我有一个计算数字的函数。这取决于一些随机条件。
所以我想做的是在这个函数中运行多个处理器，找到这个因素的处理器首先返回 value 并且所有处理器都终止。
到目前为止我所拥有的是非常错误的。处理器没有终止，我也不知道如何获取函数返回的 value
flag = False
def rho(n, processor):
while True:
x = random.randrange(1, n-1)
x2 = x
gcd = 1
c = random.randrange(1, n-1)
while gcd == 1:
x = (x**2 + c) % n
x2 = (x2**2 + c) % n
x2 = (x2**2 + c) % n
gcd = math.gcd(abs(x - x2), n)
if gcd != n:
flag = True
print("Factor was found from "+process+" and is ", gcd)
return gcd
if __name__ == "__main__":
p1 = multiprocessing.Process(target=rho, args=(91, "process 1" ))
p2 = multiprocessing.Process(target=rho, args=(91, "process 2"))
p1.start()
p2.start()
if flag:
p1.terminate()
p2.terminate()输出是：
Factor was found from process 2 and is 13
Factor was found from process 1 and is 7回答1
您可以使用 https://docs.python.org/3/library/multiprocessing.html#multiprocessing.pool.Pool 及其方法 map()、imap_unordered() 等。这些方法也会从工作函数返回 values。
示例（我使用 time.sleep() 来模拟一些时间密集型计算）：
from time import sleep
from multiprocessing import Pool
def rho(params):
n, processor = params
# your computation here
# ...
sleep(n)
print("Factor was found from " + processor + " and is 42")
return 42
if __name__ == "__main__":
with Pool() as pool:
for result in pool.imap_unordered(
rho, ((10, "process 1"), (1, "process 2"))
):
print("Result I got:", result)
break # <-- I don't want other results, so break印刷：
Factor was found from process 2 and is 42
Result I got: 42编辑：两个不同的功能：
from time import sleep
from multiprocessing import Pool
def fn1(n, p):
sleep(n)
print("Factor was found from " + p + " and is 42")
return 42
def fn2(n, p):
sleep(n)
print("Factor was found from " + p + " and is 99")
return 99
def rho(params):
what_to_call, n, processor = params
return what_to_call(n, processor)
if __name__ == "__main__":
with Pool() as pool:
for result in pool.imap_unordered(
rho, ((fn1, 10, "process 1"), (fn2, 1, "process 2"))
):
print("Result I got:", result)
break # <-- I don't want other results, so break