I’m developing a simple application in python which can be summarized in these files:

• root_dir
  • myLib (contains some modules in python)
  • libraryUnderTest (contains some modules in python that import some modules from myLib directory
    • TEST (contains the unittest program)

in particular, let’s assume that my test_program.py is :

import sys
import os
import unittest
from  unittest import mock
from unittest.mock import patch
import socket
lib_path = os.path.abspath(os.path.join(os.path.dirname(__file__), '../'))
sys.path.append(lib_path)
import libUnderTest as lut
    
class TestExample(unittest.TestCase):
    def test_1(self):
        """
        Simple Test
        """
        tmp = lut.myObject(100, "int32", "pathToFile")
        self.assertEqual(tmp.memSize, 100)
        self.assertEqual(tmp.dtype, 4)
        self.assertEqual(tmp.pathToFile, "pathToFile")
        self.assertEqual(tmp.logged, False)enter code here

And my program in libraryUnderTest/libUnderTest.py

import sys
sys.path.append("../myLib")
import helperLib as helpLib

'''
This class represents a memRegion as it is in the XML conf tag
'''
class GpuSatMemRegion:
    def __init__(self, memSize, type, defValue, pathToFile = None):
        self.memReg = helpLib.anotherObject(memSize, type)
        self.memSize  
        self.type = type
        self.defValue=defValue
        self.pathToFile = pathToFile 
        self.logged = False

Why is my test_program failing with this message?
File “C:UsersBenelliDesktopSVNGPUatSAT-SVNtrunkswGpuOnServerlibraryUnderTest libUnderTest.py”, line 3, in
import helperLib as helpLib
ModuleNotFoundError: No module named ‘helperLib ‘

How can I avoid this inconvenience?