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[ros2 launch] launch file argument to python variable

asked 2021-09-30 05:31:24 -0600

morten gravatar image

updated 2021-09-30 07:19:54 -0600

Suppose I pass an argument to a launch file by ros2 launch file.launch.py arg:='temp'. I'd like to pass this into a variable in the launch file such that I can use it, e.g.

def generate_launch_description():
    arg = DeclareLaunchArgument("arg", default_value="temp")
    output = arg.value + ".yaml"

The expectation is that, given the ros2 launch call from before, output becomes "temp.yaml". In essence I am looking for how to do this arg.value part, I can't find anything in this direction online. I have an idea that I need to use substitutions but the documentation on how it works seems to be relatively limited.

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answered 2021-10-07 20:56:31 -0600

ChuiV gravatar image

You're looking for the LaunchConfiguration Substitution:

...
from launch.substitution import LaunchConfiguration

def generate_launch_description():
    arg = LaunchConfiguration('arg')
    return LaunchDescription([
        DeclareLaunchArgument("arg", default_value="temp"),
        ExecuteProcess(cmd=['cat', (arg, '.yaml')])
    ])

Keep in mind that arg is not a python string object. Ie type(arg) == str is false. So you can't do normal string things with it. When I need to concatenate things to arg, I'll usually put them into tuples such as (arg, '.yaml'). When the substitution happens, this tuple gets concatenated together. The other thing to keep in mind, is that this arg object can only really be used when the type-hints have a SomeSubstitutionsType type. If you look at the ExecuteAction class, you'll see that the cmd argument should be of type Iterable[SomeSubstitutionsType]. That's how you know if the substitution variable will be valid or not.

Hope that gives you enough direction to get you going.

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Asked: 2021-09-30 05:31:24 -0600

Seen: 38 times

Last updated: Oct 07 '21