查找矩阵 (R) 中相邻值的数量

Question

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?

Answer 1

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