Vectorized Haversine distance calculation in Python

Question

I am trying to compute a distance matrix for a long list of locations identified by latitude and longitude using the Haversine formula , which takes two tuples of coordinate pairs to create a 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 closed loop as follows:

data.head()

   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 lot of time, given the time complexity, and works for about 20 seconds at 500 points, and I have a much longer list. This led me to vectorization, and I came across numpy.vectorize( (docs) , but cannot figure out how to apply it in this context.

Answer 1

np.vectorize(), pandas.groupby.apply, :

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

, :

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

500 :

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

Answer 2

haversine function definition . , NumPy aka broadcasting NumPy ufuncs, -

# 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))

-

np.vectorize based solution , , .

-

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))

-

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

Answer 3

itertools.product

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

, Im , , Python: