Vectorised Haversine formula with a pandas dataframe

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

In [11]:

from numpy import cos, sin, arcsin, sqrt
from math import radians

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:

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

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

In [39]:

%timeit df['distance'] = 6367 * 2 * np.arcsin(np.sqrt(np.sin((np.radians(df['LAT']) - math.radians(37.2175900))/2)**2 + math.cos(math.radians(37.2175900)) * np.cos(np.radians(df['LAT'])) * np.sin((np.radians(df['LON']) - math.radians(-56.7213600))/2)**2))
100 loops, best of 3: 12.6 ms per loop

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