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1 | initial version |
(Assumption: you know the parameter
names.)
I don't know if there's official API to achieve what you want. I've also browsed through issues and pull requests on ros_comm but couldn't find anything relevant.
You can find the parameter full names that contains the parameter name (say "baa") in Python by something like:
lislis = [n for n in rospy.get_param_names() if "baa" in n]
print(lislis)
['/namespace_foo/foobin/baa']
(Note that this snippet is obviously not robust.)
2 | No.2 Revision |
(Assumption: you know the parameter
names.names and they won't change.)
I don't know if there's official API to achieve what you want. I've also browsed through issues and pull requests on ros_comm but couldn't find anything relevant.
You can find the parameter full names that contains the parameter name (say "baa") in Python by something like:
lislis = [n for n in rospy.get_param_names() if "baa" in n]
print(lislis)
['/namespace_foo/foobin/baa']
(Note that this snippet is obviously not robust.)
3 | No.3 Revision |
(Assumption: you know the parameter
names and they won't change.)
I don't know if there's official API to achieve what you want. I've also browsed through issues and pull requests on ros_comm but couldn't find anything relevant.
You can find the parameter full names that contains the parameter name (say "baa") in Python by something like:
lislis = [n for n in rospy.get_param_names() if "baa" in n]
print(lislis)
['/namespace_foo/foobin/baa']
(Note Note that this snippet is obviously not robust.)robust, as it is not searching the param name specifically.
4 | No.4 Revision |
(Assumption: you know the parameter
names and they won't change.)
I don't know if there's official API to achieve what you want. I've also browsed through issues and pull requests on ros_comm but couldn't find anything relevant.
You can find the parameter full names that contains the parameter name (say "baa") in Python by something like:
lislis = [n for n in rospy.get_param_names() if "baa" in n]
n.rsplit('/', 1)[-1] == "baa"]
print(lislis)
['/namespace_foo/foobin/baa']
Note that this snippet is obviously not robust, as it is not searching the param name specifically.
5 | No.5 Revision |
(Assumption: you know the parameter
names and they won't change.)
I don't know if there's official API to achieve what you want. I've also browsed through issues and pull requests on ros_comm but couldn't find anything relevant.
You can find the parameter full names that contains the parameter name (say "baa") in Python by something like:
lislis = [n for n in rospy.get_param_names() if n.rsplit('/', 1)[-1] == "baa"]
print(lislis)
['/namespace_foo/foobin/baa']
Note that this snippet is obviously not robust, as it is not searching the param name specifically.
UPDATE) I too wanted a way to search parameter names downward (i.e. "from left to right"). But AFAIK there's only upward search (e.g. search_param). My guess is that there can be multiple leaves of the parameter name (the right-most portion. E.g. "baa" in the example above) can exist (while the root of namespace must be unique) so that searching a leaf name can return multiple elements and thus won't be as useful.
6 | No.6 Revision |
(Assumption: you know the parameter
names and they won't change.)
I don't know if there's official API to achieve what you want. I've also browsed through issues and pull requests on ros_comm but couldn't find anything relevant.
You can find the parameter full names that contains the parameter name (say "baa") in Python by something like:
lislis = [n for n in rospy.get_param_names() if n.rsplit('/', 1)[-1] == "baa"]
print(lislis)
['/namespace_foo/foobin/baa']
Note that this snippet is obviously not robust, as it is not searching the param name specifically.
UPDATE) I too wanted a way to search parameter names downward (i.e. "from left to right"). But AFAIK there's only upward search (e.g. search_param). My guess is that there can be multiple leaves of the parameter name (the right-most portion. E.g. "baa" in the example above) can exist (while the root of namespace must be unique) so that searching a leaf name can return multiple elements and thus won't be as useful.