How can I run roscore from python?

Question

Whenever I launch a node from python using the roslaunch script api :

#Start roslaunch
launch = roslaunch.scriptapi.ROSLaunch()
launch.start()

# start required nodes
empty_srv_node = roslaunch.core.Node('rostful_node', 'emptyService.py', name='empty_service')
empty_srv_process = launch.launch(empty_srv_node)

I get an error on rospy.init_node() because master is not started :

Unable to register with master node [http://localhost:11311]: master may not be running yet. Will keep trying.

Question : How can I run roscore ( not only rosmaster ) from python if needed ? I would expect roslaunch to do it but it seems it doesnt.

Thank you.

---

Originally posted by asmodehn on ROS Answers with karma: 131 on 2015-08-11

Post score: 5

Answer 1

I don't know the ros way to do it, and personally I use python subbprocess to run the roscore first:

import subprocess
roscore = subprocess.Popen('roscore')
time.sleep(1)  # wait a bit to be sure the roscore is really launched

then your code:

unittest.main()  # for instance in my case I need a roscore to run some node tests

---

Originally posted by lgeorge with karma: 123 on 2015-12-30

This answer was ACCEPTED on the original site

Post score: 3

Answer 2

If you want to use the subprocess approach you should make sure that all processes are killed cleanly.

import subprocess
import shlex
import sys
import signal
import psutil

def kill_child_processes(parent_pid, sig=signal.SIGTERM):
    try:
        parent = psutil.Process(parent_pid)
        print(parent)
    except psutil.NoSuchProcess:
        print("parent process not existing")
        return
    children = parent.children(recursive=True)
    print(children)
    for process in children:
        print("try to kill child: " + str(process))
        process.send_signal(sig)

class Roscore(object):
    """
    roscore wrapped into a subprocess.
    Singleton implementation prevents from creating more than one instance.
    """
    __initialized = False
    def __init__(self):
        if Roscore.__initialized:
            raise Exception("You can't create more than 1 instance of Roscore.")
        Roscore.__initialized = True
    def run(self):
        try:
            self.roscore_process = subprocess.Popen(['roscore'])
            self.roscore_pid = self.roscore_process.pid  # pid of the roscore process (which has child processes)
        except OSError as e:
            sys.stderr.write('roscore could not be run')
            raise e
    def terminate(self):
        print("try to kill child pids of roscore pid: " + str(self.roscore_pid))
        kill_child_processes(self.roscore_pid)
        self.roscore_process.terminate()
        self.roscore_process.wait()  # important to prevent from zombie process
        Roscore.__initialized = False

Usage is like follows:

roscore = Roscore()
roscore.run()
roscore.terminate()

---

Originally posted by thinwybk with karma: 468 on 2018-05-28

This answer was NOT ACCEPTED on the original site

Post score: 3

---

Original comments

Comment by archiparchi on 2022-03-23:
Thank you @thinwybk ! I got to learn a lot from this piece of code. This seems like a very good practice. Here is my upvote :)

Answer 3

I think this is the ROS way to do it:

import roslaunch
uuid = roslaunch.rlutil.get_or_generate_uuid(options_runid=None, options_wait_for_master=False)
roslaunch.configure_logging(uuid)
launch = roslaunch.parent.ROSLaunchParent(uuid, roslaunch_files=[], is_core=True)
launch.start()
#  --- your code ---
launch.shutdown()

You can specify your launch files as a list of string in the ROSLaunchParent constructor. Check the doc if needed.

---

Originally posted by qgallouedec with karma: 21 on 2020-12-09

This answer was NOT ACCEPTED on the original site

Post score: 2