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How to import a nested function in a different file in python?


By : Arieda Muço
Date : October 17 2020, 06:10 AM
To fix the issue you can do Inner functions only exist when the code in the outer function is run, due to the outer function being called. This code, on the outer function then can use the inner function normally, and, it has the option to return a reference to the inner function, or assign it to another data structure (like append it to a list it (the outer function) is modifying).
Therefore, the example you give is no-op - the inner_funciton does nothing, even if the function is called. It is created, and then destroyed when function exits after the call to print.
code :
def multiplier_factory(n):
   def inner(m):
       return n * m
   return inner
from main import multiplier_factory

mul_3 = multiplier_factory(3)

print(mul_3(2))  # will print 6


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Python Import function from file


By : Ismail Houhou
Date : March 29 2020, 07:55 AM
With these it helps Best answer: rename your file. ;)
Having a file named 8.py breaks the naming convention for files as well as Python's language grammar (your trouble importing it makes it clear why these rules are in place).
code :
def func():
    return True
>>> from 8 import func
  File "<stdin>", line 1
    from 8 import func
         ^
SyntaxError: invalid syntax
>>> x = __import__("8")
>>> x.func()
True
>>>

Import function from other file in python


By : user2961474
Date : March 29 2020, 07:55 AM
With these it helps Firstly, make sure you don't have any stale .pyc or .pyo files in the directory. Or if you're using Python 3 then remove the __pycache__ directory just to be sure. This is likely the problem.
In another.py, running from otsu import Hello should print abc. Then running Hello(5) will produce Hello 5. So your output will look like:
code :
abc
Hello 5

How to import a function from python file by Boost.Python


By : Bakhos Sayssouk
Date : March 29 2020, 07:55 AM
this will help When one needs to call Python from C++, and C++ owns the main function, then one must embed the Python interrupter within the C++ program. The Boost.Python API is not a complete wrapper around the Python/C API, so one may find the need to directly invoke parts of the Python/C API. Nevertheless, Boost.Python's API can make interoperability easier. Consider reading the official Boost.Python embedding tutorial for more information.
code :
int main()
{
  // Initialize Python.
  Py_Initialize();

  namespace python = boost::python;
  try
  {
    ... Boost.Python calls ...
  }
  catch (const python::error_already_set&)
  {
    PyErr_Print();
    return 1;
  }

  // Do not call Py_Finalize() with Boost.Python.
}
// Allow Python to load modules from the current directory.
setenv("PYTHONPATH", ".", 1);
// Initialize Python.
Py_Initialize();
#include <boost/python.hpp>
#include <cstdlib> // setenv

int main()
{
  // Allow Python to load modules from the current directory.
  setenv("PYTHONPATH", ".", 1);
  // Initialize Python.
  Py_Initialize();

  namespace python = boost::python;
  try
  {
    // >>> import MyPythonClass
    python::object my_python_class_module = python::import("MyPythonClass");

    // >>> dog = MyPythonClass.Dog()
    python::object dog = my_python_class_module.attr("Dog")();

    // >>> dog.bark("woof");
    dog.attr("bark")("woof");
  }
  catch (const python::error_already_set&)
  {
    PyErr_Print();
    return 1;
  }

  // Do not call Py_Finalize() with Boost.Python.
}
class Dog():
    def bark(self, message):
        print "The dog barks: {}".format(message)
The dog barks: woof

How to access variable inside nested @classmethod from custom import file in Python?


By : Himangshu Majhi
Date : March 29 2020, 07:55 AM
To fix this issue Let's consider the following code, , You define class variables in the class:
code :
class RX(object):
    data = 'Hello '  # here

    @classmethod
    def msg(cls, name):
        return cls.data + name

    @classmethod
    def get_msg(cls, name=None):
        return cls.msg(name)  

Python: Can't import a function from another.py file


By : Mr_Abu
Date : March 29 2020, 07:55 AM
may help you . One possible reason: there exists reference cycle between module a.py and b.py:
In a.py: import b
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