Python: speeding up geographic comparison

Wilduck picture Wilduck · Jul 11, 2011 · Viewed 7.3k times · Source

I've written some code that includes a nested loop where the inner loop is executed about 1.5 million times. I have a function in this loop that I'm trying to optimize. I've done some work, and got some results, but I need a little input to check if what I'm doing is sensible.

Some background:

I have two collections of geographic points (latitude, longitude), one relatively small collection and one relatively huge collection. For every point in the small collection, I need to find the closest point in the large collection.

The obvious way to do this would be to use the haversine formula. The benefit here is that the distances are definitely accurate.

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

def haversine(point1, point2):
    """Gives the distance between two points on earth.
    """
    earth_radius_miles = 3956
    lat1, lon1 = (radians(coord) for coord in point1)
    lat2, lon2 = (radians(coord) for coord in point2)
    dlat, dlon = (lat2 - lat1, lon2 - lon1)
    a = sin(dlat/2.0)**2 + cos(lat1) * cos(lat2) * sin(dlon/2.0)**2
    great_circle_distance = 2 * asin(min(1,sqrt(a)))
    d = earth_radius_miles * great_circle_distance
    return d

However, running this 1.5 million times takes about 9 seconds on my machine (according to timeit). Since having an accurate distance is unimportant, rather I only need to find the closest point, I decided to try some other functions.

A simple implementation of the pythagorean theorem gives me a speedup of about 30%. Thinking that I can do better, I wrote the following:

def dumb(point1, point2):
    lat1, lon1 = point1
    lat2, lon2 = point2
    d = abs((lat2 - lat1) + (lon2 - lon1))

which gives me a factor of 10 improvement. However, now I'm worried that this will not preserve the triangle inequality.

So, my final question is two fold: I'd like to have a function that runs as fast as dumb but still be correct. Will dumb work? If not, any suggestions on how to improve my haversine function?

Answer

jterrace picture jterrace · Jul 12, 2011

This is the kind of calculation that numpy is really good at. Rather than looping over the entire large set of coordinates, you can compute the distance between a single point and the entire dataset in a single calculation. With my tests below, you can get an order of magnitude speed increase.

Here's some timing tests with your haversine method, your dumb method (not really sure what that does) and my numpy haversine method. It computes the distance between two points - one in Virginia and one in California that are 2293 miles away.

from math import radians, sin, cos, asin, sqrt, pi, atan2
import numpy as np
import itertools

earth_radius_miles = 3956.0

def haversine(point1, point2):
    """Gives the distance between two points on earth.
    """
    lat1, lon1 = (radians(coord) for coord in point1)
    lat2, lon2 = (radians(coord) for coord in point2)
    dlat, dlon = (lat2 - lat1, lon2 - lon1)
    a = sin(dlat/2.0)**2 + cos(lat1) * cos(lat2) * sin(dlon/2.0)**2
    great_circle_distance = 2 * asin(min(1,sqrt(a)))
    d = earth_radius_miles * great_circle_distance
    return d

def dumb(point1, point2):
    lat1, lon1 = point1
    lat2, lon2 = point2
    d = abs((lat2 - lat1) + (lon2 - lon1))
    return d

def get_shortest_in(needle, haystack):
    """needle is a single (lat,long) tuple.
        haystack is a numpy array to find the point in
        that has the shortest distance to needle
    """
    dlat = np.radians(haystack[:,0]) - radians(needle[0])
    dlon = np.radians(haystack[:,1]) - radians(needle[1])
    a = np.square(np.sin(dlat/2.0)) + cos(radians(needle[0])) * np.cos(np.radians(haystack[:,0])) * np.square(np.sin(dlon/2.0))
    great_circle_distance = 2 * np.arcsin(np.minimum(np.sqrt(a), np.repeat(1, len(a))))
    d = earth_radius_miles * great_circle_distance
    return np.min(d)


x = (37.160316546736745, -78.75)
y = (39.095962936305476, -121.2890625)

def dohaversine():
    for i in xrange(100000):
        haversine(x,y)

def dodumb():
    for i in xrange(100000):
        dumb(x,y)

lots = np.array(list(itertools.repeat(y, 100000)))
def donumpy():
    get_shortest_in(x, lots)

from timeit import Timer
print 'haversine distance =', haversine(x,y), 'time =',
print Timer("dohaversine()", "from __main__ import dohaversine").timeit(100)
print 'dumb distance =', dumb(x,y), 'time =',
print Timer("dodumb()", "from __main__ import dodumb").timeit(100)
print 'numpy distance =', get_shortest_in(x, lots), 'time =',
print Timer("donumpy()", "from __main__ import donumpy").timeit(100)

And here's what it prints:

haversine distance = 2293.13242188 time = 44.2363960743
dumb distance = 40.6034161104 time = 5.58199882507
numpy distance = 2293.13242188 time = 1.54996609688

The numpy method takes 1.55 seconds to compute the same number of distance calculations as it takes 44.24 seconds to compute with your function method. You could probably get more of a speedup by combining some of the numpy functions into a single statement, but it would become a long, hard-to-read line.