Question

我正在做的是：

表示一个点，计算到所有其他点的距离
获取与当前距离的距离小于阈值的点数（例如500）
就所有问题而言，做同样的事情。

我的方式是完全蛮力的方法，我该如何改进呢？（因为蛮力方式太慢了）

# compute distance with latitude and longitude
def space_dist(c1, c2):
    return math.sqrt((float(c1[0])-float(c2[0]))*(float(c1[0])-float(c2[0]))+(float(c1[1])-float(c2[1]))*(float(c1[1])-float(c2[1])))

def _cluster_by_dist(i, threshold=500):
    address = address_group[i]
    threshold = threshold / 27.0 / 3600.0
    ids = [i]
    for oidx, o in enumerate(address_group):
        if oidx == i:
            continue

        dist = space_dist((address[i_x], address[i_y]), (o[i_x], o[i_y]))
        if dist >= threshold:
            continue

        ids.append(oidx)
    return ids

def cluster_check():
    pool = Pool()
    cnt = 0
    t = time.time()
    for ids in pool.imap(_cluster_by_dist, range(len(address_group))):
        cnt += 1
        if cnt % 100 == 0:
            print("{}: {}".format(cnt, time.time() - t))
            t = time.time()
        if len(ids) <= 1:
            continue

        yield([address_group[i] for i in ids])

if __name__ == "__main__":
    columns = "id,key,mode,x,y,freq".split(",")
    address_group = list(load_raw_data("input_path", columns))
    i_key, i_x, i_y = columns.index("key"), columns.index("x"), columns.index("y")

    for addresses in cluster_check():
        for address in addresses:
            print(address)
        print()

快速找到距离小于阈值的邻居？

0 个答案: