In an effort to create a simple script for another question to highlight an issue I am having, I ran into this confusing problem. My code won't run. I read several other Stack Overflow answers and ensured that I am not using a pre-defined class. I am also not doing a cyclical import. I have no idea. I am new to Python.
TestClass.py:
class TestClass:
test_number = 10000 # Default score limit
def __init__(self):
pass
def check_test_number(self):
# this needs to be an instance method
print(TestClass.test_number)
TestScript.py:
import TestClass
def main():
t1 = TestClass.TestClass()
print(TestClass.test_number)
print(t1.check_test_number())
TestClass.test_number = 500
print(TestClass.test_number)
print(t1.check_test_number())
if __name__ == "__main__":
main()
I recieve this error:
AttributeError: module 'TestClass' has no attribute 'test_number'
Thanks in advance, guys!
You need to refer to the fields test_number and score_limit on lines print(TestClass.test_number) and TestClass.test_number= 500 like this: TestClass.TestClass.test_number or use expression from *your_file* import *ClassName*. In your code you're trying to refer not to a class field, but to a method or variable in the file TestClass.py. I advise you to use snake_case to name .py files to avoid confusions. I think your code can be rewritten like this (with renaming TestClass.py):
test_script.py
from test_class import TestClass
def main():
t1 = TestClass()
print(TestClass.test_number)
print(t1.check_test_number())
TestClass.test_number= 500
print(TestClass.test_number)
print(t1.check_test_number())
if __name__ == "__main__":
main()
Given the following files:
a.py
-----
class CommonClass(object):
def do_thing(self):
pass
b.py
-----
from a import CommonClass
class SubClassA(CommonClass):
def do_thing(self):
print("I am A")
class SubClassB(CommonClass):
def do_thing(self):
print("I am B")
c.py
-----
from a import CommonClass
from b import SubClassA
if __name__ == "__main__":
for member in CommonClass.__subclasses__():
member().do_thing()
I would expect only SubClassA is imported, and visible when looping through subclasses of CommonClass, but it seems SubClassB is imported as well.
I am running Python 3.8.5 and this is the output of python3 c.py:
$ python3 c.py
I am A
I am B
How can I only import the classes I want?
You did import only SubClassA to c.py. This can be tested by doing
x = SubClassB()
or
x = b.SubClassB()
Both will result in a NameError.
The thing is, that when you import a file, it is actually being ran, even when using from x import y!
This can be easily seen by adding a print("I'm from b.py") at the end of b.py and then running c.py.
This makes both SubClassA and SubClassB be subclasses of CommonClass, which you imported. So, while you don't have access to the SubClassB name, it is still a subclass of CommonClass and you can access it from there.
In general you don't have to import a module to be able to use its objects. Importing expands your namespace to include that module so you can create an object directly. You can still use this module's objects even without importing it if you acquired them in some other way (like importing a third module which returns objects from the second one).
Anyway right now, you are not really using even the imported SubClassA. If you want to "allow" certain classes to be considered only from an external source, you can create an allowed set of classes:
from a import CommonClass
from b import SubClassA
allowed_classes = {SubClassA}
if __name__ == "__main__":
for member in CommonClass.__subclasses__():
if member in allowed_classes:
member().do_thing()
Which only prints I am A
from a import CommonClass
from b import SubClassA
if __name__ == "__main__":
h = CommonClass.__subclasses__()[0]()
h.do_thing()
You can do this.
One basic question in OOP.
test.py file content:
class test(object):
def __init__(self):
print 'INIT of test class'
obj = test()
Then I opened another file.
I just inherited from the above test class:
from test import test
class test1(test):
def __init__(self):
pass
So when I run this second file, the __init__() from the parent class is executed (the INIT got printed).
I read that I can avoid it by using
if __name__ == '__main__':
# ...
I can overcome this, but my question is why the parent class's init is executing as I am just importing this class only in my second file. Why is the object creation code executed?
Importing a module executes all module-level statements, including obj=test().
To avoid this, make an instance only when run as the main program, not when imported:
class test(object):
def __init__(self):
print 'INIT of test class'
if __name__ == '__main__':
obj=test()
The problem is not the inheritance but the import. In your case you execute obj=test() when importing:
from test import test
When you import test, its name __name__ is test.
But when you run your program on the command line as main program with python test.py, its name is __main__. So, in the import case, you skip obj=test()if you use:
if __name__ == '__main__':
obj=test()
I want to do the following:
I have a class which should provide several functions, which need different inputs. And I would like to use these functions from within other scripts, or solely from commandline.
e.g. I have the class "test". It has a function "quicktest" (which basically justs prints something). (From commandline) I want to be able to
$ python test.py quicktest "foo" "bar"
Whereas quicktest is the name of the function, and "foo" and "bar" are the variables.
