Try the construction where each process is being joined with a timeout on a separate thread. So the main program never gets stuck and as well the processes which if gets stuck, would be killed due to timeout. This technique is a combination of threading and multiprocessing modules.
Here is my way to maintain the minimum x number of threads in the memory. Its an combination of threading and multiprocessing modules. It may be unusual to other techniques like respected fellow members have explained above BUT may be worth considerable. For the sake of explanation, I am taking a scenario of crawling a minimum of 5 websites at a time.
so here it is:-
#importing dependencies.
from multiprocessing import Process
from threading import Thread
import threading
# Crawler function
def crawler(domain):
# define crawler technique here.
output.write(scrapeddata + "\n")
pass
Next is threadController function. This function will control the flow of threads to the main memory. It will keep activating the threads to maintain the threadNum "minimum" limit ie. 5. Also it won't exit until, all Active threads(acitveCount) are finished up.
It will maintain a minimum of threadNum(5) startProcess function threads (these threads will eventually start the Processes from the processList while joining them with a time out of 60 seconds). After staring threadController, there would be 2 threads which are not included in the above limit of 5 ie. the Main thread and the threadController thread itself. thats why threading.activeCount() != 2 has been used.
def threadController():
print "Thread count before child thread starts is:-", threading.activeCount(), len(processList)
# staring first thread. This will make the activeCount=3
Thread(target = startProcess).start()
# loop while thread List is not empty OR active threads have not finished up.
while len(processList) != 0 or threading.activeCount() != 2:
if (threading.activeCount() < (threadNum + 2) and # if count of active threads are less than the Minimum AND
len(processList) != 0): # processList is not empty
Thread(target = startProcess).start() # This line would start startThreads function as a seperate thread **
startProcess function, as a separate thread, would start Processes from the processlist. The purpose of this function (**started as a different thread) is that It would become a parent thread for Processes. So when It will join them with a timeout of 60 seconds, this would stop the startProcess thread to move ahead but this won't stop threadController to perform. So this way, threadController will work as required.
def startProcess():
pr = processList.pop(0)
pr.start()
pr.join(60.00) # joining the thread with time out of 60 seconds as a float.
if __name__ == '__main__':
# a file holding a list of domains
domains = open("Domains.txt", "r").read().split("\n")
output = open("test.txt", "a")
processList = [] # thread list
threadNum = 5 # number of thread initiated processes to be run at one time
# making process List
for r in range(0, len(domains), 1):
domain = domains[r].strip()
p = Process(target = crawler, args = (domain,))
processList.append(p) # making a list of performer threads.
# starting the threadController as a seperate thread.
mt = Thread(target = threadController)
mt.start()
mt.join() # won't let go next until threadController thread finishes.
output.close()
print "Done"
Besides maintaining a minimum number of threads in the memory, my aim was to also have something which could avoid stuck threads or processes in the memory. I did this using the time out function. My apologies for any typing mistake.
I hope this construction would help anyone in this world.
Regards,
Vikas Gautam
except mp.TimeoutError: print(mp.active_children())
to see that you have the opposite problem: child processes are stuck atwhile 1: pass
– Episodewhile 1: pass
is on purpose. Edited my question. – VendibleProcess.terminate()
),result.get()
will never return when called again, even after all tasks have been processed (i.e. at least tried once)? – Vendible