无法使用 ROS Noetic 从'PythonFile'导入名称'ClassName'

Question

我正在使用Python 3.8.10和Ubuntu 20.04，我得到了这个错误：

Traceback (most recent call last):
File "/home/belal/git/decision_making_pacbot/devel/lib/ros_planning/ros_plan.py", line 15, in <module>
exec(compile(fh.read(), python_script, 'exec'), context)
File "/home/belal/git/decision_making_pacbot/src/ros_planning/src/ros_planning/ros_plan.py", line 8, in <module>
from problem_handler import ProblemHandler
ImportError: cannot import name 'ProblemHandler' from 'problem_handler' (/home/belal/git/decision_making_pacbot/devel/.private/ros_planning/lib/ros_planning/problem_handler.py)

我的文件层次结构如下：

src
├── ros_planning
│   ├── CMakeLists.txt
|   ├── package.xml
|   ├── setup.py
│   └── src
│       └── ros_planning
|           ├── ros_plan.py
|           ├──problem_handler.py
|           └──__init__.py

我的CMakeLists.txt

cmake_minimum_required(VERSION 3.0.2)
project(ros_planning)
# Load catkin and all dependencies required for this package
find_package(catkin REQUIRED COMPONENTS
rospy
)
# Declare catkin package
catkin_package(
CATKIN_DEPENDS
)
catkin_install_python(PROGRAMS
src/${PROJECT_NAME}/problem_handler.py
src/${PROJECT_NAME}/ros_plan.py
DESTINATION ${CATKIN_PACKAGE_BIN_DESTINATION}
)
catkin_python_setup()
install(
DIRECTORY model downward launch
DESTINATION "${CATKIN_PACKAGE_SHARE_DESTINATION}"
)

我的package.xml

<?xml version="1.0"?>
<package format="2">
<name>ros_planning</name>
<version>0.0.0</version>
<description>The ros_planning package</description>
<maintainer email="[e-mail]">s</maintainer>
<license>MIT</license>
<buildtool_depend>catkin</buildtool_depend>
<build_depend>roscpp</build_depend>
<build_depend>rospy</build_depend>
<build_depend>std_msgs</build_depend>
<build_export_depend>roscpp</build_export_depend>
<build_export_depend>rospy</build_export_depend>
<build_export_depend>std_msgs</build_export_depend>
<exec_depend>roscpp</exec_depend>
<exec_depend>rospy</exec_depend>
<exec_depend>std_msgs</exec_depend>
</package>

我的setup.py

#!/usr/bin/python3
from distutils.core import setup
from catkin_pkg.python_setup import generate_distutils_setup
# fetch values from package.xml
setup_args = generate_distutils_setup(
scripts=['src/ros_planning/problem_handler.py', 'src/ros_planning/ros_plan.py'],
packages=['ros_planning'],
package_dir={'': 'src'},
requires=['rospy', 'std_msgs']
)
setup(**setup_args)

最后在ros_plan.py中，我导入problem_handler.py中存在的类from problem_handler import ProblemHandler，如下所示：

class ProblemHandler:
def __init__(self, path):
self.path = path

你能告诉我如何在ROS中正确导入我的python类吗？因为类似的设置在ROS旋律上对我来说效果很好！

以下是ROS:生成的devel文件夹中的problem_handler.py

#!/usr/bin/python3
# -*- coding: utf-8 -*-
# generated from catkin/cmake/template/script.py.in
# creates a relay to a python script source file, acting as that file.
# The purpose is that of a symlink
python_script = '/home/belal/git/decision_making_pacbot/src/ros_planning/src/ros_planning/problem_handler.py'
with open(python_script, 'r') as fh:
context = {
'__builtins__': __builtins__,
'__doc__': None,
'__file__': python_script,
'__name__': __name__,
'__package__': None,
}
exec(compile(fh.read(), python_script, 'exec'), context)

当我使用os.getcwd()打印ros_plan.py的执行路径时，我得到/home/$USER/.ros

Answer 1

在我的案例中，解决方案是按如下方式导入类：

from ros_package_name.script_name import className

from ros_planning.problem_handler import ProblemHandler