问题
Background
I want to add a model manager function that filters a queryset based on the proximity to coordinates. I found this blog posting with code that is doing precisely what I want.
Code
The snippet below seems to make use of geopy functions that have since been removed. It coarsely narrows down the queryset by limiting the range of latitude and longitude.
# Prune down the set of all locations to something we can quickly check precisely
rough_distance = geopy.distance.arc_degrees(arcminutes=geopy.distance.nm(miles=distance)) * 2
queryset = queryset.filter(
latitude__range=(latitude - rough_distance, latitude + rough_distance),
longitude__range=(longitude - rough_distance, longitude + rough_distance)
)
Problem
Since some of the used geopy functions have been removed/moved, I'm trying to rewrite this stanza. However, I do not understand the calculations---barely passed geometry and my research has confused me more than actually helped me.
Can anyone help? I would greatly appreciate it.
回答1:
It looks like distance in miles is being converted to nautical miles, which are each equal to a minute of arc, which are 1/60th of an arc degree each. That value is then doubled, and then added and subtracted from a given latitude and longitude. These four values can be used to form a bounding box around the coordinates.
You can lookup any needed conversion factors on Wikipedia. There's also a relevant article there titled Horizontal position representation which discusses pros and cons of alternatives to longitude and latitude positioning which avoid some of their complexities. In other words, about the considerations involved with replacing latitude and longitude with another horizontal position representation in calculations.
回答2:
In case anybody else is looking at this now, since I tried to use geopy and just hit up against it, the modern equivalent of the rough_distance snippet above is:
import geopy
rough_distance = geopy.units.degrees(arcminutes=geopy.units.nautical(miles=1))
回答3:
The Earth is not a sphere, only approximately so. If you need a more accurate calculation, use pyproj. Then you can calculate the location based a reference ellipsoid (e.g. WGS84).
回答4:
martineau's answer is right on, in terms of what the snippet actually does, but it is important to note that 1 minute of arc represents very different distances depending on location. At the equator, the query covers the least axis aligned bounding box enclosing a circle of diameter distance, but off the equator, the bounding box does not completely contain that circle.
回答5:
This code from the blog is sloppy:
def near(self, latitude=None, longitude=None, distance=None):
if not (latitude and longitude and distance):
return []
If latitude == 0 (equator) or longitude == 0 (Greenwich meridian), it returns immediately. Should be if latitude is None or longitude is None .......
@TokenMacGuy's answer is an improvement, but:
(a) The whole idea of the "bounding box" is to avoid an SQL or similar query calculating a distance to all points that otherwise satisfy the query. With appropriate indexes, the query will execute much faster. It does this at the cost of leaving the client to (1) calculate the coordinates of the bounding box (2) calculate and check the precise distance for each result returned by the query.
If step 2 is omitted, you get errors, even at the equator. For example "find all pizza shops in a 5-mile radius" means you get answers up to 7.07 miles (that's sqrt(5*2 + 5*2)) away in the corners of the box.
Note that the code that you show seems to be arbitrarily doubling the radius. This would mean you get points 14.1 miles away.
(b) As @TokenMacGuy said, away from the equator, it gets worse. The bounding box so calculated does not include all points that you are interested in -- unless of course you are overkilling by doubling the radius.
(c) If the circle of interest includes either the North or South Pole, the calculation is horribly inexact, and needs adjusting. If the circle of interest is crossed by the 180-degree meridian (i.e. the International Date Line without the zigzags), the results are a nonsense; you need to detect this case and apply a 2-part query (one part for each side of the meridian).
For solutions for problems (b) and (c), see this article.
来源：https://stackoverflow.com/questions/4325182/conversion-of-miles-to-latitude-and-longitude-degrees-using-geopy