Question

我有一个包含数百万行的数据框“数据”。每行都有坐标（'x'，'y'），我想以python可以提供的最有效方式计算连续坐标对之间的距离。并行化会有帮助吗？

我在这里看到了建议使用cython的方法。但是我想只看到python解决方案。

以下是我的数据片段

points = 
[(26406, -6869),
 (27679, -221),
 (27679, -221),
 (26416, -6156),
 (26679, -578),
 (26679, -580),
 (27813, -558),
 (26254, -1097),
 (26679, -580),
 (27813, -558),
 (28258, -893),
 (26253, -1098),
 (26678, -581),
 (27811, -558),
 (28259, -893),
 (26252, -1098),
 (27230, -481),
 (26679, -582),
 (27488, -5849),
 (27811, -558),
 (28259, -893),
 (26250, -1099),
 (27228, -481),
 (26679, -582),
 (27488, -5847),
 (28525, -1465),
 (27811, -558),
 (28259, -892)]

我相信我使用for循环的第一种方法肯定会得到改善：

    from scipy.spatial import distance
    def comp_dist(points):
        size  =len(points)
        d = 0
        i=1
        for i in range(1,size):
            if i%1000000==0:
                print i
            # print "i-1:", points[i-1]
            # print "i: ", points[i]
            dist = distance.euclidean(points[i-1],points[i])
            d= d+dist
        print d

    distance = comp_dist(points)

提前感谢您的回答。

Answer 1

你说的是python，但是因为你已经在使用scipy进行距离计算，所以我认为一个numpy解决方案是可以的。

在我的笔记本电脑上使用2800万点numpy阵列上的矢量化单线程操作只需1秒钟。使用32位整数数据类型，该阵列在内存中占用大约200MB。

import numpy as np
points = [(26406, -6869), ..., (28259, -892)]
# make test array my repeating the 28-element points list 1M times
np_points = np.array(points*1000000, dtype='int32')
# use two different slices (offset by 1) from resulting array;
# execution of next line takes ~1 second
dists = np.sqrt(np.sum((np_points[0:-2] - np_points[1:-1])**2, axis=1))

print(dists.shape)
(27999998,)

print(dists[:28])
[  6.76878372e+03   0.00000000e+00   6.06789865e+03   5.58419672e+03
   2.00000000e+00   1.13421338e+03   1.64954600e+03   6.69263775e+02
   1.13421338e+03   5.57000898e+02   2.01545280e+03   6.69263775e+02
   1.13323343e+03   5.59400572e+02   2.01744244e+03   1.15636197e+03
   5.60180328e+02   5.32876815e+03   5.30084993e+03   5.59400572e+02
   2.01953386e+03   1.15689585e+03   5.58213221e+02   5.32679134e+03
   4.50303153e+03   1.15431581e+03   5.58802291e+02   6.25764636e+03]

Answer 2

这是一个帮助您入门的简单示例：

from scipy.spatial import distance
from multiprocessing import Pool

processes = 4

# Group data into pairs in order to compute distance
pairs = [(points[i], points[i+1]) for i in range(len(points)-1)]
print pairs

# Split data into chunks
l = [pairs[i:i+processes] for i in xrange(0, len(pairs), processes)]


def worker(lst):
    return [distance.euclidean(i[0], i[1]) for i in lst]

if __name__ == "__main__":
    p = Pool(processes)
    result = p.map(worker, l)
    # Flatten list
    print [item for sublist in result for item in sublist]

用以下方法测试：

points =[(random.randint(0,1000), random.randint(0, 1000)) for i in range(1000000)]

8个过程需要大约5秒钟，1个过程需要10秒钟。

使用Python代码通过并行化计算两点之间距离的最快方法

2 个答案: