Vectorizing Haversine distance calculation in Python

I am trying to calculate a distance matrix for a long list of locations identified by Latitude & Longitude using the Haversine formula that takes two tuples of coordinate pairs to produce the distance:

def haversine(point1, point2, miles=False):
    """ Calculate the great-circle distance bewteen two points on the Earth surface.

    :input: two 2-tuples, containing the latitude and longitude of each point
    in decimal degrees.

    Example: haversine((45.7597, 4.8422), (48.8567, 2.3508))

    :output: Returns the distance bewteen the two points.
    The default unit is kilometers. Miles can be returned
    if the ``miles`` parameter is set to True.


I can calculate the distance between all points using a nested for loop as follows:


   id                      coordinates
0   1   (16.3457688674, 6.30354512503)
1   2    (12.494749307, 28.6263955635)
2   3    (27.794615136, 60.0324947881)
3   4   (44.4269923769, 110.114216113)
4   5  (-69.8540884125, 87.9468778773)

using a simple function:

distance = {}
def haver_loop(df):
    for i, point1 in df.iterrows():
        distance[i] = []
        for j, point2 in df.iterrows():
            distance[i].append(haversine(point1.coordinates, point2.coordinates))

    return pd.DataFrame.from_dict(distance, orient='index')

But this takes quite a while given the time complexity, running at around 20s for 500 points and I have a much longer list. This has me looking at vectorization, and I’ve come across numpy.vectorize ((docs), but can’t figure out how to apply it in this context.


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Method 1

From haversine's function definition, it looked pretty parallelizable. So, using one of the best tools for vectorization with NumPy aka broadcasting and replacing the math funcs with the NumPy equivalents ufuncs, here’s one vectorized solution –

# Get data as a Nx2 shaped NumPy array
data = np.array(df['coordinates'].tolist())

# Convert to radians
data = np.deg2rad(data)                     

# Extract col-1 and 2 as latitudes and longitudes
lat = data[:,0]                     
lng = data[:,1]         

# Elementwise differentiations for lattitudes & longitudes
diff_lat = lat[:,None] - lat
diff_lng = lng[:,None] - lng

# Finally Calculate haversine
d = np.sin(diff_lat/2)**2 + np.cos(lat[:,None])*np.cos(lat) * np.sin(diff_lng/2)**2
return 2 * 6371 * np.arcsin(np.sqrt(d))

Runtime tests –

The other np.vectorize based solution has shown some positive promise on performance improvement over the original code, so this section would compare the posted broadcasting based approach against that one.

Function definitions –

def vectotized_based(df):
    haver_vec = np.vectorize(haversine, otypes=[np.int16])
    return df.groupby('id').apply(lambda x: pd.Series(haver_vec(df.coordinates, x.coordinates)))

def broadcasting_based(df):
    data = np.array(df['coordinates'].tolist())
    data = np.deg2rad(data)                     
    lat = data[:,0]                     
    lng = data[:,1]         
    diff_lat = lat[:,None] - lat
    diff_lng = lng[:,None] - lng
    d = np.sin(diff_lat/2)**2 + np.cos(lat[:,None])*np.cos(lat) * np.sin(diff_lng/2)**2
    return 2 * 6371 * np.arcsin(np.sqrt(d))

Timings –

In [123]: # Input
     ...: length = 500
     ...: d1 = np.random.uniform(-90, 90, length)
     ...: d2 = np.random.uniform(-180, 180, length)
     ...: coords = tuple(zip(d1, d2))
     ...: df = pd.DataFrame({'id':np.arange(length), 'coordinates':coords})

In [124]: %timeit vectotized_based(df)
1 loops, best of 3: 1.12 s per loop

In [125]: %timeit broadcasting_based(df)
10 loops, best of 3: 68.7 ms per loop

Method 2

You would provide your function as an argument to np.vectorize(), and could then use it as an argument to pandas.groupby.apply as illustrated below:

haver_vec = np.vectorize(haversine, otypes=[np.int16])
distance = df.groupby('id').apply(lambda x: pd.Series(haver_vec(df.coordinates, x.coordinates)))

For instance, with sample data as follows:

length = 500
df = pd.DataFrame({'id':np.arange(length), 'coordinates':tuple(zip(np.random.uniform(-90, 90, length), np.random.uniform(-180, 180, length)))})

compare for 500 points:

def haver_vect(data):
    distance = data.groupby('id').apply(lambda x: pd.Series(haver_vec(data.coordinates, x.coordinates)))
    return distance

%timeit haver_loop(df): 1 loops, best of 3: 35.5 s per loop

%timeit haver_vect(df): 1 loops, best of 3: 593 ms per loop

Method 3

start by getting all combinations using itertools.product

 results= [(p1,p2,haversine(p1,p2))for p1,p2 in itertools.product(points,repeat=2)]

that said Im not sure how fast it will be this looks like it might be a duplicate of Python: speeding up geographic comparison

All methods was sourced from or, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0

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