DBSCAN for clustering data by location and density - r

I'm using the method dbscan::dbscan in order to cluster my data by location and density.
My data looks like this:
str(data)
'data.frame': 4872 obs. of 3 variables:
$ price : num ...
$ lat : num ...
$ lng : num ...
Now I'm using following code:
EPS = 7
cluster.dbscan <- dbscan(data, eps = EPS, minPts = 30, borderPoints = T,
search = "kdtree")
plot(lat ~ lng, data = data, col = cluster.dbscan$cluster + 1L, pch = 20)
but the result isn't satisfying at all, the point's aren't really clustered.
I would like to have the clusters nicely defined, something like this:
I also tried to use use a decision tree classifier tree:tree which works better, but I can't tell if it is really a good classification.
File:
http://www.file-upload.net/download-11246655/file.csv.html
Question:
is it possible to achieve what I want?
am I using the right method?
should I play more with the parameters? if yes, with which?

This is the output of a careful density-based clustering using the quite new HDBSCAN* algorithm.
Using Haversine distance, instead of Euclidean!
It identified some 50-something regions that are substantially more dense than their surroundings. In this figure, some clusters look as if they had only 3 elements, but they do have many more.
The outermost area, these are the noise points that do not belong to any cluster at all!
(Parameters used: -verbose -dbc.in file.csv -parser.labelIndices 0,1 -algorithm clustering.hierarchical.extraction.HDBSCANHierarchyExtraction -algorithm SLINKHDBSCANLinearMemory -algorithm.distancefunction geo.LatLngDistanceFunction -hdbscan.minPts 20 -hdbscan.minclsize 20)
OPTICS is another density-based algorithm, here is a result:
Again, we have a "noise" area with red dots are not dense at all.
Parameters used: -verbose -dbc.in file.csv -parser.labelIndices 0,1 -algorithm clustering.optics.OPTICSXi -opticsxi.xi 0.1 -algorithm.distancefunction geo.LatLngDistanceFunction -optics.minpts 25
The OPTICS plot for this data set looks like this:
You can see there are many small valleys that correspond to clusters. But there is no "large" structure here.
You probably were looking for a result like this:
But in fact, this is a meaningless and rather random way of breaking the data into large chunks. Sure, it minimizes variance; but it does not at all care about the structure of the data. Points within one cluster will frequently have less in common than points in different clusters. Just look at the points at the border between the red, orange, and violet clusters.
Last but not least, the oldtimers: hierarchical clustering with complete linkage:
and the dendrogram:
(Parameters used: -verbose -dbc.in file.csv -parser.labelIndices 0,1 -algorithm clustering.hierarchical.extraction.SimplifiedHierarchyExtraction -algorithm AnderbergHierarchicalClustering -algorithm.distancefunction geo.LatLngDistanceFunction -hierarchical.linkage CompleteLinkageMethod -hdbscan.minclsize 50)
Not too bad. Complete linkage works on such data rather well, too. But you could merge or split any of these clusters.

You can use something called as Hullplot
In your cases
hullplot(select(data, lng, lat), cluster.dbscan$cluster)

Related

DBSCAN Clustering returning single cluster with noise points

I am trying to perform DBSCAN clustering on the data https://www.kaggle.com/arjunbhasin2013/ccdata. I have cleaned the data and applied the algorithm.
data1 <- read.csv('C:\\Users\\write\\Documents\\R\\data\\Project\\Clustering\\CC GENERAL.csv')
head(data1)
data1 <- data1[,2:18]
dim(data1)
colnames(data1)
head(data1,2)
#to check if data has empty col or rows
library(purrr)
is_empty(data1)
#to check if data has duplicates
library(dplyr)
any(duplicated(data1))
#to check if data has NA values
any(is.na(data1))
data1 <- na.omit(data1)
any(is.na(data1))
dim(data1)
Algorithm was applied as follows.
#DBSCAN
data1 <- scale(data1)
library(fpc)
library(dbscan)
set.seed(500)
#to find optimal eps
kNNdistplot(data1, k = 34)
abline(h = 4, lty = 3)
The figure shows the 'knee' to identify the 'eps' value. Since there are 17 attributes to be considered for clustering, I have taken k=17*2 =34.
db <- dbscan(data1,eps = 4,minPts = 34)
db
The result I obtained is "The clustering contains 1 cluster(s) and 147 noise points."
No matter whatever values I change for eps and minPts the result is same.
Can anyone tell where I have gone wrong?
Thanks in advance.
You have two options:
Increase the radius of your center points (given by the epsilon parameter)
Decrease the minimum number of points (minPts) to define a center point.
I would start by decreasing the minPts parameter, since I think it is very high and since it does not find points within that radius, it does not group more points within a group
A typical problem with using DBSCAN (and clustering in general) is that real data typically does not fall into nice clusters, but forms one connected point cloud. In this case, DBSCAN will always find only a single cluster. You can check this with several methods. The most direct method would be to use a pairs plot (a scatterplot matrix):
plot(as.data.frame(data1))
Since you have many variables, the scatterplot pannels are very small, but you can see that the points are very close together in almost all pannels. DBSCAN will connect all points in these dense areas into a single cluster. k-means will just partition the dense area.
Another option is to check for clusterability with methods like VAT or iVAT (https://link.springer.com/chapter/10.1007/978-3-642-13657-3_5).
library("seriation")
## calculate distances for a small sample
d <- dist(data1[sample(seq(nrow(data1)), size = 1000), ])
iVAT(d)
You will see that the plot shows no block structure around the diagonal indicating that clustering will not find much.
To improve clustering, you need to work on the data. You can remove irrelevant variables, you may have very skewed variables that should be transformed first. You could also try non-linear embedding before clustering.

R - DBSCAN fviz_cluster - get coordinates of elements with dim1 and dim2

I'm a noob with R, and I'm trying to do clustering on some data samples.
I've tried a PCA,
res.pca <- PCA(df,
ncp = 5, # nb composantes principales.
graph = TRUE,
)
and I can get the full elements list with new coordinates using
res.pca$ind
This is great and works perfectly
for info using the 2 first axis with the PCA, I've 80% of variability on one axis and a bit more than 10% on the Second axis. I was quite proud of the result considering that I've 30 variables ... and in the End the PCA implicitly says that 2 dimension will be enough.
Still working on those data I tried the DBSCAN Clustering method fpc::dbscan :
library (factoextra)
db <- fpc::dbscan(df, eps = 22, MinPts = 3)
and after doing the dbscan and graphing the clusters using fviz_cluster, the Two dimensions display says : 92.8% on axis 1 and 6.7% on axis 2!!!! (more than 99% of the total variance explained with 2 axis !
In short, the DBSCAN has transformed my 30 variables data in a way that looks to be better than the PCA. The overall clustering of DBSCAN is rubbish for my data, but the transformation that has been used is absolutely excellent.
My issue is that I would like to get access to those new coordinates ... but no way at this time...
the only accessible variables I can see are :
db$cluster, db$eps, db$Minpts, db$isseed
BUT I suspect that some data are accessible otherwize how fviz_cluster, could present the data.
Any Idea ?
The projection is not done by dbscan. fviz_cluster uses the first two components obtained via stats::prcomp on the data.

plot result of kmeans clustering in r

I have a dataset containing 1599 observations and 10 attributes on which iIneed to do kmeans clustering. I have done the kmeans with 6 clusters and I can see the cluster centers, size, etc. and which observation lies in which cluster. Now, I need to plot these results such that I have in a single plot the following information: On x-axis, I want 1 of the 10 attributes of my original data, on y-axis I want another attribute and in the plot, I want all 1599 observations, but I want them in 6 different colors for each cluster they belong. So, I will have 10C2 = 45 plots. Basically, this should give me the information that cluster 1 is high/medium/low in terms of a particular attribute while cluster 2 is so and so.....for all 6 clusters.
I tried the function plotcluster from fpc package but from what I understood, it maps the data into 2D, using PCA, and then plots the clusters in terms of 2 dimensions which are different from the original attributes. So now when I will say cluster 1 is low, in dim1, it wouldn't really make much sense.
Is there a function to do what I want, or should I just append the '$cluster' information from the kmeans output with my original data and try to plot taking 2 columns from my data at a time using the basic function plot()?
I suggest one solution, probably not the simplest one (with a for loop) but it seems to answer what you need:
df=mtcars
df$cluster = factor( kmeans(df, centers=6)$clust )
mycomb <- combn(1:ncol(df), 2)
for (xy in 1:45 ) {
plot(x=df[, mycomb[1,xy]],
y=df[, mycomb[2,xy]],
col=as.numeric(df$clust),
xlab=names(df)[mycomb[1,xy]],
ylab=names(df)[mycomb[2,xy]])
}

options to allow heavily-weighted points on a map to overwhelm other points with low weights

