you will immediately execute all the statements in that module. As such, you have a problem with your dependencies (aka imports), because b is importing a before the value of Test is 'corrected' and then immediately acting on this.
Consider two changes. First, introduce a third file config.py that contains this configuration information and that b does not import a. Second, move all your statements that require this config in b into functions that are called/bootstrapped by a, as obviously intended.
Previous answer:
I have a solution demonstrating the issue, by only modifying b.py
def fun(): # Added because your main calls this. An error?
pass
from a import Test, main
import a
print Test # prints 'abc'
print a.Test # prints 'abc'
main()
print Test # prints 'abc'
print a.Test # prints 'klm'
Within the python interpretor, I can produce the following:
>>> import b
abc
abc
abc
klm
In your code, you create a new variable called Test with the command from a import Test that points to the original string object. You actually want to access the Test variable owned by the module a.
In a.py you run main in the if statement:
if __name__ == "__main__":
main()
Only executes main() if that is the main script. When you import the module all the code in the if block is not run because it is not the main script. To have the main method be called remove the if statement or just call main in b.py.
I'm still in the learning phase and I have this question.
So in order to execute a class, we use if __name__ == '__main__': and call the class as the following
class Example():
def test(self):
print "Hello There"
if __name__ == '__main__':
Example()
However, I saw some classes that use def launch(): instead of if __name__ == '__main__':, so the question here: Are they similar so I can both ways or def launch(): has a special propose?
Thank you.
Python runs anything in the top level this is why we use classes and functions to separate jobs (among other reasons).
So for example here
Script a.py
def main():
pass
main()
The interpreter will define a function called main() but when it reaches the main() call in the top level (aligned left most)
it will execute the main function.
Now in the case of your launch()
if __name__ == '__main__':
Example()
vs
__name__ = __main__
This is used in the case where someone wants to import a program or class but doesn't want it to run when the interpreter runs into it.
Import a will call the main() at that point and time
however let's say b.py is structurally similar but instead of main() it has __name__ = __main__, b.py won't run unless directly called.
The reason I bring this is up is because as #harshil9968 pointed out, Python has no "launch" method. What likely was happening is they defined a launch() method instead of main()
Then put it under a class
class A():
def launch(self):
#actions
if __name__ == '__main__':
A()
Call to A() calls the launch() method within the A class.
I wanted to mix a config.py approach and ConfigParser to set some default values in config.py which could be overridden by the user in its root folder:
import ConfigParser
import os
CACHE_FOLDER = 'cache'
CSV_FOLDER = 'csv'
def main():
cp = ConfigParser.ConfigParser()
cp.readfp(open('defaults.cfg'))
cp.read(os.path.expanduser('~/.python-tools.cfg'))
CACHE_FOLDER = cp.get('folders', 'cache_folder')
CSV_FOLDER = cp.get('folders', 'csv_folder')
if __name__ == '__main__':
main()
When running this module I can see the value of CACHE_FOLDER being changed. However when in another module I do the following:
import config
def main()
print config.CACHE_FOLDER
This will print the original value of the variable ('cache').
Am I doing something wrong ?
The main function in the code you show only gets run when that module is run as a script (due to the if __name__ == '__main__' block). If you want that turn run any time the module is loaded, you should get rid of that restriction. If there's extra code that actually does something useful in the main function, in addition to setting up the configuration, you might want to split that part out from the setup code:
def setup():
# the configuration stuff from main in the question
def main():
# other stuff to be done when run as a script
setup() # called unconditionally, so it will run if you import this module
if __name__ == "__main__":
main() # this is called only when the module is run as a script
In Python I have a module myModule.py where I define a few functions and a main(), which takes a few command line arguments.
I usually call this main() from a bash script. Now, I would like to put everything into a small package, so I thought that maybe I could turn my simple bash script into a Python script and put it in the package.
So, how do I actually call the main() function of myModule.py from the main() function of MyFormerBashScript.py? Can I even do that? How do I pass any arguments to it?
It's just a function. Import it and call it:
import myModule
myModule.main()
If you need to parse arguments, you have two options:
Parse them in main(), but pass in sys.argv as a parameter (all code below in the same module myModule):
def main(args):
# parse arguments using optparse or argparse or what have you
if __name__ == '__main__':
import sys
main(sys.argv[1:])
Now you can import and call myModule.main(['arg1', 'arg2', 'arg3']) from other another module.
