Question

基本上有一个脚本，可以对节点/点的数据集进行梳理以除去重叠的节点/点。实际的脚本更为复杂，但我将其简化为一个简单的重叠检查，该重叠检查不进行任何演示。

我尝试了几种带有锁，队列，池的变体，一次添加一个作业，而不是批量添加。一些最严重的罪犯的速度下降了两个数量级。最终，我以最快的速度做到了。

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)

和测试功能：

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)

不过，非多重处理始终在不到0.4s的时间内运行，而上面所说的多重处理我可以在3.0s内得到。我知道这里的算法可能太简单了，无法真正获得收益，但是考虑到上面的案例有近一百万次迭代，而实际案例的迭代次数明显更多，因此我感到奇怪的是，多处理的速度慢了一个数量级。 >

我缺少什么/我该怎么做？

Answer 1

构建O(N**2) 3序列式中未使用的字典，并通过进程间管道进行传输，是一种确保多处理无济于事的好方法；-)没有什么是免费的-一切费用。

以下是重写，无论执行串行还是多处理模式，它都执行相同的代码。没有新的字典等。通常，len(coords)越大，它从多处理中获得的好处就越大。在我的盒子上，在20000时，多处理运行大约需要挂钟时间的三分之一。

关键是所有进程都有自己的coords副本。这是通过在创建池时仅传输一次来完成的。那应该适用于所有平台。在Linux-y系统上，它可能是“魔术”发生的，而不是通过派生的进程继承发生的。将跨进程的数据发送量从O(N**2)减少到O(N)是一个巨大的进步。

从多处理中获得更多收益将需要更好的负载平衡。照原样，对check_overlap(i)的调用会将coords[i]与coords[i+1:]中的每个值进行比较。 i越大，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并将结果传输回去的成本就会使 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)

Python多处理比常规处理慢。我该如何改善？

1 个答案: