基本上有一个脚本，它可以梳理节点/点的数据集，以删除重叠的节点/点。实际的脚本更复杂，但我把它简化为一个简单的重叠检查，对它不做任何演示。
我尝试了一些锁、队列、池的变体，每次添加一个作业，而不是批量添加。一些最严重的罪犯速度慢了几个数量级。最终我以最快的速度完成了比赛。
发送到各个进程的重叠检查算法：
def check_overlap(args):
tolerance = args['tolerance']
this_coords = args['this_coords']
that_coords = args['that_coords']
overlaps = False
distance_x = this_coords[0] - that_coords[0]
if distance_x <= tolerance:
distance_x = pow(distance_x, 2)
distance_y = this_coords[1] - that_coords[1]
if distance_y <= tolerance:
distance = pow(distance_x + pow(distance_y, 2), 0.5)
if distance <= tolerance:
overlaps = True
return overlaps
处理功能：
def process_coords(coords, num_processors=1, tolerance=1):
import multiprocessing as mp
import time
if num_processors > 1:
pool = mp.Pool(num_processors)
start = time.time()
print "Start script w/ multiprocessing"
else:
num_processors = 0
start = time.time()
print "Start script w/ standard processing"
total_overlap_count = 0
# outer loop through nodes
start_index = 0
last_index = len(coords) - 1
while start_index <= last_index:
# nature of the original problem means we can process all pairs of a single node at once, but not multiple, so batch jobs by outer loop
batch_jobs = []
# inner loop against all pairs for this node
start_index += 1
count_overlapping = 0
for i in range(start_index, last_index+1, 1):
if num_processors:
# add job
batch_jobs.append({
'tolerance': tolerance,
'this_coords': coords[start_index],
'that_coords': coords[i]
})
else:
# synchronous processing
this_coords = coords[start_index]
that_coords = coords[i]
distance_x = this_coords[0] - that_coords[0]
if distance_x <= tolerance:
distance_x = pow(distance_x, 2)
distance_y = this_coords[1] - that_coords[1]
if distance_y <= tolerance:
distance = pow(distance_x + pow(distance_y, 2), 0.5)
if distance <= tolerance:
count_overlapping += 1
if num_processors:
res = pool.map_async(check_overlap, batch_jobs)
results = res.get()
for r in results:
if r:
count_overlapping += 1
# stuff normally happens here to process nodes connected to this node
total_overlap_count += count_overlapping
print total_overlap_count
print " time: {0}".format(time.time() - start)
以及测试功能：
from random import random
coords = []
num_coords = 1000
spread = 100.0
half_spread = 0.5*spread
for i in range(num_coords):
coords.append([
random()*spread-half_spread,
random()*spread-half_spread
])
process_coords(coords, 1)
process_coords(coords, 4)
尽管如此，非多处理的运行时间始终不到0.4秒，而我可以获得的多处理时间刚好不到3.0秒。我知道，也许这里的算法太简单了，无法真正获得好处，但考虑到上面的情况有近50万次迭代，而实际情况有更多的迭代，我觉得奇怪的是，多处理慢了一个数量级。
我缺少什么/我能做些什么来改进？
构建序列化代码中未使用的O(N**2)三元素dict，并通过进程间管道传输它们，是保证多处理的一种很好的方法没有什么是免费的——一切都要花钱。
下面是一个重写，它执行许多相同的代码，无论它是在串行模式还是多处理模式下运行。没有新的dict等。一般来说，len(coords)越大，它从多处理中获得的好处就越多。在我的盒子上，在20000的时候，多处理运行所需的时间大约是墙上时钟的三分之一。
关键是所有进程都有自己的coords副本。这是通过在创建池时只传输一次来完成的。这应该适用于所有平台。在Linux-y系统上，它可以"魔术般"地通过分叉进程继承来实现。将跨进程发送的数据量从O(N**2)减少到O(N)是一个巨大的改进。
从多处理中获得更多需要更好的负载平衡。按原样，对check_overlap(i)的调用将coords[i]与coords[i+1:]中的每个值进行比较。i越大，它要做的工作就越少，而对于i的最大值，仅在进程之间传输i和将结果传输回来的成本就占用了在check_overlap(i)中花费的时间。
def init(*args):
global _coords, _tolerance
_coords, _tolerance = args
def check_overlap(start_index):
coords, tolerance = _coords, _tolerance
tsq = tolerance ** 2
overlaps = 0
start0, start1 = coords[start_index]
for i in range(start_index + 1, len(coords)):
that0, that1 = coords[i]
dx = abs(that0 - start0)
if dx <= tolerance:
dy = abs(that1 - start1)
if dy <= tolerance:
if dx**2 + dy**2 <= tsq:
overlaps += 1
return overlaps
def process_coords(coords, num_processors=1, tolerance=1):
global _coords, _tolerance
import multiprocessing as mp
_coords, _tolerance = coords, tolerance
import time
if num_processors > 1:
pool = mp.Pool(num_processors, initializer=init, initargs=(coords, tolerance))
start = time.time()
print("Start script w/ multiprocessing")
else:
num_processors = 0
start = time.time()
print("Start script w/ standard processing")
N = len(coords)
if num_processors:
total_overlap_count = sum(pool.imap_unordered(check_overlap, range(N)))
else:
total_overlap_count = sum(check_overlap(i) for i in range(N))
print(total_overlap_count)
print(" time: {0}".format(time.time() - start))
if __name__ == "__main__":
from random import random
coords = []
num_coords = 20000
spread = 100.0
half_spread = 0.5*spread
for i in range(num_coords):
coords.append([
random()*spread-half_spread,
random()*spread-half_spread
])
process_coords(coords, 1)
process_coords(coords, 4)