Python 对面向队列的函数使用多处理后没有性能提升
我想要优化的实际代码太复杂，无法包含在这里，因此这里有一个简化的示例：Python 对面向队列的函数使用多处理后没有性能提升,python,performance,optimization,queue,multiprocessing,Python,Performance,Optimization,Queue,Multiprocessing,我想要优化的实际代码太复杂，无法包含在这里，因此这里有一个简化的示例： def enumerate_paths(n, k): """ John want to go up a flight of stairs that has N steps. He can take up to K steps each time. This function enumerate all different ways he can go up this flight of sta
def enumerate_paths(n, k):
"""
John want to go up a flight of stairs that has N steps. He can take
up to K steps each time. This function enumerate all different ways
he can go up this flight of stairs.
"""
paths = []
to_analyze = [(0,)]
while to_analyze:
path = to_analyze.pop()
last_step = path[-1]
if last_step >= n:
# John has reach the top
paths.append(path)
continue
for i in range(1, k + 1):
# possible paths from this point
extended_path = path + (last_step + i,)
to_analyze.append(extended_path)
return paths
>>> enumerate_paths(3, 2)
[(0, 2, 4), (0, 2, 3), (0, 1, 3), (0, 1, 2, 4), (0, 1, 2, 3)]
（0,1,2,4）多处理import multiprocessing
def enumerate_paths_worker(n, k, queue):
paths = []
while not queue.empty():
path = queue.get()
last_step = path[-1]
if last_step >= n:
# John has reach the top
paths.append(path)
continue
for i in range(1, k + 1):
# possible paths from this point
extended_path = path + (last_step + i,)
queue.put(extended_path)
return paths
def enumerate_paths(n, k):
pool = multiprocessing.Pool()
manager = multiprocessing.Manager()
queue = manager.Queue()
path_init = (0,)
queue.put(path_init)
apply_result = pool.apply_async(enumerate_paths_worker, (n, k, queue))
return apply_result.get()
to_analysis使用
apply\u asyncapply_asyncJoinableQueuebusy\u foofrom multiprocessing import Process
from multiprocessing import JoinableQueue as Queue
import time
SENTINEL = 'SENTINEL'
def busy_foo(x = 10e6):
for _ in range(int(x)):
x -= 1
def enumerate_paths(q_analyze, q_result, n, k):
"""
John want to go up a flight of stairs that has N steps. He can take
up to K steps each time. This function enumerate all different ways
he can go up this flight of stairs.
"""
for path in iter(q_analyze.get, SENTINEL):
last_step = path[-1]
if last_step >= n:
busy_foo()
# John has reach the top
q_result.put(path)
q_analyze.task_done()
continue
else:
busy_foo()
for i in range(1, k + 1):
# possible paths from this point
extended_path = path + (last_step + i,)
q_analyze.put(extended_path)
q_analyze.task_done()
if __name__ == '__main__':
N_CORES = 4
N = 6
K = 2
start = time.perf_counter()
q_analyze = Queue()
q_result = Queue()
q_analyze.put((0,))
pool = []
for _ in range(N_CORES):
pool.append(
Process(target=enumerate_paths, args=(q_analyze, q_result, N, K))
)
for p in pool:
p.start()
q_analyze.join() # block until everything is processed
for p in pool:
q_analyze.put(SENTINEL) # let the processes exit gracefully
results = []
while not q_result.empty():
results.append(q_result.get())
for p in pool:
p.join()
print(f'elapsed: {time.perf_counter() - start: .2f} s')
busy\u foo- ~208sN_核心=4
- 2.78s连续原稿
busy\u foo- 6.45sN_芯=4
- 30.46s连续原始
Numpy可以多次加速矢量化操作，但您可能会在这一次看到Numpy的性能损失。Numpy使用连续的内存块作为它的数组。当您更改数组大小时，整个数组将不得不重新构建，这与使用python列表不同。“但我认为它们只能处理彼此完全独立的任务”事实并非如此，例如，进程可以在列表上协同工作。对于你的奖金问题，如果不实施它，就不可能知道；可能是相同的速度，可能是1000倍faster@roganjosh我明白你的意思，但我的意思是，你不能用
map（0，1，2，6）枚举路径（3，4）