How to start a new thread when old one finishes? - python

I have a large dataset in a list that I need to do some work on.
I want to start x amounts of threads to work on the list at any given time, until everything in that list has been popped.
I know how to start x amounts of threads (lets say 20) at a given time (by using thread1....thread20.start())
but how do I make it start a new thread when one of the first 20 threads finish? so at any given time there are 20 threads running, until the list is empty.
what I have so far:
class queryData(threading.Thread):
def __init__(self,threadID):
threading.Thread.__init__(self)
self.threadID = threadID
def run(self):
global lst
#Get trade from list
trade = lst.pop()
tradeId=trade[0][1][:6]
print tradeId
thread1 = queryData(1)
thread1.start()
Update
I have something going with the following code:
for i in range(20):
threads.append(queryData(i))
for thread in threads:
thread.start()
while len(lst)>0:
for iter,thread in enumerate(threads):
thread.join()
lock.acquire()
threads[iter] = queryData(i)
threads[iter].start()
lock.release()
Now it starts 20 threads in the beginning...and then keeps starting a new thread when one finishes.
However, it is not efficient, as it waits for the first one in the list to finish, and then the second..and so on.
Is there a better way of doing this?
Basically I need:
-Start 20 threads:
-While list is not empty:
-wait for 1 of the 20 threads to finish
-reuse or start a new thread

As I suggested in a comment, I think using a multiprocessing.pool.ThreadPool would be appropriate — because it would handle much of the thread management you're manually doing in your code automatically. Once all the threads are queued-up for processing via ThreadPool's apply_async() method calls, the only thing that needs to be done is wait until they've all finished execution (unless there's something else your code could be doing, of course).
I've translated the code in my linked answer to another related question so it's more similar to what you appear to be doing to make it easier to understand in the current context.
from multiprocessing.pool import ThreadPool
from random import randint
import threading
import time
MAX_THREADS = 5
print_lock = threading.Lock() # Prevent overlapped printing from threads.
def query_data(trade):
trade_id = trade[0][1][:6]
time.sleep(randint(1, 3)) # Simulate variable working time for testing.
with print_lock:
print(trade_id)
def process_trades(trade_list):
pool = ThreadPool(processes=MAX_THREADS)
results = []
while(trade_list):
trade = trade_list.pop()
results.append(pool.apply_async(query_data, (trade,)))
pool.close() # Done adding tasks.
pool.join() # Wait for all tasks to complete.
def test():
trade_list = [[['abc', ('%06d' % id) + 'defghi']] for id in range(1, 101)]
process_trades(trade_list)
if __name__ == "__main__":
test()

You can wait for a thread to complete with : thread.join(). This call will block until that thread completes, at which point you can create a new one.
However, instead of respawning a Thread each time, why not recycle your existing threads ?
This can be done by the use of tasks for example. You keep a list of tasks in a shared collection, and when one of your threads finishes a task, it retrieves another one from that collection.

