I have a matrix with values 0 or 1 and I would like to obtain a list of groups of adjacent 1's.
For example, the matrix
mat = rbind(c(1,0,0,0,0),
c(1,0,0,1,0),
c(0,0,1,0,0),
c(0,0,0,0,0),
c(1,1,1,1,1))
> mat
[,1] [,2] [,3] [,4] [,5]
[1,] 1 0 0 0 0
[2,] 1 0 0 1 0
[3,] 0 0 1 0 0
[4,] 0 0 0 0 0
[5,] 1 1 1 1 1
should return the following 4 connected components:
C1 = {(1,1);(2,1)}
C2 = {(2,4)}
C3 = {(3,3)}
C4 = {(5,1);(5,2);(5,3);(5,4);(5,5)}
Does anybody has an idea of how to do it fast in R? My real matrix is indeed rather large, like 2000x2000 (but I expect that the number of connected components to be reasonably small, i.e. 200).
With the update, you can turn your binary matrix into a raster object and use the clumps function. Then it is just data management to return the exact format you want. Example below:
library(igraph)
library(raster)
mat = rbind(c(1,0,0,0,0),
c(1,0,0,1,0),
c(0,0,1,0,0),
c(0,0,0,0,0),
c(1,1,1,1,1))
Rmat <- raster(mat)
Clumps <- as.matrix(clump(Rmat, directions=4))
#turn the clumps into a list
tot <- max(Clumps, na.rm=TRUE)
res <- vector("list",tot)
for (i in 1:max(Clumps, na.rm=TRUE)){
res[i] <- list(which(Clumps == i, arr.ind = TRUE))
}
Which then res
prints out at the console:
> res
[[1]]
row col
[1,] 1 1
[2,] 2 1
[[2]]
row col
[1,] 2 4
[[3]]
row col
[1,] 3 3
[[4]]
row col
[1,] 5 1
[2,] 5 2
[3,] 5 3
[4,] 5 4
[5,] 5 5
I wouldn't be surprised if there is a better way to go from the raster object to your end goal though. Again a 2000 by 2000 matrix should not be a big deal for this.
Old (wrong answer) but should be useful for people who want connected components of a graph.
You can use the igraph package to turn your adjacency matrix into a network and return the components. Your example graph is one component, so I removed one edge for illustration.
library(igraph)
mat = rbind(c(1,0,0,0,0),
c(1,0,0,1,0),
c(0,0,1,0,0),
c(0,0,0,0,0),
c(1,1,1,1,1))
g <- graph.adjacency(mat) %>% delete_edges("5|3")
plot(g)
clu <- components(g)
groups(clu)
The final line then returns at the prompt:
> groups(clu)
$`1`
[1] 1 2 4 5
$`2`
[1] 3
My experience with this algorithm it is pretty fast - so I don't think 2,000 by 2,000 will be a problem.