Node not able to get its namespace parameters

asked 2017-06-05 09:13:02 -0600

gnurf
20 ●3 ●4 ●6

My code looks like:

rospy.init_node('images_feeder', anonymous=True)
params = rospy.get_param('~')

I'm pretty sure it used to work "before", but now it gives:

KeyError: '~'

The parameters server seems to be running because I can for example successfully:

print rospy.get_param('/rosdistro')

(which is indigo by the way). Between "before" and now, I performed several ros (system) packages updates, and maybe changed the way I run things (currently "roscore" in one terminal, "rosrun x y" in another).

Could this be a regression, or is there something more to do, so that the node can "get its namespace" ? Thanks.

answered 2017-06-05 09:41:20 -0600

jarvisschultz

9031 ●58 ●128 ●143 https://www.linkedin.c...

updated 2017-06-05 09:51:11 -0600

Are you sure that code worked previously? I'm not aware of any recent changes to rospy that would have changed this behavior, but I suppose I could have missed something. Also note that the KeyError being thrown is exactly what should happen if the parameter you are trying to get ("~") doesn't exist and you don't provide a default value (documentation here).

Regardless, there are a variety of functions specifically for getting names from the ROS graph in the rospy.names module. You could instead be using rospy.get_namespace() or perhaps rospy.get_name() (depending on what you are trying to do). Additionally rospy is slightly different from C++ in that you don't need a "private node handle" to be able to access private parameters. In other words, depending on what you are doing, you might not even need to get the namespace directly. You might be able to use something like rospy.get_param("~private_param1").