what are some good kriging/interpolation idea/options that will allow heavily-weighted points to bleed over lightly-weighted points on a plotted R map?
the state of connecticut has eight counties. i found the centroid and want to plot poverty rates of each of these eight counties. three of the counties are very populated (about 1 million people) and the other five counties are sparsely populated (about 100,000 people). since the three densely-populated counties have more than 90% of the total state population, i would like those the three densely-populated counties to completely "overwhelm" the map and impact other points across the county borders.
the Krig function in the R fields package has a lot of parameters and also covariance functions that can be called, but i'm not sure where to start?
here is reproducible code to quickly produce a hard-bordered map and then three differently-weighted maps. hopefully i can just make changes to this code, but perhaps it requires something more complex like the geoRglm package? two of the three weighted maps look almost identical, despite one being 10x as weighted as the other..
https://raw.githubusercontent.com/davidbrae/swmap/master/20141001%20how%20to%20modify%20the%20Krig%20function%20so%20a%20huge%20weight%20overwhelms%20nearby%20points.R
thanks!!
edit: here's a picture example of the behavior i want-
disclaimer - I am not an expert on Krigging. Krigging is complex and takes a good understanding of the underlying data, the method and the purpose to achieve the correct result. You may wish to try to get input from #whuber [on the GIS Stack Exchange or contact him through his website (http://www.quantdec.com/quals/quals.htm)] or another expert you know.
That said, if you just want to achieve the visual effect you requested and are not using this for some sort of statistical analysis, I think there are some relatively simple solutions.
EDIT:
As you commented, though the suggestions below to use theta and smoothness arguments do even out the prediction surface, they apply equally to all measurements and thus do not extend the "sphere of influence" of more densely populated counties relative to less-densely populated. After further consideration, I think there are two ways to achieve this: by altering the covariance function to depend on population density or by using weights, as you have. Your weighting approach, as I wrote below, alters the error term of the krigging function. That is, it inversely scales the nugget variance.
As you can see in the semivariogram image, the nugget is essentially the y-intercept, or the error between measurements at the same location. Weights affect the nugget variance (sigma2) as sigma2/weight. Thus, greater weights mean less error at small-scale distances. This does not, however, change the shape of the semivariance function or have much effect on the range or sill.
I think that the best solution would be to have your covariance function depend on population. however, I'm not sure how to accomplish that and I don't see any arguments to Krig to do so. I tried playing with defining my own covariance function as in the Krig example, but only got errors.
Sorry I couldn't help more!
Another great resource to help understand Krigging is: http://www.epa.gov/airtrends/specialstudies/dsisurfaces.pdf
As I said in my comment, the sill and nugget values as well as the range of the semivariogram are things you can alter to affect the smoothing. By specifying weights in the call to Krig, you are altering the variance of the measurement errors. That is, in a normal use, weights are expected to be proportional to the accuracy of the measurement value so that higher weights represent more accurate measurements, essentially. This isn't actually true with your data, but it may be giving you the effect you desire.
To alter the way your data is interpolated, you can adjust two (and many more) parameters in the simple Krig call you are using: theta and smoothness. theta adjusts the semivariance range, meaning that measured points farther away contribute more to the estimates as you increase theta. Your data range is
range <- data.frame(lon=range(ct.data$lon),lat=range(ct.data$lat))
range[2,]-range[1,]
lon lat
2 1.383717 0.6300484
so, your measurement points vary by ~1.4 degrees lon and ~0.6 degrees lat. Thus, you can play with specifying your theta value in that range to see how that affects your result. In general, a larger theta leads to more smoothing since you are drawing from more values for each prediction.
Krig.output.wt <- Krig( cbind(ct.data$lon,ct.data$lat) , ct.data$county.poverty.rate ,
weights=c( size , 1 , 1 , 1 , 1 , size , size , 1 ),Covariance="Matern", theta=.8)
r <- interpolate(ras, Krig.output.wt)
r <- mask(r, ct.map)
plot(r, col=colRamp(100) ,axes=FALSE,legend=FALSE)
title(main="Theta = 0.8", outer = FALSE)
points(cbind(ct.data$lon,ct.data$lat))
text(ct.data$lon, ct.data$lat-0.05, ct.data$NAME, cex=0.5)
Gives:
