有效计算Haversine距离的最小值

Question

我有一个数据帧与> 2.7mm的坐标,和一个单独的列表中〜2000坐标.我试图返回的坐标之间的最小距离每个单排相比,列表中的每个坐标.以下代码适用于小规模(具有200行的数据帧),但是当计算超过2.7MM的行时,它似乎永远运行.

from haversine import haversine

df
Latitude   Longitude
39.989    -89.980
39.923    -89.901
39.990    -89.987
39.884    -89.943
39.030    -89.931

end_coords_list = [(41.342,-90.423),(40.349,-91.394),(38.928,-89.323)]

for row in df.itertuples():
    def min_distance(row):
        beg_coord = (row.Latitude, row.Longitude)
        return min(haversine(beg_coord, end_coord) for end_coord in end_coords_list)
    df['Min_Distance'] = df.apply(min_distance, axis=1)

我知道问题在于发生的大量计算(5.7MM*2,000 = ~11.4BN),并且运行这么多循环的事实非常低效.

根据我的研究,似乎矢量化NumPy函数可能是更好的方法,但我是Python和NumPy的新手,所以我不太确定如何在这种特殊情况下实现它.

理想输出:

df
Latitude   Longitude  Min_Distance
39.989    -89.980     3.7
39.923    -89.901     4.1
39.990    -89.987     4.2
39.884    -89.943     5.9
39.030    -89.931     3.1

提前致谢!

Answer 1

的haversine func本质是:

# convert all latitudes/longitudes from decimal degrees to radians
lat1, lng1, lat2, lng2 = map(radians, (lat1, lng1, lat2, lng2))

# calculate haversine
lat = lat2 - lat1
lng = lng2 - lng1

d = sin(lat * 0.5) ** 2 + cos(lat1) * cos(lat2) * sin(lng * 0.5) ** 2
h = 2 * AVG_EARTH_RADIUS * asin(sqrt(d))

这是一个利用强大的矢量化方法NumPy broadcasting并NumPy ufuncs替换那些数学模块函数,以便我们可以一次性操作整个数组 -

# Get array data; convert to radians to simulate 'map(radians,...)' part    
coords_arr = np.deg2rad(coords_list)
a = np.deg2rad(df.values)

# Get the differentiations
lat = coords_arr[:,0] - a[:,0,None]
lng = coords_arr[:,1] - a[:,1,None]

# Compute the "cos(lat1) * cos(lat2) * sin(lng * 0.5) ** 2" part.
# Add into "sin(lat * 0.5) ** 2" part.
add0 = np.cos(a[:,0,None])*np.cos(coords_arr[:,0])* np.sin(lng * 0.5) ** 2
d = np.sin(lat * 0.5) ** 2 +  add0

# Get h and assign into dataframe
h = 2 * AVG_EARTH_RADIUS * np.arcsin(np.sqrt(d))
df['Min_Distance'] = h.min(1)

为了进一步提升性能,我们可以利用numexpr模块来取代超越函数.

运行时测试和验证

方法 -

def loopy_app(df, coords_list):
    for row in df.itertuples():
        df['Min_Distance1'] = df.apply(min_distance, axis=1)

def vectorized_app(df, coords_list):   
    coords_arr = np.deg2rad(coords_list)
    a = np.deg2rad(df.values)

    lat = coords_arr[:,0] - a[:,0,None]
    lng = coords_arr[:,1] - a[:,1,None]

    add0 = np.cos(a[:,0,None])*np.cos(coords_arr[:,0])* np.sin(lng * 0.5) ** 2
    d = np.sin(lat * 0.5) ** 2 +  add0

    h = 2 * AVG_EARTH_RADIUS * np.arcsin(np.sqrt(d))
    df['Min_Distance2'] = h.min(1)

验证 -

In [158]: df
Out[158]: 
   Latitude  Longitude
0    39.989    -89.980
1    39.923    -89.901
2    39.990    -89.987
3    39.884    -89.943
4    39.030    -89.931

In [159]: loopy_app(df, coords_list)

In [160]: vectorized_app(df, coords_list)

In [161]: df
Out[161]: 
   Latitude  Longitude  Min_Distance1  Min_Distance2
0    39.989    -89.980     126.637607     126.637607
1    39.923    -89.901     121.266241     121.266241
2    39.990    -89.987     126.037388     126.037388
3    39.884    -89.943     118.901195     118.901195
4    39.030    -89.931      53.765506      53.765506

计时 -

In [163]: df
Out[163]: 
   Latitude  Longitude
0    39.989    -89.980
1    39.923    -89.901
2    39.990    -89.987
3    39.884    -89.943
4    39.030    -89.931

In [164]: %timeit loopy_app(df, coords_list)
100 loops, best of 3: 2.41 ms per loop

In [165]: %timeit vectorized_app(df, coords_list)
10000 loops, best of 3: 96.8 µs per loop