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You should be able to load a YAML file into a python data structure using the pyyaml module.
You can then use rospy.set_param('/prefix', data) to push the entire contents of your data structure to a prefix on the parameter server.
It's not the same as doing "rosparam load" . I've tried what you propose, but it does not make the namespace structure as "rosparam load" does. I mean, with "rosparam load file.yaml" I load all the namespaces in the file, that is, the data I can access is just the leaves of the structure. and all the internal nodes in the file are loaded as namespaces. With rospy.set_param, I just set the "/prefix" with a dictionary, so I can't access the data with rospy.get_param("/prefix/namespace1/namespace2").
That's a bit surprising. What sort of structure/namespaces do you get on the parameter server after uploading a dict with set_param? It might be worth adding what you've tried to your question, particularly a small piece of python and the resulting `rosparam list /prefix` and `rosparam get /prefix`
In response to @ahendrix, this an use-case
the data: "asdf.yaml"
namespace1: {data1: 1, data2: 2}
namespace2: {data3: 3, data4: 4.5}
the use-case:
> rosparam load asdf.yaml main
> rosparam list
/main/namespace1/data1
/main/namespace1/data2
/main/namespace2/data3
/main/namespace2/data4
...and other ros stuff
> rosparam get main/namespace1/data1
1
In an IPython console
> data = rosparam.load_file("asdf.yaml")
[({'main': {'namespace1': {'data1': 1, 'data2': 2}, 'namespace2': {'data3': 3, 'data4': 4.5}}},'/')]
# The parameter server is not touched, it just returns a list
> rospy.set_param("main", data)
> rospy.get_param("main")
[[{'namespace1': {'data1': 1, 'data2': 2}, 'namespace2': {'data3': 3, 'data4': 4.5}}, '/']]
Outside, in another terminal:
> rosparam list
/main
> rosparam get main/namespace1/data1
ERROR: Parameter [/main/namespace1/data1] is not set
So it doesn't load the namespace structure I think.
This function does what I expected it to do:
def load_params_from_yaml(complete_file_path):
from rosparam import upload_params
from yaml import load
f = open(full_path, 'r')
yamlfile = load(f)
f.close()
upload_params('/', yamlfile)
Have you tried rosparam set with -t option? http://wiki.ros.org/rosparam#rosparam...
It's possible from .launch file as well (example).
That is not what I'm looking for. What I want to do is something like this: rosparam.load_something("filename.yaml") and then I want to see the parameters of filename.yaml in the parameter server. Currently the rosparam.load_file method just prints the content of the file in stdout, without touching the parameter server.
Asked: 2014-05-26 06:50:01 -0600
Seen: 11,898 times
Last updated: May 27 '14
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This is duplicate of http://answers.ros.org/question/58819... , there is an answer there.