问题
I have this simple data.frame
lat<-c(1,2,3,10,11,12,20,21,22,23)
lon<-c(5,6,7,30,31,32,50,51,52,53)
data=data.frame(lat,lon)
The idea is to find the spatial clusters based on the distance
First, I plot the map (lon,lat) :
plot(data$lon,data$lat)
so clearly I have three clusters based in the distance between the position of points.
For this aim, I've tried this code in R :
d= as.matrix(dist(cbind(data$lon,data$lat))) #Creat distance matrix
d=ifelse(d<5,d,0) #keep only distance < 5
d=as.dist(d)
hc<-hclust(d) # hierarchical clustering
plot(hc)
data$clust <- cutree(hc,k=3) # cut the dendrogram to generate 3 clusters
This gives :
Now I try to plot the same points but with colors from clusters
plot(data$x,data$y, col=c("red","blue","green")[data$clust],pch=19)
Here the results
Which is not what I'm looking for.
Actually, I want to find something like this plot
Thank you for help.
回答1:
What about something like this:
lat<-c(1,2,3,10,11,12,20,21,22,23)
lon<-c(5,6,7,30,31,32,50,51,52,53)
km <- kmeans(cbind(lat, lon), centers = 3)
plot(lon, lat, col = km$cluster, pch = 20)
回答2:
Here's a different approach. First it assumes that the coordinates are WGS-84 and not UTM (flat). Then it clusters all neighbors within a given radius to the same cluster using hierarchical clustering (with method = single, which adopts a 'friends of friends' clustering strategy).
In order to compute the distance matrix, I'm using the rdist.earth method from the package fields. The default earth radius for this package is 6378.388 (the equatorial radius) which might not be what one is looking for, so I've changed it to 6371. See this article for more info.
library(fields)
lon = c(31.621785, 31.641773, 31.617269, 31.583895, 31.603284)
lat = c(30.901118, 31.245008, 31.163886, 30.25058, 30.262378)
threshold.in.km <- 40
coors <- data.frame(lon,lat)
#distance matrix
dist.in.km.matrix <- rdist.earth(coors,miles = F,R=6371)
#clustering
fit <- hclust(as.dist(dist.in.km.matrix), method = "single")
clusters <- cutree(fit,h = threshold.in.km)
plot(lon, lat, col = clusters, pch = 20)
This could be a good solution if you don't know the number of clusters (like the k-means option), and is somewhat related to the dbscan option with minPts = 1.
---EDIT---
With the original data:
lat<-c(1,2,3,10,11,12,20,21,22,23)
lon<-c(5,6,7,30,31,32,50,51,52,53)
data=data.frame(lat,lon)
dist <- rdist.earth(data,miles = F,R=6371) #dist <- dist(data) if data is UTM
fit <- hclust(as.dist(dist), method = "single")
clusters <- cutree(fit,h = 1000) #h = 2 if data is UTM
plot(lon, lat, col = clusters, pch = 20)
回答3:
As you have a spatial data to cluster, so DBSCAN is best suited for you data.
You can do this clustering using dbscan() function provided by fpc, a R package.
library(fpc)
lat<-c(1,2,3,10,11,12,20,21,22,23)
lon<-c(5,6,7,30,31,32,50,51,52,53)
DBSCAN <- dbscan(cbind(lat, lon), eps = 1.5, MinPts = 3)
plot(lon, lat, col = DBSCAN$cluster, pch = 20)
来源：https://stackoverflow.com/questions/28672399/spatial-clustering-in-r-simple-example