我正在努力理解Python中的线程。我看过文档和示例,但坦率地说,许多示例过于复杂,我很难理解它们。
如何清楚地显示为多线程划分的任务?
我正在努力理解Python中的线程。我看过文档和示例,但坦率地说,许多示例过于复杂,我很难理解它们。
如何清楚地显示为多线程划分的任务?
当前回答
这很容易理解。这里有两种简单的线程处理方法。
import time
from concurrent.futures import ThreadPoolExecutor, as_completed
import threading
def a(a=1, b=2):
print(a)
time.sleep(5)
print(b)
return a+b
def b(**kwargs):
if "a" in kwargs:
print("am b")
else:
print("nothing")
to_do=[]
executor = ThreadPoolExecutor(max_workers=4)
ex1=executor.submit(a)
to_do.append(ex1)
ex2=executor.submit(b, **{"a":1})
to_do.append(ex2)
for future in as_completed(to_do):
print("Future {} and Future Return is {}\n".format(future, future.result()))
print("threading")
to_do=[]
to_do.append(threading.Thread(target=a))
to_do.append(threading.Thread(target=b, kwargs={"a":1}))
for threads in to_do:
threads.start()
for threads in to_do:
threads.join()
其他回答
我在这里看到了很多没有执行实际工作的示例,它们大多是CPU限制的。这里是一个CPU绑定任务的示例,它计算1000万到1005万之间的所有素数。我在这里使用了所有四种方法:
import math
import timeit
import threading
import multiprocessing
from concurrent.futures import ThreadPoolExecutor, ProcessPoolExecutor
def time_stuff(fn):
"""
Measure time of execution of a function
"""
def wrapper(*args, **kwargs):
t0 = timeit.default_timer()
fn(*args, **kwargs)
t1 = timeit.default_timer()
print("{} seconds".format(t1 - t0))
return wrapper
def find_primes_in(nmin, nmax):
"""
Compute a list of prime numbers between the given minimum and maximum arguments
"""
primes = []
# Loop from minimum to maximum
for current in range(nmin, nmax + 1):
# Take the square root of the current number
sqrt_n = int(math.sqrt(current))
found = False
# Check if the any number from 2 to the square root + 1 divides the current numnber under consideration
for number in range(2, sqrt_n + 1):
# If divisible we have found a factor, hence this is not a prime number, lets move to the next one
if current % number == 0:
found = True
break
# If not divisible, add this number to the list of primes that we have found so far
if not found:
primes.append(current)
# I am merely printing the length of the array containing all the primes, but feel free to do what you want
print(len(primes))
@time_stuff
def sequential_prime_finder(nmin, nmax):
"""
Use the main process and main thread to compute everything in this case
"""
find_primes_in(nmin, nmax)
@time_stuff
def threading_prime_finder(nmin, nmax):
"""
If the minimum is 1000 and the maximum is 2000 and we have four workers,
1000 - 1250 to worker 1
1250 - 1500 to worker 2
1500 - 1750 to worker 3
1750 - 2000 to worker 4
so let’s split the minimum and maximum values according to the number of workers
"""
nrange = nmax - nmin
threads = []
for i in range(8):
start = int(nmin + i * nrange/8)
end = int(nmin + (i + 1) * nrange/8)
# Start the thread with the minimum and maximum split up to compute
# Parallel computation will not work here due to the GIL since this is a CPU-bound task
t = threading.Thread(target = find_primes_in, args = (start, end))
threads.append(t)
t.start()
# Don’t forget to wait for the threads to finish
for t in threads:
t.join()
@time_stuff
def processing_prime_finder(nmin, nmax):
"""
Split the minimum, maximum interval similar to the threading method above, but use processes this time
"""
nrange = nmax - nmin
processes = []
for i in range(8):
start = int(nmin + i * nrange/8)
end = int(nmin + (i + 1) * nrange/8)
p = multiprocessing.Process(target = find_primes_in, args = (start, end))
processes.append(p)
p.start()
for p in processes:
p.join()
@time_stuff
def thread_executor_prime_finder(nmin, nmax):
"""
Split the min max interval similar to the threading method, but use a thread pool executor this time.
This method is slightly faster than using pure threading as the pools manage threads more efficiently.
This method is still slow due to the GIL limitations since we are doing a CPU-bound task.
"""
nrange = nmax - nmin
with ThreadPoolExecutor(max_workers = 8) as e:
for i in range(8):
start = int(nmin + i * nrange/8)
end = int(nmin + (i + 1) * nrange/8)
e.submit(find_primes_in, start, end)
@time_stuff
def process_executor_prime_finder(nmin, nmax):
"""
Split the min max interval similar to the threading method, but use the process pool executor.
This is the fastest method recorded so far as it manages process efficiently + overcomes GIL limitations.
RECOMMENDED METHOD FOR CPU-BOUND TASKS
"""
nrange = nmax - nmin
with ProcessPoolExecutor(max_workers = 8) as e:
for i in range(8):
start = int(nmin + i * nrange/8)
end = int(nmin + (i + 1) * nrange/8)
e.submit(find_primes_in, start, end)
def main():
nmin = int(1e7)
nmax = int(1.05e7)
print("Sequential Prime Finder Starting")
sequential_prime_finder(nmin, nmax)
print("Threading Prime Finder Starting")
threading_prime_finder(nmin, nmax)
print("Processing Prime Finder Starting")
processing_prime_finder(nmin, nmax)
print("Thread Executor Prime Finder Starting")
thread_executor_prime_finder(nmin, nmax)
print("Process Executor Finder Starting")
process_executor_prime_finder(nmin, nmax)
if __name__ == "__main__":
main()
以下是我的Mac OS X四核计算机的结果
Sequential Prime Finder Starting
9.708213827005238 seconds
Threading Prime Finder Starting
9.81836523200036 seconds
Processing Prime Finder Starting
3.2467174359990167 seconds
Thread Executor Prime Finder Starting
10.228896902000997 seconds
Process Executor Finder Starting
2.656402041000547 seconds
这里是多线程,有一个简单的例子会很有帮助。您可以运行它并轻松了解多线程在Python中的工作方式。我使用了一个锁来防止访问其他线程,直到前面的线程完成它们的工作。通过使用这行代码,
t锁定=线程。有界信号量(值=4)
您可以一次允许多个进程,并保留将在稍后或完成之前的进程后运行的其余线程。
import threading
import time
#tLock = threading.Lock()
tLock = threading.BoundedSemaphore(value=4)
def timer(name, delay, repeat):
print "\r\nTimer: ", name, " Started"
tLock.acquire()
print "\r\n", name, " has the acquired the lock"
while repeat > 0:
time.sleep(delay)
print "\r\n", name, ": ", str(time.ctime(time.time()))
repeat -= 1
print "\r\n", name, " is releaseing the lock"
tLock.release()
print "\r\nTimer: ", name, " Completed"
def Main():
t1 = threading.Thread(target=timer, args=("Timer1", 2, 5))
t2 = threading.Thread(target=timer, args=("Timer2", 3, 5))
t3 = threading.Thread(target=timer, args=("Timer3", 4, 5))
t4 = threading.Thread(target=timer, args=("Timer4", 5, 5))
t5 = threading.Thread(target=timer, args=("Timer5", 0.1, 5))
t1.start()
t2.start()
t3.start()
t4.start()
t5.start()
print "\r\nMain Complete"
if __name__ == "__main__":
Main()
大多数文档和教程都使用Python的“线程和队列”模块,对于初学者来说,它们可能会让人不知所措。
也许可以考虑Python 3的concurrent.futures.ThreadPoolExecutor模块。
结合子句和列表理解,这可能是一个真正的魅力。
from concurrent.futures import ThreadPoolExecutor, as_completed
def get_url(url):
# Your actual program here. Using threading.Lock() if necessary
return ""
# List of URLs to fetch
urls = ["url1", "url2"]
with ThreadPoolExecutor(max_workers = 5) as executor:
# Create threads
futures = {executor.submit(get_url, url) for url in urls}
# as_completed() gives you the threads once finished
for f in as_completed(futures):
# Get the results
rs = f.result()
使用全新的concurrent.futures模块
def sqr(val):
import time
time.sleep(0.1)
return val * val
def process_result(result):
print(result)
def process_these_asap(tasks):
import concurrent.futures
with concurrent.futures.ProcessPoolExecutor() as executor:
futures = []
for task in tasks:
futures.append(executor.submit(sqr, task))
for future in concurrent.futures.as_completed(futures):
process_result(future.result())
# Or instead of all this just do:
# results = executor.map(sqr, tasks)
# list(map(process_result, results))
def main():
tasks = list(range(10))
print('Processing {} tasks'.format(len(tasks)))
process_these_asap(tasks)
print('Done')
return 0
if __name__ == '__main__':
import sys
sys.exit(main())
执行器方法对于所有以前接触过Java的人来说似乎都很熟悉。
还有一个附带说明:为了保持宇宙的正常,如果你不使用上下文,不要忘记关闭你的池/执行器(这是如此棒,它为你做了)
这里是使用线程导入CSV的一个非常简单的示例。(图书馆的收录可能因不同的目的而有所不同。)
助手函数:
from threading import Thread
from project import app
import csv
def import_handler(csv_file_name):
thr = Thread(target=dump_async_csv_data, args=[csv_file_name])
thr.start()
def dump_async_csv_data(csv_file_name):
with app.app_context():
with open(csv_file_name) as File:
reader = csv.DictReader(File)
for row in reader:
# DB operation/query
驾驶员功能:
import_handler(csv_file_name)