How can I use rospy through PyCall directly?
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Beschreibung
The code I have can only be able to run the callback once.
```julia
using PyCall
function solve_callback(data)
println("solve_callback.")
end
rospy = pyimport("rospy")
msg = pyimport("geometry_msgs.msg")
WrenchStamped = msg.WrenchStamped
rospy.init_node("solver_interface_sub", disable_signals = false)
println("init_node solver_interface_sub.")
solver_sub = rospy.Subscriber("/solver_input", WrenchStamped, solve_callback, queue_size=1)
if rospy.is_shutdown()
@warn "rospy.is_shutdown()"
end
rospy.spin()
println("stop.")
```
I am aware that RobotOS.jl is one method to accomplish this, but I want to know how can I call rospy directly using PyCall.
Thank you very much.
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Rechercherichtung
Start with the PyCall callback and rospy.Subscriber/rospy.spin calls shown in the report, and compare them with RobotOS.jl's handling of ROS callbacks. Reproduce the one-callback behavior using the supplied Julia snippet and investigate the callback lifetime or threading behavior involved. Done means a verified direct-PyCall approach that invokes solve_callback repeatedly, or a clear explanation of why it cannot.
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