"有效地从 Python 中的集合中找到最近的坐标对"

Question

The Problem

Imagine I am stood in an airport. Given a geographic coordinate pair, how can one efficiently determine which airport I am stood in?

Inputs

• A coordinate pair (x,y) representing the location I am stood at.
• A set of coordinate pairs [(a1,b1), (a2,b2)...] where each coordinate pair represents one airport.

Desired Output

A coordinate pair (a,b) from the set of airport coordinate pairs representing the closest airport to the point (x,y) .

Inefficient Solution

Here is my inefficient attempt at solving this problem. It is clearly linear in the length of the set of airports.

shortest_distance = None
shortest_distance_coordinates = None

point = (50.776435, -0.146834)

for airport in airports:
    distance = compute_distance(point, airport)
    if distance < shortest_distance or shortest_distance is None:
        shortest_distance = distance
        shortest_distance_coordinates = airport

The Question

How can this solution be improved? This might involve some way of pre-filtering the list of airports based on the coordinates of the location we are currently stood at, or sorting them in a certain order beforehand.

Answer 1

Using a k-dimensional tree:

>>> from scipy import spatial
>>> airports = [(10,10),(20,20),(30,30),(40,40)]
>>> tree = spatial.KDTree(airports)
>>> tree.query([(21,21)])
(array([ 1.41421356]), array([1]))

Where 1.41421356 is the distance between the queried point and the nearest neighbour and 1 is the index of the neighbour.

See: http://docs.scipy.org/doc/scipy/reference/generated/scipy.spatial.KDTree.query.html#scipy.spatial.KDTree.query

Answer 2

From this SO question :

import numpy as np
def closest_node(node, nodes):
    nodes = np.asarray(nodes)
    deltas = nodes - node
    dist_2 = np.einsum('ij,ij->i', deltas, deltas)
    return np.argmin(dist_2)

where node is a tuple with two values (x, y) and nodes is an array of tuples with two values ( [(x_1, y_1), (x_2, y_2),] )

Answer 3

If your coordinates are unsorted, your search can only be improved slightly assuming it is (latitude,longitude) by filtering on latitude first as for earth

1 degree of latitude on the sphere is 111.2 km or 69 miles

but that would not give a huge speedup.

If you sort the airports by latitude first then you can use a binary search for finding the first airport that could match ( airport_lat >= point_lat-tolerance ) and then only compare up to the last one that could match ( airport_lat <= point_lat+tolerance ) - but take care of 0 degrees equaling 360. While you cannot use that library directly, the sources of bisect are a good start for implementing a binary search.

While technically this way the search is still O(n), you have much fewer actual distance calculations (depending on tolerance) and few latitude comparisons. So you will have a huge speedup.

Answer 4

The answer of @Juddling is great, but KDTree does not support haversine distance, which is better suited for latitude/longitude coordinates. For the haversine distance you can use BallTree. Please note, that you need to convert your coordinates to radians first.

from math import radians
from sklearn.neighbors import BallTree
import numpy as np

airports = [(10,10),(20,20),(30,30),(40,40)]
airports_rad = np.array([[radians(x[0]), radians(x[1])] for x in airports ])
tree = BallTree(airports_rad , metric = 'haversine')
result = tree.query([(radians(21),radians(21))])
print(result)

gives

(array([[0.02391369]]), array([[1]], dtype=int64))

To convert the distance to meters you need to multiply by the earth radius (in meters).

earth_radius = 6371000 # meters in earth
print(result[0][0] * earth_radius)
[152354.11114795]