Select config file based upon launch argument

asked 2021-11-01 11:58:48 -0500

nealtanner
21 ●3 ●4 ●5

I am trying to select between two configuration files based upon a boolean launch file argument. The example below works, but it seems awfully complicated. Is there a simpler way to do this?

def generate_launch_description():
    red_arg = launch.actions.DeclareLaunchArgument('red', default_value='false')
    launch_description = launch.LaunchDescription([red_arg])
    config_file_path = os.path.join(get_package_share_directory('my_package'), 'config')
    config_file = launch.substitutions.PythonExpression(
        ['"', config_file_path, '" + ("/red.yml" if "',
         launch.substitutions.LaunchConfiguration("color"), '".lower() == "true" else "/blue.yml")'])
    launch_description.add_action(launch.add_action(launch.actions.LogInfo(
        msg=('Config file: ', config_file)

    my_node = Node(package='my_package', executable='my_executable.py', parameters=[config_file])
    launch_description.add_action(my_node)

    return launch_description

For readability, the cryptic PythonExpression evaluates to

"/opt/ros_ws/install/my_package/share/my_package/config" + ("/red.yml" if "<launch arg>".lower() == "true" else "/blue.yml")

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... from launch.actions import OpaqueFunction def start_my_executable(context, *args, **kwargs) -> List[LaunchDescriptionEntity]: package_path = get_package_share_directory('my_package') color_param = context.perform_substitution(LaunchConfiguration('color')) config_file_name = 'red.yml' if color_param else 'blue.yml' config_file = os.path.join(package_path, 'config', config_file_name) return [Node(... , parameters=[config_file])] def generate_launch_description(): ... launch_description.add_action(OpaqueFunction(function=start_my_executable)

Select config file based upon launch argument

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Select config file based upon launch argument