Find the nearest point in distance for all the points in the dataset - Python

Question

I have a dataset as follows,

Id     Latitude      longitude
1      25.42         55.47
2      25.39         55.47
3      24.48         54.38
4      24.51             


        

Answer 1

You can do this very efficiently by calling a library that implements smart algorithms for this, one example would be sklearn which has a NearestNeighbors method that does exactly this.

Example of the code modified to do this:

from sklearn.neighbors import NearestNeighbors
import numpy as np

def distance(p1, p2):
    """
    Calculate the great circle distance between two points
    on the earth (specified in decimal degrees)
    """
    lon1, lat1 = p1
    lon2, lat2 = p2
    # convert decimal degrees to radians
    lon1, lat1, lon2, lat2 = map(np.radians, [lon1, lat1, lon2, lat2])
    # haversine formula
    dlon = lon2 - lon1
    dlat = lat2 - lat1
    a = np.sin(dlat/2)**2 + np.cos(lat1) * np.cos(lat2) * np.sin(dlon/2)**2
    c = 2 * np.arcsin(np.sqrt(a))
    km = 6367 * c
    return km

points = [[25.42, 55.47],
          [25.39, 55.47],
          [24.48, 54.38],
          [24.51, 54.54]]

nbrs = NearestNeighbors(n_neighbors=2, metric=distance).fit(points)

distances, indices = nbrs.kneighbors(points)

result = distances[:, 1]

which gives

>>> result
array([  1.889697  ,   1.889697  ,  17.88530556,  17.88530556])

Answer 2

The brute force method of finding the nearest of N points to a given point is O(N) -- you'd have to check each point. In contrast, if the N points are stored in a KD-tree, then finding the nearest point is on average O(log(N)). There is also the additional one-time cost of building the KD-tree, which requires O(N) time.

If you need to repeat this process N times, then the brute force method is O(N**2) and the kd-tree method is O(N*log(N)). Thus, for large enough N, the KD-tree will beat the brute force method.

See here for more on nearest neighbor algorithms (including KD-tree).

---

Below (in the function using_kdtree) is a way to compute the great circle arclengths of nearest neighbors using scipy.spatial.kdtree.

scipy.spatial.kdtree uses the Euclidean distance between points, but there is a formula for converting Euclidean chord distances between points on a sphere to great circle arclength (given the radius of the sphere). So the idea is to convert the latitude/longitude data into cartesian coordinates, use a KDTree to find the nearest neighbors, and then apply the great circle distance formula to obtain the desired result.

---

Here are some benchmarks. Using N = 100, using_kdtree is 39x faster than the orig (brute force) method.

In [180]: %timeit using_kdtree(data)
100 loops, best of 3: 18.6 ms per loop

In [181]: %timeit using_sklearn(data)
1 loop, best of 3: 214 ms per loop

In [179]: %timeit orig(data)
1 loop, best of 3: 728 ms per loop

For N = 10000:

In [5]: %timeit using_kdtree(data)
1 loop, best of 3: 2.78 s per loop

In [6]: %timeit using_sklearn(data)
1 loop, best of 3: 1min 15s per loop

In [7]: %timeit orig(data)
# untested; too slow

Since using_kdtree is O(N log(N)) and orig is O(N**2), the factor by which using_kdtree is faster than orig will grow as N, the length of data, grows.

---

import numpy as np
import scipy.spatial as spatial
import pandas as pd
import sklearn.neighbors as neighbors
from math import radians, cos, sin, asin, sqrt

