Re: [R] Spatial: number of independent components?
I was missing the spam library.
I did some testing with m x m matrices (see below). Computing 'a' is the
villain. The computation time for 'a' is exponential in m. For a 100 by 100
matrix, the predicted time is about 20 seconds. Thus, 100,000 runs, would
take about 23 days.
library(igraph)
library(spdep)
library(spam)
d=5:40
tim1=NULL
tim2=NULL
tim3=NULL
for(i in 1:length(d)){
x=rbinom(d[i]^2,1,0.5)
dim(x)=c(d[i],d[i])
tim1[i]=system.time(
a <- as.spam.listw(nb2listw(cell2nb(nrow(x),ncol(x),torus=T),
style="B"))
)[3]
tim2[i]=system.time(
g <- delete.vertices(graph.adjacency(a), which(x!=1)-1)
)[3]
tim3[i]=system.time(
clusters(g)
)[3]
}
reg=lm(log(tim1)~log(d))
par(mfcol=c(1,2))
plot(log(tim1)~log(d))
lines(x=log(d),y=coef(reg)[1]+coef(reg)[2]*log(d),type="l")
plot(tim1~d)
lines(x=d,y=exp(coef(reg)[1]+coef(reg)[2]*log(d)),type="l")
#estimated time for a 100 by 100 matrix
exp(predict(reg,newdata=data.frame(d=100)))
#observed time for a 100 by 100 matrix
d=100
x=rbinom(d^2,1,0.5)
dim(x)=c(d,d)
system.time(a <- as.spam.listw(nb2listw(cell2nb(nrow(x),ncol(x),torus=T),
style="B")))
Best,
Daniel
--
View this message in context:
http://r.789695.n4.nabble.com/Spatial-number-of-independent-components-tp2262018p2263168.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Spatial: number of independent components?
I just updated spdep and I see that as.spam.listw works. Below is sessionInfo Furthermore, it may be straightforward to condense the adjacency matrix *before* converting to graph which may help a little bit. You can profile the code and see which part needs speeding up. library(spdep) library(igraph) x=matrix(c(0,1,0,0,0, 0,1,1,0,0, 0,0,0,0,0, 0,0,0,1,0, 0,0,0,1,0),nrow=5) a <- as.spam.listw(nb2listw(cell2nb(nrow(x),ncol(x),torus=T), style="B")) ind <- which(x>0) b <- a[ind, ind] g1 <- graph.adjacency(b) clusters(g1)$no --- R version 2.11.1 (2010-05-31) i386-apple-darwin9.8.0 locale: [1] en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8 attached base packages: [1] stats graphics grDevices utils [5] datasets methods base other attached packages: [1] igraph_0.5.3 spam_0.22-0 [3] spdep_0.5-11 coda_0.13-5 [5] deldir_0.0-12 maptools_0.7-34 [7] foreign_0.8-40 nlme_3.1-96 [9] MASS_7.3-6 Matrix_0.999375-39 [11] lattice_0.18-8 boot_1.2-42 [13] sp_0.9-64 loaded via a namespace (and not attached): [1] grid_2.11.1 tools_2.11.1 Nikhil Kaza Asst. Professor, City and Regional Planning University of North Carolina nikhil.l...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Spatial: number of independent components?
I have spdep 4.58. Perhaps it is deprecated in the new version. Try looking for sparse matrix representation in the help files for spdep Nikhil On Mon, Jun 21, 2010 at 6:10 AM, Daniel Malter wrote: > > as.spam.listw is an unknown function. Is it in a different package? > > Daniel > > other attached packages: > [1] spdep_0.5-11coda_0.13-5 deldir_0.0-12 > maptools_0.7-34 foreign_0.8-38 nlme_3.1-96 MASS_7.3-3 > [8] Matrix_0.999375-31 lattice_0.17-26 boot_1.2-41 sp_0.9-64 > igraph_0.5.3RandomFields_1.3.41 svSocket_0.9-48 > [15] TinnR_1.0.3 R2HTML_1.59-1 Hmisc_3.7-0 > survival_2.35-7 > > loaded via a namespace (and not attached): > [1] cluster_1.12.1 grid_2.10.0svMisc_0.9-56 tools_2.10.0 > > -- > View this message in context: > http://r.789695.n4.nabble.com/Spatial-number-of-independent-components-tp2262018p2262422.html > Sent from the R help mailing list archive at Nabble.com. > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Spatial: number of independent components?
as.spam.listw is an unknown function. Is it in a different package? Daniel other attached packages: [1] spdep_0.5-11coda_0.13-5 deldir_0.0-12 maptools_0.7-34 foreign_0.8-38 nlme_3.1-96 MASS_7.3-3 [8] Matrix_0.999375-31 lattice_0.17-26 boot_1.2-41 sp_0.9-64 igraph_0.5.3RandomFields_1.3.41 svSocket_0.9-48 [15] TinnR_1.0.3 R2HTML_1.59-1 Hmisc_3.7-0 survival_2.35-7 loaded via a namespace (and not attached): [1] cluster_1.12.1 grid_2.10.0svMisc_0.9-56 tools_2.10.0 -- View this message in context: http://r.789695.n4.nabble.com/Spatial-number-of-independent-components-tp2262018p2262422.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Spatial: number of independent components?
Instead of nb2mat try as.spam.listw(nb2listw(cell2nb(...))) this will coerce the adjacency matrix into a sparse matrix representation saving lot of memory. Nikhil On Sun, Jun 20, 2010 at 10:27 PM, Daniel Malter wrote: > > Hi, thanks much. This works in principle. The corrected code is below: > > a <- nb2mat(cell2nb(nrow(x),ncol(x),torus=T), style="B") > g <- delete.vertices(graph.adjacency(a), which(x!=1)-1) > > plot(g) > clusters(g) > > the $no argument of the clusters(g) function is the sought after number. > However, the function is very slow, and my machine runs out of memory (1G) > for a 101 by 101 matrix. > > Daniel > -- > View this message in context: > http://r.789695.n4.nabble.com/Spatial-number-of-independent-components-tp2262018p2262090.html > Sent from the R help mailing list archive at Nabble.com. > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Spatial: number of independent components?
Hi, thanks much. This works in principle. The corrected code is below: a <- nb2mat(cell2nb(nrow(x),ncol(x),torus=T), style="B") g <- delete.vertices(graph.adjacency(a), which(x!=1)-1) plot(g) clusters(g) the $no argument of the clusters(g) function is the sought after number. However, the function is very slow, and my machine runs out of memory (1G) for a 101 by 101 matrix. Daniel -- View this message in context: http://r.789695.n4.nabble.com/Spatial-number-of-independent-components-tp2262018p2262090.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Spatial: number of independent components?
I am sure this can be written in a much more elegant and faster code. One way I can think of, is to modify cell2nb code and create a new function that creates the neighbour lists of only cells that are not 0. These are best directed to R-sig-Geo list. However, the following might work. library(spdep) library(igraph) x=matrix(c(0,1,0,0,0, 0,1,1,0,0, 0,0,0,0,0, 0,0,0,1,0, 0,0,0,1,0),nrow=5) a <- nb2mat(cell2nb(nrow(x),ncol(x)), style="B", torus="TRUE") g <- delete.vertices(graph.adjacency(a), which(x!=1)-1) plot(g) clusters(g) Nikhil --- Nikhil Kaza Asst. Professor, City and Regional Planning University of North Carolina nikhil.l...@gmail.com On Jun 20, 2010, at 7:17 PM, Daniel Malter wrote: Hi all, I am sorry if this is a very basic quesion, but I have no experience with analyzing spatial data and could not find the right function/ package quickly. Any hints would be much appreciated. I have a matrix of spatial point patterns like the one below and want to find the number of independent components (if that's the right term) in that matrix (or in that image). x=matrix(c(0,1,0,0,0, 0,1,1,0,0, 0,0,0,0,0, 0,0,0,1,0, 0,0,0,1,0),nrow=5) image(x) I can find the number of populated points easily table(x) #or more generally sum(x!=0) But I want to find the number of independent components. The answer in this example should be 2. There are three criteria to the function I am seeking: 1. Points that have a neighboring nonzero point should be counted as one contiguous component. 2. The function should respect that the matrix is projected on a torso. That is, points in the leftmost column border points in the rightmost column and points in the top row border points in the bottom row (if they are contiguous when you wrap the image around a cylinder). 3. The function should be fast/efficient since I need to run this over thousands of images/matrices. Is anyone aware of an implementation of such a function? Thanks much for your help, Daniel -- View this message in context: http://r.789695.n4.nabble.com/Spatial-number-of-independent-components-tp2262018p2262018.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
