blocks|key|449312|text|是的，试试geopy。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|449313|import+geopy
import+geopy.distance

pt1+=+geopy.Point(48.853,+2.349)
pt2+=+geopy.Point(52.516,+13.378)

dist+=+geopy.distance.distance(pt1,+pt2).km
#+878.25|code-block|syntax|javascript|449314|之后，您可以查询您的点列表：|449315|[pt+for+pt+in+points+if+geopy.distance.distance(orig,+pt).km+<+5.]|449316|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/geopy/geopy^0|5|5|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

Yes, try <a href="https://github.com/geopy/geopy" rel="noreferrer">geopy</a>.

<pre><code>import geopy
import geopy.distance

pt1 = geopy.Point(48.853, 2.349)
pt2 = geopy.Point(52.516, 13.378)

dist = geopy.distance.distance(pt1, pt2).km
# 878.25
</code></pre>

afterwards you can query your lists of points:

<pre><code>[pt for pt in points if geopy.distance.distance(orig, pt).km &lt; 5.]
</code></pre>

blocks|key|449374|text|我知道这不是您想要的，但是您可以将GeoDjango与内存中的SQLite数据库一起使用。它是作为Web应用程序公开的一整套GIS工具，这使其成为快速开发GIS应用程序的瑞士军刀，特别是用于小型即席查询。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|449375|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.djangoproject.com/en/dev/ref/contrib/gis/^0|H|9|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

I know this isn't exactly what you meant, but you could use <a href="https://docs.djangoproject.com/en/dev/ref/contrib/gis/">GeoDjango</a> with an in-memory SQLite database. It's a full set of GIS tools exposed as a Web application, which makes it a Swiss Army knife for rapidly developing GIS applications, especially for small ad hoc queries.

blocks|key|663684|text|你的字典想法听起来并没有那么糟糕，尽管你也需要检查属于“相邻”字典关键字的点。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|663685|如果你找不到合适的工具，就像编码算法一样，你可以实现一个二进制空间分区树，afaik是实现类似事情的一种不那么麻烦的方法。|663686|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

Your dictionary idea doesn't sound that bad, though you will need to check points that fall under 'neighbouring' dictionary keys as well.

If you can't find the right tool, and like coding algorithms, you could implement a binary space partition tree which afaik is a less hacky way of achieving a similar thing.

blocks|key|449505|text|您可以使用具有Rtree扩展的SQLite来执行这种存储和查询。如果您的数据大于您要使用的内存，或者您希望在程序运行之间保存和操作数据，则此方法非常有用。实际的存储和查询代码是用C语言编写的，这意味着必须对其进行编译，但其好处是比纯Python解决方案(如geopy+)具有更高的性能。pysqlite或APSW都可以用于SQLite访问。(披露:我是APSW的作者。)|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|449506|entityMap|0|LINK|mutability|MUTABLE|url|http://www.sqlite.org/rtree.html^0|7|5|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

You can use SQLite which has an <a href="http://www.sqlite.org/rtree.html" rel="nofollow">Rtree</a> extension for doing exactly that kind of storage and querying. This approach is useful if your data is larger than the memory you want to use or you want to save and manipulate data between program runs. The actual storage and query code is in C which means it has to be compiled, but the benefit is extra performance over pure Python solutions like geopy. Either pysqlite or APSW will work for the SQLite access. (Disclosure: I am the APSW author.)

