带有熊猫数据帧的Vectorised Haversine公式

I know that to find the distance between two latitude, longitude points I need to use the haversine function:

我知道要找到两个纬度,经度点之间的距离我需要使用hasrsine函数:

def haversine(lon1, lat1, lon2, lat2):
    lon1, lat1, lon2, lat2 = map(radians, [lon1, lat1, lon2, lat2])
    dlon = lon2 - lon1 
    dlat = lat2 - lat1 
    a = sin(dlat/2)**2 + cos(lat1) * cos(lat2) * sin(dlon/2)**2
    c = 2 * asin(sqrt(a)) 
    km = 6367 * c
    return km

I have a DataFrame where one column is latitude and another column is longitude. I want to find out how far these points are from a set point, -56.7213600, 37.2175900. How do I take the values from the DataFrame and put them into the function?

我有一个DataFrame,其中一列是纬度,另一列是经度。我想知道这些点距设定点有多远,-56.7213600,37.2175900。如何从DataFrame中获取值并将它们放入函数中?

example DataFrame:

     SEAZ     LAT          LON
1    296.40,  58.7312210,  28.3774110  
2    274.72,  56.8148320,  31.2923240
3    192.25,  52.0649880,  35.8018640
4     34.34,  68.8188750,  67.1933670
5    271.05,  56.6699880,  31.6880620
6    131.88,  48.5546220,  49.7827730
7    350.71,  64.7742720,  31.3953780
8    214.44,  53.5192920,  33.8458560
9      1.46,  67.9433740,  38.4842520
10   273.55,  53.3437310,   4.4716664

1 个解决方案

#1

I can't confirm if the calculations are correct but the following worked:

我无法确认计算是否正确但是以下工作:

In [11]:

def haversine(row):
    lon1 = -56.7213600
    lat1 = 37.2175900
    lon2 = row['LON']
    lat2 = row['LAT']
    lon1, lat1, lon2, lat2 = map(radians, [lon1, lat1, lon2, lat2])
    dlon = lon2 - lon1 
    dlat = lat2 - lat1 
    a = sin(dlat/2)**2 + cos(lat1) * cos(lat2) * sin(dlon/2)**2
    c = 2 * arcsin(sqrt(a)) 
    km = 6367 * c
    return km

df['distance'] = df.apply(lambda row: haversine(row), axis=1)
df
Out[11]:
         SEAZ        LAT        LON     distance
index                                           
1      296.40  58.731221  28.377411  6275.791920
2      274.72  56.814832  31.292324  6509.727368
3      192.25  52.064988  35.801864  6990.144378
4       34.34  68.818875  67.193367  7357.221846
5      271.05  56.669988  31.688062  6538.047542
6      131.88  48.554622  49.782773  8036.968198
7      350.71  64.774272  31.395378  6229.733699
8      214.44  53.519292  33.845856  6801.670843
9        1.46  67.943374  38.484252  6418.754323
10     273.55  53.343731   4.471666  4935.394528

The following code is actually slower on such a small dataframe but I applied it to a 100,000 row df:

以下代码在如此小的数据帧上实际上较慢,但我将其应用于100,000行df:

In [35]:

%%timeit
df['LAT_rad'], df['LON_rad'] = np.radians(df['LAT']), np.radians(df['LON'])
df['dLON'] = df['LON_rad'] - math.radians(-56.7213600)
df['dLAT'] = df['LAT_rad'] - math.radians(37.2175900)
df['distance'] = 6367 * 2 * np.arcsin(np.sqrt(np.sin(df['dLAT']/2)**2 + math.cos(math.radians(37.2175900)) * np.cos(df['LAT_rad']) * np.sin(df['dLON']/2)**2))

1 loops, best of 3: 17.2 ms per loop

Compared to the apply function which took 4.3s so nearly 250 times quicker, something to note in the future

与应用功能相比,4.3s的速度提高了近2.5倍,将来需要注意

If we compress all the above in to a one-liner:

如果我们将上述所有内容压缩为单行:

In [39]:

%timeit df['distance'] = 6367 * 2 * np.arcsin(np.sqrt(np.sin((np.radians(df['LAT']) - math.radians(37.21759))/2)**2 + math.cos(math.radians(37.21759)) * np.cos(np.radians(df['LAT']) * np.sin((np.radians(df['LON']) - math.radians(-56.72136))/2)**2)))
100 loops, best of 3: 12.6 ms per loop

We observe further speed ups now a factor of ~341 times quicker.

我们观察到现在进一步加速的速度提高了约341倍。

#1