Krig.output.wt <- Krig( cbind(ct.data$lon,ct.data$lat) , ct.data$county.poverty.rate ,
weights=c( size , 1 , 1 , 1 , 1 , size , size , 1 ),Covariance="Matern", theta=1.6)
r <- interpolate(ras, Krig.output.wt)
r <- mask(r, ct.map)
plot(r, col=colRamp(100) ,axes=FALSE,legend=FALSE)
title(main="Theta = 1.6", outer = FALSE)
points(cbind(ct.data$lon,ct.data$lat))
text(ct.data$lon, ct.data$lat-0.05, ct.data$NAME, cex=0.5)
Gives:
Adding the smoothness argument, will change the order of the function used to smooth your predictions. The default is 0.5 leading to a second-order polynomial.
Krig.output.wt <- Krig( cbind(ct.data$lon,ct.data$lat) , ct.data$county.poverty.rate ,
weights=c( size , 1 , 1 , 1 , 1 , size , size , 1 ),
Covariance="Matern", smoothness = 0.6)
r <- interpolate(ras, Krig.output.wt)
r <- mask(r, ct.map)
plot(r, col=colRamp(100) ,axes=FALSE,legend=FALSE)
title(main="Theta unspecified; Smoothness = 0.6", outer = FALSE)
points(cbind(ct.data$lon,ct.data$lat))
text(ct.data$lon, ct.data$lat-0.05, ct.data$NAME, cex=0.5)
Gives:
This should give you a start and some options, but you should look at the manual for fields. It is pretty well-written and explains the arguments well.
Also, if this is in any way quantitative, I would highly recommend talking to someone with significant spatial statistics know how!
Kriging is not what you want. (It is a statistical method for accurate--not distorted!--interpolation of data. It requires preliminary analysis of the data--of which you do not have anywhere near enough for this purpose--and cannot accomplish the desired map distortion.)
The example and the references to "bleed over" suggest considering an anamorph or area cartogram. This is a map which will expand and shrink the areas of the county polygons so that they reflect their relative population while retaining their shapes. The link (to the SE GIS site) explains and illustrates this idea. Although its answers are less than satisfying, a search of that site will reveal some effective solutions.
lot's of interesting comments and leads above.
I took a look at the Harvard dialect survey to get a sense for what you are trying to do first. I must say really cool maps. And before I start in on what I came up with...I've looked at your work on survey analysis before and have learned quite a few tricks. Thanks.
So my first take pretty quickly was that if you wanted to do spatial smoothing by way of kernel density estimation then you need to be thinking in terms of point process models. I'm sure there are other ways, but that's where I went.
So what I do below is grab a very generic US map and convert it into something I can use as a sampling window. Then I create random samples of points within that region, just pretend those are your centroids. After I attach random values to those points and plot it up.
I just wanted to test this conceptually, which is why I didn't go through the extra steps to grab cbsa's and also sorry for not projecting, but I think these are the fundamentals. Oh and the smoothing in the dialect study is being done over the whole country. I think. That is the author is not stratifying his smoothing procedure within polygons....so I just added states at the end.
code:
library(sp)
library(spatstat)
library(RColorBrewer)
library(maps)
library(maptools)
# grab us map from R maps package
usMap <- map("usa")
usIds <- usMap$names
# convert to spatial polygons so this can be used as a windo below
usMapPoly <- map2SpatialPolygons(usMap,IDs=usIds)
# just select us with no islands
usMapPoly <- usMapPoly[names(usMapPoly)=="main",]
# create a random sample of points on which to smooth over within the map
pts <- spsample(usMapPoly, n=250, type='random')
# just for a quick check of the map and sampling locations
plot(usMapPoly)
points(pts)
# create values associated with points, be sure to play aroud with
# these after you get the map it's fun
vals <-rnorm(250,100,25)
valWeights <- vals/sum(vals)
ptsCords <- data.frame(pts#coords)
# create window for the point pattern object (ppp) created below
usWindow <- as.owin(usMapPoly)
# create spatial point pattern object
usPPP <- ppp(ptsCords$x,ptsCords$y,marks=vals,window=usWindow)
# create colour ramp
col <- colorRampPalette(brewer.pal(9,"Reds"))(20)
# the plots, here is where the gausian kernal density estimation magic happens
# if you want a continuous legend on one of the sides get rid of ribbon=FALSE
# and be sure to play around with sigma
plot(Smooth(usPPP,sigma=3,weights=valWeights),col=col,main=NA,ribbon=FALSE)
map("state",add=TRUE,fill=FALSE)
example no weights:
example with my trivial weights
There is obviously a lot of work in between this and your goal of making this type of map reproducible at various levels of spatial aggregation and sample data, but good luck it seems like a cool project.
p.s. initially I did not use any weighting, but I suppose you could provide weights directly to the Smooth function. Two example maps above.

