How can I keep the ROS Publisher publishing the messages while calling a sub-process:
import subprocess
import rospy
class Pub():
def __init__(self):
pass
def updateState(self, msg):
cmd = ['python3', planner_path, "--alias", search_options, "--plan-file", plan_path, domain_path, problem_path]
subprocess.run(cmd, shell=False, stdout=subprocess.PIPE)
self.plan_pub.publish(msg)
def myPub(self):
rospy.init_node('problem_formulator', anonymous=True)
self.plan_pub = rospy.Publisher("plan", String, queue_size=10)
rate = rospy.Rate(10) # 10hz
rospy.Subscriber('model', String, updateState)
rospy.sleep(1)
rospy.spin()
if __name__ == "__main__":
p_ = Pub()
p_.myPub()
Since subprocess.call is a blocking call your subscription callback may take a long time.
Run the command described by args. Wait for command to complete, then return the returncode attribute.
ROS itself will not call the callback again while it is executed already. This means you are blocking this and potentially also other callbacks to be called in time.
The most simple solution would be to replace subprocess.call by subprocess.Popen which
Execute a child program in a new process
nonblocking.
But keep in mind that this potentially starts the process multiple times quite fast.
Think about starting the process only conditionally if not already running. This can be achieved by checking the process to be finished in another thread. Simple but effective, use boolean flag. Here is a small prototype:
def updateState(self, msg):
#Start the process if not already running
if not self._process_running:
p = subprocess.Popen(...)
self._process_running = True
def wait_process():
while p.poll() is None:
time.sleep(0.1)
self._process_running = False
threading.Thread(target=wait_process).start()
#Other callback code
self.plan_pub.publish(msg)