R = 6367

def using_kdtree(data):
    "Based on https://stackoverflow.com/q/43020919/190597"
    def dist_to_arclength(chord_length):
        """
        https://en.wikipedia.org/wiki/Great-circle_distance
        Convert Euclidean chord length to great circle arc length
        """
        central_angle = 2*np.arcsin(chord_length/(2.0*R)) 
        arclength = R*central_angle
        return arclength

    phi = np.deg2rad(data['Latitude'])
    theta = np.deg2rad(data['Longitude'])
    data['x'] = R * np.cos(phi) * np.cos(theta)
    data['y'] = R * np.cos(phi) * np.sin(theta)
    data['z'] = R * np.sin(phi)
    tree = spatial.KDTree(data[['x', 'y','z']])
    distance, index = tree.query(data[['x', 'y','z']], k=2)
    return dist_to_arclength(distance[:, 1])

def orig(data):
    def distance(lon1, lat1, lon2, lat2):
        """
        Calculate the great circle distance between two points 
        on the earth (specified in decimal degrees)
        """
        # convert decimal degrees to radians 
        lon1, lat1, lon2, lat2 = map(radians, [lon1, lat1, lon2, lat2])
        # haversine formula 
        dlon = lon2 - lon1 
        dlat = lat2 - lat1 
        a = sin(dlat/2.0)**2 + cos(lat1) * cos(lat2) * sin(dlon/2.0)**2
        c = 2 * asin(sqrt(a)) 
        km = R * c
        return km

    shortest_distance = []
    for i in range(len(data)):
        distance1 = []
        for j in range(len(data)):
            if i == j: continue
            distance1.append(distance(data['Longitude'][i], data['Latitude'][i], 
                                      data['Longitude'][j], data['Latitude'][j]))
        shortest_distance.append(min(distance1))
    return shortest_distance


def using_sklearn(data):
    """
    Based on https://stackoverflow.com/a/45127250/190597 (Jonas Adler)
    """
    def distance(p1, p2):
        """
        Calculate the great circle distance between two points
        on the earth (specified in decimal degrees)
        """
        lon1, lat1 = p1
        lon2, lat2 = p2
        # convert decimal degrees to radians
        lon1, lat1, lon2, lat2 = map(np.radians, [lon1, lat1, lon2, lat2])
        # haversine formula
        dlon = lon2 - lon1
        dlat = lat2 - lat1
        a = np.sin(dlat/2)**2 + np.cos(lat1) * np.cos(lat2) * np.sin(dlon/2)**2
        c = 2 * np.arcsin(np.sqrt(a))
        km = R * c
        return km
    points = data[['Longitude', 'Latitude']]
    nbrs = neighbors.NearestNeighbors(n_neighbors=2, metric=distance).fit(points)
    distances, indices = nbrs.kneighbors(points)
    result = distances[:, 1]
    return result

np.random.seed(2017)
N = 1000
data = pd.DataFrame({'Latitude':np.random.uniform(-90,90,size=N), 
                     'Longitude':np.random.uniform(0,360,size=N)})

expected = orig(data)
for func in [using_kdtree, using_sklearn]:
    result = func(data)
    assert np.allclose(expected, result)

Answer 3

You can use a dictionary to hash some calculations. Your code calculates the distance A to B many times (A and B being 2 arbitrary points in your dataset).

Either implement your own cache:

from math import radians, cos, sin, asin, sqrt

dist_cache = {}
def distance(lon1, lat1, lon2, lat2):
    """
    Calculate the great circle distance between two points 
    on the earth (specified in decimal degrees)
    """

    try:
        return dist_cache[(lon1, lat1, lon2, lat2)]
    except KeyError:
        # convert decimal degrees to radians 
        lon1, lat1, lon2, lat2 = map(radians, [lon1, lat1, lon2, lat2])
        # haversine formula 
        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
        dist_cache[(lon1, lat1, lon2, lat2)] = km
        return km

Or use lru_cache:

from math import radians, cos, sin, asin, sqrt
from functools import lru_cache

@lru_cache
def distance(lon1, lat1, lon2, lat2):
    """
    Calculate the great circle distance between two points 
    on the earth (specified in decimal degrees)
    """
    # convert decimal degrees to radians 
    lon1, lat1, lon2, lat2 = map(radians, [lon1, lat1, lon2, lat2])
    # haversine formula 
    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