How to generate medoid plots

Hi I am using partitioning around medoids algorithm for clustering using the pam function in clustering package. I have 4 attributes in the dataset that I clustered and they seem to give me around 6 clusters and I want to generate a a plot of these clusters across those 4 attributes like this 1: http://www.flickr.com/photos/52099123#N06/7036003411/in/photostream/lightbox/ "Centroid plot"
But the only way I can draw the clustering result is either using a dendrogram or using
plot (data, col = result$clustering) command which seems to generate a plot similar to this
[2] : http://www.flickr.com/photos/52099123#N06/7036003777/in/photostream "pam results".
Although the first image is a centroid plot I am wondering if there are any tools available in R to do the same with a medoid plot Note that it also prints the size of each cluster in the plot. It would be great to know if there are any packages/solutions available in R that facilitate to do this or if not what should be a good starting point in order to achieve plots similar to that in Image 1.
Thanks
Hi All,I was trying to work out the problem the way Joran told but I think I did not understand it correctly and have not done it the right way as it is supposed to be done. Anyway this is what I have done so far. Following is how the file looks like that I tried to cluster
geneID RPKM-base RPKM-1cm RPKM+4cm RPKMtip
GRMZM2G181227 3.412444267 3.16437442 1.287909035 0.037320722
GRMZM2G146885 14.17287135 11.3577013 2.778514642 2.226818648
GRMZM2G139463 6.866752401 5.373925806 1.388843962 1.062745344
GRMZM2G015295 1349.446347 447.4635291 29.43627879 29.2643755
GRMZM2G111909 47.95903081 27.5256729 1.656555758 0.949824883
GRMZM2G078097 4.433627458 0.928492841 0.063329249 0.034255945
GRMZM2G450498 36.15941083 9.45235616 0.700105077 0.194759794
GRMZM2G413652 25.06985426 15.91342458 5.372151214 3.618914949
GRMZM2G090087 21.00891969 18.02318412 17.49531186 10.74302155
following is the Pam clustering output
GRMZM2G181227
1
GRMZM2G146885
2
GRMZM2G139463
2
GRMZM2G015295
2
GRMZM2G111909
2
GRMZM2G078097
3
GRMZM2G450498
3
GRMZM2G413652
2
GRMZM2G090087
2
AC217811.3_FG003
2
Using the above two files I generated a third file that somewhat looks like this and has cluster information in the form of cluster type K1,K2,etc
geneID RPKM-base RPKM-1cm RPKM+4cm RPKMtip Cluster_type
GRMZM2G181227 3.412444267 3.16437442 1.287909035 0.037320722 K1
GRMZM2G146885 14.17287135 11.3577013 2.778514642 2.226818648 K2
GRMZM2G139463 6.866752401 5.373925806 1.388843962 1.062745344 K2
GRMZM2G015295 1349.446347 447.4635291 29.43627879 29.2643755 K2
GRMZM2G111909 47.95903081 27.5256729 1.656555758 0.949824883 K2
GRMZM2G078097 4.433627458 0.928492841 0.063329249 0.034255945 K3
GRMZM2G450498 36.15941083 9.45235616 0.700105077 0.194759794 K3
GRMZM2G413652 25.06985426 15.91342458 5.372151214 3.618914949 K2
GRMZM2G090087 21.00891969 18.02318412 17.49531186 10.74302155 K2
I certainly don't think that this is the file that joran would have wanted me to create but I could not think of anything else thus I ran lattice on the above file using the following code.
clusres<- read.table("clusinput.txt",header=TRUE,sep="\t");
jpeg(filename = "clusplot.jpeg", width = 800, height = 1078,
pointsize = 12, quality = 100, bg = "white",res=100);
parallel(~clusres[2:5]|Cluster_type,clusres,horizontal.axis=FALSE);
dev.off();
and I get a picture like this
Since I want one single line as the representative of the whole cluster at four different points this output is wrong moreover I tried playing with lattice but I can not figure out how to make it accept the Rpkm values as the X coordinate It always seems to plot so many lines against a maximum or minimum value at the Y coordinate which I don't understand what it is.
It will be great if anybody can help me out. Sorry If my question still seems absurd to you.
I do not know of any pre-built functions that generate the plot you indicate, which looks to me like a sort of parallel coordinates plot.
But generating such a plot would be a fairly trivial exercise.
Add a column of cluster labels (K1,K2, etc.) to your original data set, based on your clustering algorithm's output.
Use one of the many, many tools in R for aggregating data (plyr, aggregate, etc.) to calculate the relevant summary statistics by cluster on each of the four variables. (You haven't said what the first graph is actually plotting. Mean and sd? Median and MAD?)
Since you want the plots split into six separate panels, or facets, you will probably want to plot the data using either ggplot or lattice, both of which provide excellent support for creating the same plot, split across a single grouping vector (i.e. the clusters in your case).
But that's about as specific as anyone can get, given that you've provided so little information (i.e. no minimal runnable example, as recommended here).
How about using clusplot from package cluster with partitioning around medoids? Here is a simple example (from the example section):
require(cluster)
#generate 25 objects, divided into 2 clusters.
x <- rbind(cbind(rnorm(10,0,0.5), rnorm(10,0,0.5)),
cbind(rnorm(15,5,0.5), rnorm(15,5,0.5)))
clusplot(pam(x, 2)) #`pam` does you partitioning

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