Related

Starting n number of threads from a loop

So basically, I've this function th() which counts till certain number and then prints "done".
I'd want to start n number of such threads at the same time, running simultaneously.
So I wrote:
thread_num = 3 #here n is 3, but I'd normally want something way higher
thrds = []
i = 0
while i < thread_num:
thr = Thread(target=th, args=())
thrds.append(thr)
i += 1
print("thread", str(i), "added")
for t in thrds:
t.start()
t.join()
I'd want all the threads to print "done" at the same time, but they have a noticeable lag in between of them. They print "thread i started" at seemingly the same time, but print "done" with quite a bit of time lag.
Why is this happening?
Edit: Since someone asked me to add th() function as well, here it is:
def th():
v = 0
num = 10**7
while v < num:
v += 1
print("done")
This is happening because of the t.join() method that you are calling on each thread before start the next one. t.join() blocks the execution of the current thread until the thread t has completed execution. So, each thread is starting after the previous one has finished.
You first have to start all the threads, then join all the threads in separate for loops; otherwise, each thread starts but runs to completion due to join before starting another thread.
for t in thrds: # start all the threads
t.start()
for t in thrds: # wait for all threads to finish
t.join()
If you only have a simple counting thread, you may need to add some short sleep to actually see the threads output intermingle as they may still run fast enough to complete before another thread starts.
Because you start and join each thread sequentially, one thread will run to completion before the next even starts. You'd be better off running a thread pool which is a more comprehensive implementation that handles multiple issues in multithreading.
Because of memory management and object reference count issues, python only lets a single thread execute byte code at a time. Periodically, each thread will release and reacquire the Global Interpreter Lock (GIL) to let other threads run. Exactly which thread runs at any given time is up to the operating system and you may find one gets more slices than another, causing staggered results.
To get them all to print "done" at the same time, you could use a control structure like a barrier for threads to wait until all are done. With a barrier, all threads must call wait before any can continue.
thread_num = 3 #here n is 3, but I'd normally want something way higher
wait_done = threading.Barrier(thread_num)
def th(waiter):
x = 1 # to what you want
waiter.wait()
print("done")
thrds = []
i = 0
while i < thread_num:
thr = Thread(target=th, args=(wait_done,))
thrds.append(thr)
i += 1
print("thread", str(i), "added")
for t in thrds:
t.start()
for t in thrds:
t.join()

Threads are not doing their job when there is a print instruction [duplicate]

I have the following script (don't refer to the contents):
import _thread
def func1(arg1, arg2):
print("Write to CLI")
def verify_result():
func1()
for _ in range (4):
_thread.start_new_thread(func1, (DUT1_CLI, '0'))
verify_result()
I want to concurrently execute (say 4 threads) func1() which in my case includes a function call that can take time to execute. Then, only after the last thread finished its work I want to execute verify_result().
Currently, the result I get is that all threads finish their job, but verify_result() is executed before all threads finish their job.
I have even tried to use the following code (of course I imported threading) under the for loop but that didn't do the work (don't refer to the arguments)
t = threading.Thread(target = Enable_WatchDog, args = (URL_List[x], 180, Terminal_List[x], '0'))
t.start()
t.join()
Your last threading example is close, but you have to collect the threads in a list, start them all at once, then wait for them to complete all at once. Here's a simplified example:
import threading
import time
# Lock to serialize console output
output = threading.Lock()
def threadfunc(a,b):
for i in range(a,b):
time.sleep(.01) # sleep to make the "work" take longer
with output:
print(i)
# Collect the threads
threads = []
for i in range(10,100,10):
# Create 9 threads counting 10-19, 20-29, ... 90-99.
thread = threading.Thread(target=threadfunc,args=(i,i+10))
threads.append(thread)
# Start them all
for thread in threads:
thread.start()
# Wait for all to complete
for thread in threads:
thread.join()
Say you have a list of threads.
You loop(each_thread) over them -
for each_thread in thread_pool:
each_thread.start()
within the loop to start execution of the run function within each thread.
The same way, you write another loop after you start all threads and have
for each_thread in thread_pool:
each_thread.join()
what join does is that it will wait for thread i to finish execution before letting i+1th thread to finish execution.
The threads would run concurrently, join() would just synchronize the way each thread returns its results.
In your case specifically, you can the join() loop and the run verify_result() function.

