在Python中给定纬度和经度数据计算距离矩阵的有效方法

Question

我有纬度和经度的数据,我需要计算包含位置的两个数组之间的距离矩阵.我用这本获得给定的纬度和经度两个位置之间的距离.

这是我的代码示例:

import numpy as np
import math

def get_distances(locs_1, locs_2):
    n_rows_1 = locs_1.shape[0]
    n_rows_2 = locs_2.shape[0]
    dists = np.empty((n_rows_1, n_rows_2))
    # The loops here are inefficient
    for i in xrange(n_rows_1):
        for j in xrange(n_rows_2):
            dists[i, j] = get_distance_from_lat_long(locs_1[i], locs_2[j])
    return dists


def get_distance_from_lat_long(loc_1, loc_2):

    earth_radius = 3958.75

    lat_dif = math.radians(loc_1[0] - loc_2[0])
    long_dif = math.radians(loc_1[1] - loc_2[1])
    sin_d_lat = math.sin(lat_dif / 2)
    sin_d_long = math.sin(long_dif / 2)
    step_1 = (sin_d_lat ** 2) + (sin_d_long ** 2) * math.cos(math.radians(loc_1[0])) * math.cos(math.radians(loc_2[0])) 
    step_2 = 2 * math.atan2(math.sqrt(step_1), math.sqrt(1-step_1))
    dist = step_2 * earth_radius

    return dist

我的预期输出是这样的:

>>> locations_1 = np.array([[34, -81], [32, -87], [35, -83]])
>>> locations_2 = np.array([[33, -84], [39, -81], [40, -88], [30, -80]])
>>> get_distances(locations_1, locations_2)
array([[ 186.13522573,  345.46610882,  566.23466349,  282.51056676],
       [ 187.96657622,  589.43369894,  555.55312473,  436.88855214],
       [ 149.5853537 ,  297.56950329,  440.81203371,  387.12153747]])

性能对我来说很重要,我能做的一件事是Cython用来加速循环,但如果我不必去那里就会很好.

有没有一个模块可以做这样的事情？或任何其他解决方案？

Answer 1

你正在使用的Haversine方程中有许多次优的东西.您可以修剪其中一些并最小化您需要计算的正弦,余弦和平方根的数量.以下是我能够提出的最好的,并且在我的系统运行速度比Ophion的代码快了大约5倍(在矢量化方面大致相同),在1000和2000个元素的两个随机数组上:

def spherical_dist(pos1, pos2, r=3958.75):
    pos1 = pos1 * np.pi / 180
    pos2 = pos2 * np.pi / 180
    cos_lat1 = np.cos(pos1[..., 0])
    cos_lat2 = np.cos(pos2[..., 0])
    cos_lat_d = np.cos(pos1[..., 0] - pos2[..., 0])
    cos_lon_d = np.cos(pos1[..., 1] - pos2[..., 1])
    return r * np.arccos(cos_lat_d - cos_lat1 * cos_lat2 * (1 - cos_lon_d))

如果你"按原样"提供你的两个数组它会抱怨,但这不是一个错误,它是一个功能.基本上,此函数计算球体在最后一个维度上的距离,并在其余维度上进行广播.所以你可以得到你所追求的:

>>> spherical_dist(locations_1[:, None], locations_2)
array([[ 186.13522573,  345.46610882,  566.23466349,  282.51056676],
       [ 187.96657622,  589.43369894,  555.55312473,  436.88855214],
       [ 149.5853537 ,  297.56950329,  440.81203371,  387.12153747]])

但它也可用于计算两个点列表之间的距离,即:

>>> spherical_dist(locations_1, locations_2[:-1])
array([ 186.13522573,  589.43369894,  440.81203371])

或两个单点之间:

>>> spherical_dist(locations_1[0], locations_2[0])
186.1352257300577

这取决于gufunc的工作原理,一旦你习惯了它,我发现它是一种美妙的"瑞士军刀"编码风格,它允许你在许多不同的设置中重复使用单个功能.

Answer 2

使用meshgrid代替双for循环会更高效：

import numpy as np

earth_radius = 3958.75

def get_distances(locs_1, locs_2):
   lats1, lats2 = np.meshgrid(locs_1[:,0], locs_2[:,0])
   lons1, lons2 = np.meshgrid(locs_1[:,1], locs_2[:,1])

   lat_dif = np.radians(lats1 - lats2)
   long_dif = np.radians(lons1 - lons2)

   sin_d_lat = np.sin(lat_dif / 2.)
   sin_d_long = np.sin(long_dif / 2.)

   step_1 = (sin_d_lat ** 2) + (sin_d_long ** 2) * np.cos(np.radians(lats1[0])) * np.cos(np.radians(lats2[0])) 
   step_2 = 2 * np.arctan2(np.sqrt(step_1), np.sqrt(1-step_1))

   dist = step_2 * earth_radius

   return dist

Answer 3

这只是对代码进行矢量化:

def new_get_distances(loc1, loc2):
    earth_radius = 3958.75

    locs_1 = np.deg2rad(loc1)
    locs_2 = np.deg2rad(loc2)

    lat_dif = (locs_1[:,0][:,None]/2 - locs_2[:,0]/2)
    lon_dif = (locs_1[:,1][:,None]/2 - locs_2[:,1]/2)

    np.sin(lat_dif, out=lat_dif)
    np.sin(lon_dif, out=lon_dif)

    np.power(lat_dif, 2, out=lat_dif)
    np.power(lon_dif, 2, out=lon_dif)

    lon_dif *= ( np.cos(locs_1[:,0])[:,None] * np.cos(locs_2[:,0]) )
    lon_dif += lat_dif

    np.arctan2(np.power(lon_dif,.5), np.power(1-lon_dif,.5), out = lon_dif)
    lon_dif *= ( 2 * earth_radius )

    return lon_dif

locations_1 = np.array([[34, -81], [32, -87], [35, -83]])
locations_2 = np.array([[33, -84], [39, -81], [40, -88], [30, -80]])
old = get_distances(locations_1, locations_2)

new = new_get_distances(locations_1,locations_2)

np.allclose(old,new)
True

如果我们看一下时间:

%timeit new_get_distances(locations_1,locations_2)
10000 loops, best of 3: 80.6 µs per loop

%timeit get_distances(locations_1,locations_2)
10000 loops, best of 3: 74.9 µs per loop

对于一个小例子来说,它实际上更慢; 但是,让我们看一个更大的例子:

locations_1 = np.random.rand(1000,2)

locations_2 = np.random.rand(1000,2)

%timeit get_distances(locations_1,locations_2)
1 loops, best of 3: 5.84 s per loop

%timeit new_get_distances(locations_1,locations_2)
10 loops, best of 3: 149 ms per loop

我们现在的加速比为40倍.可能会在一些地方挤出更多的速度.

编辑:做了一些更新,以删除多余的位置,并明确表示我们不会改变原始位置数组.