Also (from within another script) I want to
from test import test
# this
t = test()
t.quicktest(["foo1", "bar1"])
# or this
test().quicktest(["foo2", "bar2"])
I just can't bring that to work. I managed to write a class for the first request and one for the second, but not for both of them. The problem is that I sometimes have to call the functions via (self), sometimes not, and also I have to provide the given parameters at any time, which is also kinda complicated.
So, does anybody have an idea for that?
This is what I already have:
Works only from commandline:
class test:
def quicktest(params):
pprint(params)
if (__name__ == '__main__'):
if (sys.argv[1] == "quicktest"):
quicktest(sys.argv)
else:
print "Wrong call."
Works only from within other scripts:
class test:
_params = sys.argv
def quicktest(self, params):
pprint(params)
pprint(self._params)
if (__name__ == '__main__'):
if (sys.argv[1] == "quicktest"):
quicktest()
else:
print "Wrong call"
try the following (note that the different indentation, the if __name__ part is not part of class test anymore):
class test:
def quicktest(params):
pprint(params)
if __name__ == '__main__':
if sys.argv[1] == "quicktest":
testObj = test()
testObj.quicktest(sys.argv)
else:
print "Wrong call."
from other scripts:
from test import test
testObj = test()
testObj.quicktest(...)
The if __name__ == '__main__': block needs to be at the top level:
class Test(object): # Python class names are capitalized and should inherit from object
def __init__(self, *args):
# parse args here so you can import and call with options too
self.args = args
def quicktest(self):
return 'ret_value'
if __name__ == '__main__':
test = Test(sys.argv[1:])
You can parse the command line with the help of argparse to parse the value from the command line.
Your class which has the method and associate methods to arguments.
Say I have a module with the following:
def main():
pass
if __name__ == "__main__":
main()
I want to write a unit test for the bottom half (I'd like to achieve 100% coverage). I discovered the runpy builtin module that performs the import/__name__-setting mechanism, but I can't figure out how to mock or otherwise check that the main() function is called.
This is what I've tried so far:
import runpy
import mock
#mock.patch('foobar.main')
def test_main(self, main):
runpy.run_module('foobar', run_name='__main__')
main.assert_called_once_with()
I will choose another alternative which is to exclude the if __name__ == '__main__' from the coverage report , of course you can do that only if you already have a test case for your main() function in your tests.
As for why I choose to exclude rather than writing a new test case for the whole script is because if as I stated you already have a test case for your main() function the fact that you add an other test case for the script (just for having a 100 % coverage) will be just a duplicated one.
For how to exclude the if __name__ == '__main__' you can write a coverage configuration file and add in the section report:
[report]
exclude_lines =
if __name__ == .__main__.:
More info about the coverage configuration file can be found here.
Hope this can help.
You can do this using the imp module rather than the import statement. The problem with the import statement is that the test for '__main__' runs as part of the import statement before you get a chance to assign to runpy.__name__.
For example, you could use imp.load_source() like so:
import imp
runpy = imp.load_source('__main__', '/path/to/runpy.py')
The first parameter is assigned to __name__ of the imported module.
Whoa, I'm a little late to the party, but I recently ran into this issue and I think I came up with a better solution, so here it is...
I was working on a module that contained a dozen or so scripts all ending with this exact copypasta:
if __name__ == '__main__':
if '--help' in sys.argv or '-h' in sys.argv:
print(__doc__)
else:
sys.exit(main())
Not horrible, sure, but not testable either. My solution was to write a new function in one of my modules:
def run_script(name, doc, main):
"""Act like a script if we were invoked like a script."""
if name == '__main__':
if '--help' in sys.argv or '-h' in sys.argv:
sys.stdout.write(doc)
else:
sys.exit(main())
and then place this gem at the end of each script file:
run_script(__name__, __doc__, main)
Technically, this function will be run unconditionally whether your script was imported as a module or ran as a script. This is ok however because the function doesn't actually do anything unless the script is being ran as a script. So code coverage sees the function runs and says "yes, 100% code coverage!" Meanwhile, I wrote three tests to cover the function itself:
#patch('mymodule.utils.sys')
def test_run_script_as_import(self, sysMock):
"""The run_script() func is a NOP when name != __main__."""
mainMock = Mock()
sysMock.argv = []
run_script('some_module', 'docdocdoc', mainMock)
self.assertEqual(mainMock.mock_calls, [])
self.assertEqual(sysMock.exit.mock_calls, [])
self.assertEqual(sysMock.stdout.write.mock_calls, [])
#patch('mymodule.utils.sys')
def test_run_script_as_script(self, sysMock):
"""Invoke main() when run as a script."""
mainMock = Mock()
sysMock.argv = []
run_script('__main__', 'docdocdoc', mainMock)
mainMock.assert_called_once_with()
sysMock.exit.assert_called_once_with(mainMock())
self.assertEqual(sysMock.stdout.write.mock_calls, [])
#patch('mymodule.utils.sys')
def test_run_script_with_help(self, sysMock):
"""Print help when the user asks for help."""
mainMock = Mock()
for h in ('-h', '--help'):
sysMock.argv = [h]
run_script('__main__', h*5, mainMock)
self.assertEqual(mainMock.mock_calls, [])
self.assertEqual(sysMock.exit.mock_calls, [])
sysMock.stdout.write.assert_called_with(h*5)
Blam! Now you can write a testable main(), invoke it as a script, have 100% test coverage, and not need to ignore any code in your coverage report.
Python 3 solution:
import os
from importlib.machinery import SourceFileLoader
from importlib.util import spec_from_loader, module_from_spec
from importlib import reload
from unittest import TestCase
from unittest.mock import MagicMock, patch
class TestIfNameEqMain(TestCase):
def test_name_eq_main(self):
loader = SourceFileLoader('__main__',
os.path.join(os.path.dirname(os.path.dirname(__file__)),
'__main__.py'))
with self.assertRaises(SystemExit) as e:
loader.exec_module(module_from_spec(spec_from_loader(loader.name, loader)))
Using the alternative solution of defining your own little function:
# module.py
def main():
if __name__ == '__main__':
return 'sweet'
return 'child of mine'
You can test with:
# Override the `__name__` value in your module to '__main__'
with patch('module_name.__name__', '__main__'):
import module_name
self.assertEqual(module_name.main(), 'sweet')
with patch('module_name.__name__', 'anything else'):
reload(module_name)
del module_name
import module_name
self.assertEqual(module_name.main(), 'child of mine')
I did not want to exclude the lines in question, so based on this explanation of a solution, I implemented a simplified version of the alternate answer given here...
I wrapped if __name__ == "__main__": in a function to make it easily testable, and then called that function to retain logic:
# myapp.module.py
def main():
pass
def init():
if __name__ == "__main__":
main()
init()
I mocked the __name__ using unittest.mock to get at the lines in question:
from unittest.mock import patch, MagicMock
from myapp import module
def test_name_equals_main():
# Arrange
with patch.object(module, "main", MagicMock()) as mock_main:
with patch.object(module, "__name__", "__main__"):
# Act
module.init()
# Assert
mock_main.assert_called_once()
If you are sending arguments into the mocked function, like so,
if __name__ == "__main__":
main(main_args)
then you can use assert_called_once_with() for an even better test:
expected_args = ["expected_arg_1", "expected_arg_2"]
mock_main.assert_called_once_with(expected_args)
If desired, you can also add a return_value to the MagicMock() like so:
with patch.object(module, "main", MagicMock(return_value='foo')) as mock_main:
One approach is to run the modules as scripts (e.g. os.system(...)) and compare their stdout and stderr output to expected values.
I found this solution helpful. Works well if you use a function to keep all your script code.
The code will be handled as one code line. It doesn't matter if the entire line was executed for coverage counter (though this is not what you would actually actually expect by 100% coverage)
The trick is also accepted pylint. ;-)
if __name__ == '__main__': \
main()
If it's just to get the 100% and there is nothing "real" to test there, it is easier to ignore that line.
If you are using the regular coverage lib, you can just add a simple comment, and the line will be ignored in the coverage report.
if __name__ == '__main__':
main() # pragma: no cover
https://coverage.readthedocs.io/en/coverage-4.3.3/excluding.html
Another comment by # Taylor Edmiston also mentions it
My solution is to use imp.load_source() and force an exception to be raised early in main() by not providing a required CLI argument, providing a malformed argument, setting paths in such a way that a required file is not found, etc.
import imp
import os
import sys
def mainCond(testObj, srcFilePath, expectedExcType=SystemExit, cliArgsStr=''):
sys.argv = [os.path.basename(srcFilePath)] + (
[] if len(cliArgsStr) == 0 else cliArgsStr.split(' '))
testObj.assertRaises(expectedExcType, imp.load_source, '__main__', srcFilePath)
Then in your test class you can use this function like this:
def testMain(self):
mainCond(self, 'path/to/main.py', cliArgsStr='-d FailingArg')
To import your "main" code in pytest in order to test it you can import main module like other functions thanks to native importlib package :
def test_main():
import importlib
loader = importlib.machinery.SourceFileLoader("__main__", "src/glue_jobs/move_data_with_resource_partitionning.py")
runpy_main = loader.load_module()
assert runpy_main()