Python concurrent.futures starvation

I'm using Python concurrent.futures, executes parent multi threads and each parent thread execute child threads.
When ThreadPoolExecutor is less than number of required parent threads I got starvation and program stuck.
What is the best approach to:
1. Use const ThreadPoolExecutor
2. Do not get into starvation
Please find below example code:
import time
import sys
import concurrent.futures
MAX_THREAD_EXECUTORS = 5
threadPool = concurrent.futures.ThreadPoolExecutor(MAX_THREAD_EXECUTORS)
threads = []
command_threads = []
def main():
start_tests()
join_threads()
def start_tests():
for i in range(1,14):
threads.append(threadPool.submit(start_test_flow, i))
def start_test_flow(test):
print(f"Start test flow for: {test}")
execute_commands()
join_command_threads()
def execute_commands():
for i in range(1,5):
command_threads.append(threadPool.submit(start_command, i))
def start_command(command):
print(f"Start command for: {command}")
time.sleep(120)
def join_threads():
for thread in threads:
result = thread.result()
print(f"test result={result}")
def join_command_threads():
for thread in command_threads:
result = thread.result()
print(f"command result={result}")
if __name__ == '__main__':
main()
sys.exit(0)
Best Regards, Moshe
The minimum number of threads you actually need is non-deterministic and depends on timing, although there is a number (13 + 1, i.e. one thread for each of the parent threads and at least one thread to run a child thread) that will guarantee that you will never stall. What is most likely happening is that you are quickly creating 5 parent threads and then waiting to create further parent threads and child threads because you only have 5 worker threads. But until you are able to create 4 child threads (in execute_commands) and run them to completion, a parent thread will not complete and thus you are stuck.
Now, for example, insert a call to time.sleep(1) in function start_tests as follows:
def start_tests():
for i in range(1,14):
threads.append(threadPool.submit(start_test_flow, i))
time.sleep(1)
This will allow the 4 child threads to be created and there will be some progress. But depending on timing, you may eventually stall. To guarantee that you never stall, you would have to sleep long enough to allow all 4 child threads to complete before attempting to start the next parent thread.
The bottom line is that you just don't have enough worker threads (13 + 1) to guarantee that you won't stall.

Asynchronous multiprocessing with a worker pool in Python: how to keep going after timeout?

I would like to run a number of jobs using a pool of processes and apply a given timeout after which a job should be killed and replaced by another working on the next task.
I have tried to use the multiprocessing module which offers a method to run of pool of workers asynchronously (e.g. using map_async), but there I can only set a "global" timeout after which all processes would be killed.
Is it possible to have an individual timeout after which only a single process that takes too long is killed and a new worker is added to the pool again instead (processing the next task and skipping the one that timed out)?
Here's a simple example to illustrate my problem:
def Check(n):
import time
if n % 2 == 0: # select some (arbitrary) subset of processes
print "%d timeout" % n
while 1:
# loop forever to simulate some process getting stuck
pass
print "%d done" % n
return 0
from multiprocessing import Pool
pool = Pool(processes=4)
result = pool.map_async(Check, range(10))
print result.get(timeout=1)
After the timeout all workers are killed and the program exits. I would like instead that it continues with the next subtask. Do I have to implement this behavior myself or are there existing solutions?
Update
It is possible to kill the hanging workers and they are automatically replaced. So I came up with this code:
jobs = pool.map_async(Check, range(10))
while 1:
try:
print "Waiting for result"
result = jobs.get(timeout=1)
break # all clear
except multiprocessing.TimeoutError:
# kill all processes
for c in multiprocessing.active_children():
c.terminate()
print result
The problem now is that the loop never exits; even after all tasks have been processed, calling get yields a timeout exception.
The pebble Pool module has been built for solving these types of issue. It supports timeout on given tasks allowing to detect them and easily recover.
from pebble import ProcessPool
from concurrent.futures import TimeoutError
with ProcessPool() as pool:
future = pool.schedule(function, args=[1,2], timeout=5)
try:
result = future.result()
except TimeoutError:
print "Function took longer than %d seconds" % error.args[1]
For your specific example:
from pebble import ProcessPool
from concurrent.futures import TimeoutError
results = []
with ProcessPool(max_workers=4) as pool:
future = pool.map(Check, range(10), timeout=5)
iterator = future.result()
# iterate over all results, if a computation timed out
# print it and continue to the next result
while True:
try:
result = next(iterator)
results.append(result)
except StopIteration:
break
except TimeoutError as error:
print "function took longer than %d seconds" % error.args[1]
print results
Currently the Python does not provide native means to the control execution time of each distinct task in the pool outside the worker itself.
So the easy way is to use wait_procs in the psutil module and implement the tasks as subprocesses.
If nonstandard libraries are not desirable, then you have to implement own Pool on base of subprocess module having the working cycle in the main process, poll() - ing the execution of each worker and performing required actions.
As for the updated problem, the pool becomes corrupted if you directly terminate one of the workers (it is the bug in the interpreter implementation, because such behavior should not be allowed): the worker is recreated, but the task is lost and the pool becomes nonjoinable.
You have to terminate all the pool and then recreate it again for another tasks:
from multiprocessing import Pool
while True:
pool = Pool(processes=4)
jobs = pool.map_async(Check, range(10))
print "Waiting for result"
try:
result = jobs.get(timeout=1)
break # all clear
except multiprocessing.TimeoutError:
# kill all processes
pool.terminate()
pool.join()
print result
UPDATE
Pebble is an excellent and handy library, which solves the issue. Pebble is designed for the asynchronous execution of Python functions, where is PyExPool is designed for the asynchronous execution of modules and external executables, though both can be used interchangeably.
One more aspect is when 3dparty dependencies are not desirable, then PyExPool can be a good choice, which is a single-file lightweight implementation of Multi-process Execution Pool with per-Job and global timeouts, opportunity to group Jobs into Tasks and other features.
PyExPool can be embedded into your sources and customized, having permissive Apache 2.0 license and production quality, being used in the core of one high-loaded scientific benchmarking framework.
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

