My matrice is like this:
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
[1,] 0 0 0 0 0 0 0 0 0
[2,] 0 0 0 0 0 0 0 0 0
[3,] 0 0 0 0 0 0 0 0 0
[4,] 0 0 0 0 0 0 0 1 1
[5,] 0 0 0 0 0 0 1 1 0
[6,] 0 0 0 0 0 1 0 0 0
[7,] 0 0 0 0 1 1 0 0 0
[8,] 0 0 0 0 1 0 0 0 0
[9,] 0 0 0 0 1 0 0 0 0
[10,] 0 0 0 0 1 1 0 0 0
[11,] 0 0 0 0 0 1 0 0 0
[12,] 0 0 0 0 0 1 1 1 1
[13,] 0 0 0 0 0 0 0 0 0
[14,] 0 0 0 0 0 0 0 0 0
[15,] 0 0 0 0 0 0 0 0 0
[16,] 0 0 0 0 0 0 0 0 0
[17,] 0 0 0 0 0 0 0 0 0
[,10] [,11] [,12] [,13] [,14] [,15] [,16] [,17]
[1,] 0 0 0 0 0 0 0 0
[2,] 0 0 0 0 0 0 0 0
[3,] 0 0 0 0 0 0 0 0
[4,] 1 1 0 0 0 0 0 0
[5,] 0 1 0 0 0 0 0 0
[6,] 0 1 0 0 0 0 0 0
[7,] 0 1 0 0 0 0 0 0
[8,] 0 1 0 0 0 1 0 0
[9,] 0 1 0 0 0 1 0 0
[10,] 1 1 0 0 0 1 0 0
[11,] 1 1 0 1 1 0 0 0
[12,] 1 1 1 1 0 0 0 0
[13,] 0 0 0 0 0 0 0 0
[14,] 0 0 0 0 0 0 0 0
[15,] 0 0 0 0 0 0 0 0
[16,] 0 0 0 0 0 0 0 0
[17,] 0 0 0 0 0 0 0 0
[,18]
[1,] 0
[2,] 0
[3,] 0
[4,] 0
[5,] 0
[6,] 0
[7,] 0
[8,] 0
[9,] 0
[10,] 0
[11,] 0
[12,] 0
[13,] 0
[14,] 0
[15,] 0
[16,] 0
[17,] 0
How can I find the number of values which has 1 in neighbour? (neighbour of a pixel is the value above the value, below the value, to the right, to the left, top right, top left, below left, below right).
I just need to get a way of devising how I can even find the number of values which has 1 above/below it. If I get that, I'll be able to solve the other variables of the problem (top right and such).
I've been experimenting around with which such as which(imageMatrix == 1, arr.ind = TRUE)[1,1]. But I cannot figure it out. (ImageMatrix is the name of the matrix)
Can anyone lend me a hand on how I can begin with the problem so I get a jump?
CodePudding user response：
In the following we use the example matrix m generated reproducibly in the Note at the end.
1) Append a row and column of zeros on each end and then apply rollsum , transpose, apply it again and transpose again. Finally subtract the original matrix so that only neighbors are counted. This solution is the most compact of those here.
library(zoo)
m2 <- rbind(0, cbind(0, m, 0), 0)
t(rollsum(t(rollsum(m2, 3)), 3)) - m
## [,1] [,2] [,3] [,4] [,5]
## [1,] 2 3 5 3 2
## [2,] 3 5 6 4 2
## [3,] 2 6 5 5 2
## [4,] 2 5 2 2 1
## [5,] 1 3 2 2 1
2) A second approach is the following. It would be much faster than the others here.
nr <- nrow(m)
nc <- ncol(m)
mm <- cbind(m[, -1], 0) m cbind(0, m[, -nc])
rbind(mm[-1, ], 0) mm rbind(0, mm[-nr, ]) - m
## [,1] [,2] [,3] [,4] [,5]
## [1,] 2 3 5 3 2
## [2,] 3 5 6 4 2
## [3,] 2 6 5 5 2
## [4,] 2 5 2 2 1
## [5,] 1 3 2 2 1
3) Using loops we can write the following. It is probably the most straight forward. Note that subscripting by 0 omits that element and i %% 6 equals i if i is in 1, 2, 3, 4, 5 and equals 0 if i equals 0 or 6.
nr <- nrow(m); nr1 <- nr 1
nc <- ncol(m); nc1 <- nc 1
mm <- 0 * m # initialize result
for(i in seq_len(nr))
for(j in seq_len(nc))
mm[i, j] <- sum(m[seq(i-1, i 1) %% nr1, seq(j-1, j 1) %% nc1]) - m[i,j]
mm
## [,1] [,2] [,3] [,4] [,5]
## [1,] 2 3 5 3 2
## [2,] 3 5 6 4 2
## [3,] 2 6 5 5 2
## [4,] 2 5 2 2 1
## [5,] 1 3 2 2 1
Note
set.seed(123)
m <- matrix(rnorm(25) > 0, 5)
m
## [,1] [,2] [,3] [,4] [,5]
## [1,] 0 1 1 1 0
## [2,] 0 1 1 1 0
## [3,] 1 0 1 0 0
## [4,] 1 0 1 1 0
## [5,] 1 0 0 0 0