blocks|key|598331|text|也许是像这样的|type|unstyled|depth|inlineStyleRanges|entityRanges|data|598332|def+dist(lat1,lon1,lat2,lon2):
++++#just+return+normal+x,y+dist
++++return+sqrt((lat1-lat2)**2%2B(lon1-lon2)**2)

def+sitesDist(site1,site2):
++++#just+a+helper+to+shorted+list+comprehensioin+below
++++return+dist(site1.lat,site1.lon,site2.lat,site2.lon)
sites_dict+=+{}
threshhold_dist=5+#example+dist
for+site+in+sites:
++++#for+each+site+put+it+in+a+dictionarry+with+its+value+being+an+array+of+neighbors
++++sites_dict[site]+=+[x+for+x+in+sites+if+x+!=+site+and+sitesDist(site,x)+<+threshhold_dist]
print+"\n".join(sites_dict)|code-block|syntax|javascript|598333|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

maybe something like

<pre><code>def dist(lat1,lon1,lat2,lon2):
 #just return normal x,y dist
 return sqrt((lat1-lat2)**2+(lon1-lon2)**2)

def sitesDist(site1,site2):
 #just a helper to shorted list comprehensioin below
 return dist(site1.lat,site1.lon,site2.lat,site2.lon)
sites_dict = {}
threshhold_dist=5 #example dist
for site in sites:
 #for each site put it in a dictionarry with its value being an array of neighbors
 sites_dict[site] = [x for x in sites if x != site and sitesDist(site,x) &lt; threshhold_dist]
print "\n".join(sites_dict)
</code></pre>

blocks|key|382537|text|我经常这样做，首先创建一个距离矩阵，然后在它上运行集群。这是我的代码。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|382538|library(geosphere)
library(cluster)
clusteramounts+<-+10
distance.matrix+<-+(distm(points.to.group[,c("lon","lat")]))
clustersx+<-+as.hclust(agnes(distance.matrix,+diss+=+T))
points.to.group$group+<-+cutree(clustersx,+k=clusteramounts)|code-block|syntax|javascript|382539|我不确定它是否完全解决了你的问题。你可能想用不同的k进行测试，也许还会对第一个集群中的一些集群进行第二次聚类，以防它们太大，比如你在明尼苏达州有一个点，在加利福尼亚州有一千个点。有了points.to.group$group后，您可以通过查找每个组的max和min+lat来获得边界框。|382540|如果你想让X等于20，你在纽约有18分，在达拉斯有22分，你必须决定是想要一个小盒子和一个真正的大盒子(每个盒子20分)，还是让达拉斯盒子包括22分更好，或者你想把达拉斯的22分分成两组。在其中一些情况下，基于距离的聚类可能是很好的。但这当然取决于您为什么要对这些点进行分组。|382541|/Chris|382542|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

I'm doing this on a regular basis by first creating a distance matrix and then running clustering on it. Here is my code.

<pre><code>library(geosphere)
library(cluster)
clusteramounts &lt;- 10
distance.matrix &lt;- (distm(points.to.group[,c("lon","lat")]))
clustersx &lt;- as.hclust(agnes(distance.matrix, diss = T))
points.to.group$group &lt;- cutree(clustersx, k=clusteramounts)
</code></pre>

I'm not sure if it completely solves your problem. You might want to test with different k, and also perhaps do a second run of clustering of some of the first clusters in case they are too big, like if you have one point in Minnesota and a thousand in California.
When you have the points.to.group$group, you can get the bounding boxes by finding max and min lat lon per group.

If you want X to be 20, and you have 18 points in New York and 22 in Dallas, you must decide if you want one small and one really big box (20 points each), if it is better to have have the Dallas box include 22 points, or if you want to split the 22 points in Dallas to two groups. Clustering based on distance can be good in some of these cases. But it of course depend on why you want to group the points.

/Chris

blocks|key|382612|text|如果使用shapely，可以通过shapely几何体的.bounds属性扩展我的cluster_points+function以返回簇的边界框，例如：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|382613|clusterlist.append(cluster,+(poly.buffer(-b)).bounds)|code-block|syntax|javascript|382614|entityMap|0|LINK|mutability|MUTABLE|url|http://sptl.eu/2015/01/03/finding-geographical-clusters/^0|14|N|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

if you use shapely, you could extend my <a href="http://sptl.eu/2015/01/03/finding-geographical-clusters/" rel="nofollow">cluster_points function</a>
to return the bounding box of the cluster via the .bounds property of the shapely geometry , for example like this:

<pre><code>clusterlist.append(cluster, (poly.buffer(-b)).bounds)
</code></pre>

I have a R data.frame containing longitude, latitude which spans over the entire USA map. When X number of entries are all within a small geographic region of say a few degrees longitude &amp; a few degrees latitude, I want to be able to detect this and then have my program then return the coordinates for the geographic bounding box. Is there a Python or R CRAN package that already does this? If not, how would I go about ascertaining this information?

Detecting geographic clusters

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有一个包含经度和纬度的R data.frame，它跨越整个美国地图。当X个条目都在一个很小的地理区域内时，比如几度经度&几度纬度，我希望能够检测到这一点，然后让我的程序返回地理边界框的坐标。有没有Python或R CRAN包已经做到了这一点？如果不是，我该如何确定此信息？

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