Return whichever expression returns first

I have two different functions f, and g that compute the same result with different algorithms. Sometimes one or the other takes a long time while the other terminates quickly. I want to create a new function that runs each simultaneously and then returns the result from the first that finishes.
I want to create that function with a higher order function
h = firstresult(f, g)
What is the best way to accomplish this in Python?
I suspect that the solution involves threading. I'd like to avoid discussion of the GIL.
I would simply use a Queue for this. Start the threads and the first one which has a result ready writes to the queue.
Code
from threading import Thread
from time import sleep
from Queue import Queue
def firstresult(*functions):
queue = Queue()
threads = []
for f in functions:
def thread_main():
queue.put(f())
thread = Thread(target=thread_main)
threads.append(thread)
thread.start()
result = queue.get()
return result
def slow():
sleep(1)
return 42
def fast():
return 0
if __name__ == '__main__':
print firstresult(slow, fast)
Live demo
http://ideone.com/jzzZX2
Notes
Stopping the threads is an entirely different topic. For this you need to add some state variable to the threads which needs to be checked in regular intervals. As I want to keep this example short I simply assumed that part and assumed that all workers get the time to finish their work even though the result is never read.
Skipping the discussion about the Gil as requested by the questioner. ;-)
Now - unlike my suggestion on the other answer, this piece of code does exactly what you are requesting:
from multiprocessing import Process, Queue
import random
import time
def firstresult(func1, func2):
queue = Queue()
proc1 = Process(target=func1,args=(queue,))
proc2 = Process(target=func2, args=(queue,))
proc1.start();proc2.start()
result = queue.get()
proc1.terminate(); proc2.terminate()
return result
def algo1(queue):
time.sleep(random.uniform(0,1))
queue.put("algo 1")
def algo2(queue):
time.sleep(random.uniform(0,1))
queue.put("algo 2")
print firstresult(algo1, algo2)
Run each function in a new worker thread, the 2 worker threads send the result back to the main thread in a 1 item queue or something similar. When the main thread receives the result from the winner, it kills (do python threads support kill yet? lol.) both worker threads to avoid wasting time (one function may take hours while the other only takes a second).
Replace the word thread with process if you want.
You will need to run each function in another process (with multiprocessing) or in a different thread.
If both are CPU bound, multithread won help much - exactly due to the GIL -
so multiprocessing is the way.
If the return value is a pickleable (serializable) object, I have this decorator I created that simply runs the function in background, in another process:
https://bitbucket.org/jsbueno/lelo/src
It is not exactly what you want - as both are non-blocking and start executing right away. The tirck with this decorator is that it blocks (and waits for the function to complete) as when you try to use the return value.
But on the other hand - it is just a decorator that does all the work.

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