blocks|key|664003|text|我也有类似的问题，似乎使用SciPy的cKDTree进行快速最近点查找，结合使用GeoPy进行地理距离计算效果很好。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|664004|In+[1]:+import+numpy+as+np
In+[2]:+from+scipy.spatial+import+cKDTree
In+[3]:+from+geopy+import+Point,+distance
In+[4]:+points+=+np.random.sample((100000,+2))+*+180+-+90+++#+make+100k+random+lat-long+points
In+[5]:+index+=+cKDTree(points)
In+[6]:+%25time+lat_long_dist,+inds+=+index.query(points[234],+20)
CPU+times:+user+118+µs,+sys:+164+µs,+total:+282+µs
Wall+time:+248+µs
In+[7]:+points_geopy+=+[Point(*p)+for+p+in+points]
In+[8]:+%25time+geo_dists+=+[distance.great_circle(points_geopy[234],+points_geopy[i])+for+i+in+inds]
CPU+times:+user+244+µs,+sys:+218+µs,+total:+462+µs
Wall+time:+468+µs
In+[9]:+geo_dists
Out[9]:+
[Distance(0.0),
+Distance(29.661520907955524),
+...
+Distance(156.5471729956897),
+Distance(144.7528417712309)]|code-block|syntax|javascript|664005|要得到半径内的所有点，需要做一些额外的工作。|664006|我尝试了Shapely的STRtree，但性能要差得多(我是用pip+install+shapely[vectorized]安装的)。|style|CODE|664007|entityMap|0|LINK|mutability|MUTABLE|url|https://www.scipy.org/|1|https://github.com/geopy/geopy^0|D|5|0|14|5|1|0|0|0|V|V|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]|$A|12|B|13|1|14]]|C|$]]|$1|D|3|E|5|F|7|15|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|16|8|@]|9|@]|C|$]]|$1|K|3|L|5|6|7|17|8|@$A|18|B|19|M|N]]|9|@]|C|$]]|$1|O|3|-4|5|6|7|1A|8|@]|9|@]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]|W|$5|R|S|T|C|$U|X]]]]

I have a similar problem and it seems that using <a href="https://www.scipy.org/" rel="nofollow noreferrer">SciPy</a>'s cKDTree for fast nearest-points lookups together with <a href="https://github.com/geopy/geopy" rel="nofollow noreferrer">GeoPy</a> for geographic distance calculation works fine.

<pre><code>In [1]: import numpy as np
In [2]: from scipy.spatial import cKDTree
In [3]: from geopy import Point, distance
In [4]: points = np.random.sample((100000, 2)) * 180 - 90 # make 100k random lat-long points
In [5]: index = cKDTree(points)
In [6]: %time lat_long_dist, inds = index.query(points[234], 20)
CPU times: user 118 µs, sys: 164 µs, total: 282 µs
Wall time: 248 µs
In [7]: points_geopy = [Point(*p) for p in points]
In [8]: %time geo_dists = [distance.great_circle(points_geopy[234], points_geopy[i]) for i in inds]
CPU times: user 244 µs, sys: 218 µs, total: 462 µs
Wall time: 468 µs
In [9]: geo_dists
Out[9]: 
[Distance(0.0),
 Distance(29.661520907955524),
 ...
 Distance(156.5471729956897),
 Distance(144.7528417712309)]
</code></pre>

A bit of extra work is necessary to get all points within a radius.

I tried Shapely's STRtree, but got far worse performance (I installed with <code>pip install shapely[vectorized]</code>).

blocks|key|663730|text|你看过Shapely吗？它有一些方法可以查询距离内的对象。看一看Binary+Spatial+Predicates。这可能就是你要找的。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|663731|entityMap|0|LINK|mutability|MUTABLE|url|http://pypi.python.org/pypi/Shapely|1|http://gispython.org/shapely/docs/1.2/manual.html#binary-predicates^0|3|7|0|W|P|1|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@$A|O|B|P|1|Q]|$A|R|B|S|1|T]]|C|$]]|$1|D|3|-4|5|6|7|U|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]|L|$5|G|H|I|C|$J|M]]]]

Have you looked at <a href="http://pypi.python.org/pypi/Shapely" rel="nofollow">Shapely</a>? It has some methods to query objects within a distance. Take a look at the <a href="http://gispython.org/shapely/docs/1.2/manual.html#binary-predicates" rel="nofollow">Binary Spatial Predicates</a>. It might just be what you are looking for.

Is there a Python module where I can create objects with a geographical location coordinate (latitude and longitude), and query all the objects for ones which are within a 5km distance (i.e. radius) of a given coordinate?

I've been trying to store the latitude and longitude as keys in dictionaries (as they're indexed by key) and use some distance finding algorithms to query them. But this feels like a horrible hack.

Essentially something like PostGIS for PostgreSQL, but all within my Python app's memory.

Python module for storing and querying geographical coordinates

PostgreSQL

数据库

SQLite

Python

有没有一个Python模块，我可以在其中创建具有地理位置坐标(纬度和经度)的对象，并在所有对象中查询距离给定坐标5 5km (即半径)的对象？我一直在尝试将纬度和经度作为键存储在字典中(因为它们是通过键进行索引的)，并使用一些距离查找算法来查询它们。但这感觉像是一次可怕的黑客攻击。本质上类似于PostgreSQL的PostGIS，但都在我的Python应用程序的记忆中。

问用于存储和查询地理坐标的Python模块
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