Have main() accept parameters that are already parsed (again all code in the myModule module):
def main(foo, bar, baz='spam'):
# run with already parsed arguments
if __name__ == '__main__':
import sys
# parse sys.argv[1:] using optparse or argparse or what have you
main(foovalue, barvalue, **dictofoptions)
and import and call myModule.main(foovalue, barvalue, baz='ham') elsewhere and passing in python arguments as needed.
The trick here is to detect when your module is being used as a script; when you run a python file as the main script (python filename.py) no import statement is being used, so python calls that module "__main__". But if that same filename.py code is treated as a module (import filename), then python uses that as the module name instead. In both cases the variable __name__ is set, and testing against that tells you how your code was run.
Martijen's answer makes sense, but it was missing something crucial that may seem obvious to others but was hard for me to figure out.
In the version where you use argparse, you need to have this line in the main body.
args = parser.parse_args(args)
Normally when you are using argparse just in a script you just write
args = parser.parse_args()
and parse_args find the arguments from the command line. But in this case the main function does not have access to the command line arguments, so you have to tell argparse what the arguments are.
Here is an example
import argparse
import sys
def x(x_center, y_center):
print "X center:", x_center
print "Y center:", y_center
def main(args):
parser = argparse.ArgumentParser(description="Do something.")
parser.add_argument("-x", "--xcenter", type=float, default= 2, required=False)
parser.add_argument("-y", "--ycenter", type=float, default= 4, required=False)
args = parser.parse_args(args)
x(args.xcenter, args.ycenter)
if __name__ == '__main__':
main(sys.argv[1:])
Assuming you named this mytest.py
To run it you can either do any of these from the command line
python ./mytest.py -x 8
python ./mytest.py -x 8 -y 2
python ./mytest.py
which returns respectively
X center: 8.0
Y center: 4
or
X center: 8.0
Y center: 2.0
or
X center: 2
Y center: 4
Or if you want to run from another python script you can do
import mytest
mytest.main(["-x","7","-y","6"])
which returns
X center: 7.0
Y center: 6.0
It depends. If the main code is protected by an if as in:
if __name__ == '__main__':
...main code...
then no, you can't make Python execute that because you can't influence the automatic variable __name__.
But when all the code is in a function, then might be able to. Try
import myModule
myModule.main()
This works even when the module protects itself with a __all__.
from myModule import * might not make main visible to you, so you really need to import the module itself.
I had the same need using argparse too.
The thing is parse_args function of an argparse.ArgumentParser object instance implicitly takes its arguments by default from sys.args. The work around, following Martijn line, consists of making that explicit, so you can change the arguments you pass to parse_args as desire.
def main(args):
# some stuff
parser = argparse.ArgumentParser()
# some other stuff
parsed_args = parser.parse_args(args)
# more stuff with the args
if __name__ == '__main__':
import sys
main(sys.argv[1:])
The key point is passing args to parse_args function.
Later, to use the main, you just do as Martijn tell.
The answer I was searching for was answered here: How to use python argparse with args other than sys.argv?
If main.py and parse_args() is written in this way, then the parsing can be done nicely
# main.py
import argparse
def parse_args():
parser = argparse.ArgumentParser(description="")
parser.add_argument('--input', default='my_input.txt')
return parser
def main(args):
print(args.input)
if __name__ == "__main__":
parser = parse_args()
args = parser.parse_args()
main(args)
Then you can call main() and parse arguments with parser.parse_args(['--input', 'foobar.txt']) to it in another python script:
# temp.py
from main import main, parse_args
parser = parse_args()
args = parser.parse_args([]) # note the square bracket
# to overwrite default, use parser.parse_args(['--input', 'foobar.txt'])
print(args) # Namespace(input='my_input.txt')
main(args)
Assuming you are trying to pass the command line arguments as well.
import sys
import myModule
def main():
# this will just pass all of the system arguments as is
myModule.main(*sys.argv)
# all the argv but the script name
myModule.main(*sys.argv[1:])
I hit this problem and I couldn't call a files Main() method because it was decorated with these click options, eg:
# #click.command()
# #click.option('--username', '-u', help="Username to use for authentication.")
When I removed these decorations/attributes I could call the Main() method successfully from another file.
from PyFileNameInSameDirectory import main as task
task()