在 R 中找到矩阵的相邻元素

Question

Edit: huge thanks to the users below for great contributions and to Gregor for benchmarking.

Say I have a matrix filled with integer values like this...

    mat <- matrix(1:100, 10, 10)

I can create a list of x, y coordinates of each element like this...

    addresses <- expand.grid(x = 1:10, y = 1:10)

Now for each of these coordinates (ie for each element in mat) i would like to find the neighboring elements (including diagonals this should make 8 neighbors).

I'm sure there is an easy way, can anyone help?

What I have tried so far is to loop through and for each element record the neighboring elements as follows;

    neighbours <- list()
    for(i in 1:dim(addresses)[1]){
      x <- addresses$x[i]
      y <- addresses$y[i]
      neighbours[[i]] <- c(mat[y-1, x  ],
                           mat[y-1, x+1],
                           mat[y  , x+1],
                           mat[y+1, x+1],
                           mat[y+1, x  ],
                           mat[y+1, x-1],
                           mat[y  , x-1],
                           mat[y-1, x-1])
    }

This runs into problems when it hits the edge of the matrix, particularly when the index is greater than the edge of the matrix.

Answer 1

Here's a nice example. I did 4x4 so we can see it easily, but it's all adjustable by n . It's also fully vectorized so should have good speed.

n = 4
mat = matrix(1:n^2, nrow = n)
mat.pad = rbind(NA, cbind(NA, mat, NA), NA)

With the padded matrix, the neighbors are just n by n submatrices, shifting around. Using compass directions as labels:

ind = 2:(n + 1) # row/column indices of the "middle"
neigh = rbind(N  = as.vector(mat.pad[ind - 1, ind    ]),
              NE = as.vector(mat.pad[ind - 1, ind + 1]),
              E  = as.vector(mat.pad[ind    , ind + 1]),
              SE = as.vector(mat.pad[ind + 1, ind + 1]),
              S  = as.vector(mat.pad[ind + 1, ind    ]),
              SW = as.vector(mat.pad[ind + 1, ind - 1]),
              W  = as.vector(mat.pad[ind    , ind - 1]),
              NW = as.vector(mat.pad[ind - 1, ind - 1]))

mat
#      [,1] [,2] [,3] [,4]
# [1,]    1    5    9   13
# [2,]    2    6   10   14
# [3,]    3    7   11   15
# [4,]    4    8   12   16

  neigh[, 1:6]
#    [,1] [,2] [,3] [,4] [,5] [,6]
# N    NA    1    2    3   NA    5
# NE   NA    5    6    7   NA    9
# E     5    6    7    8    9   10
# SE    6    7    8   NA   10   11
# S     2    3    4   NA    6    7
# SW   NA   NA   NA   NA    2    3
# W    NA   NA   NA   NA    1    2
# NW   NA   NA   NA   NA   NA    1

So you can see for the first element mat[1,1] , starting at North and going clockwise the neighbors are the first column of neigh . The next element is mat[2,1] , and so on down the columns of mat . (You can also compare to @mrip's answer and see that our columns have the same elements, just in a different order.)

Benchmarking

Small matrix

mat = matrix(1:16, nrow = 4)
mbm(gregor(mat), mrip(mat), marat(mat), u20650(mat), times = 100)
# Unit: microseconds
#         expr     min       lq      mean   median       uq      max neval  cld
#  gregor(mat)  25.054  30.0345  34.04585  31.9960  34.7130   61.879   100 a   
#    mrip(mat) 420.167 443.7120 482.44136 466.1995 483.4045 1820.121   100   c 
#   marat(mat) 746.462 784.0410 812.10347 808.1880 832.4870  911.570   100    d
#  u20650(mat) 186.843 206.4620 220.07242 217.3285 230.7605  269.850   100  b  

On a larger matrix I had to take out user20650's function because it tried to allocate a 232.8 Gb vector, and I also took out Marat's answer after waiting for about 10 minutes.

mat = matrix(1:500^2, nrow = 500)

mbm(gregor(mat), mrip(mat), times = 100)
# Unit: milliseconds
#         expr       min        lq      mean    median        uq      max neval cld
#  gregor(mat) 19.583951 21.127883 30.674130 21.656866 22.433661 127.2279   100   b
#    mrip(mat)  2.213725  2.368421  8.957648  2.758102  2.958677 104.9983   100  a 

So it looks like in any case where time matters, @mrip's solutions is by far the fastest.

Functions used:

gregor = function(mat) {
    n = nrow(mat)
    mat.pad = rbind(NA, cbind(NA, mat, NA), NA)
    ind = 2:(n + 1) # row/column indices of the "middle"
    neigh = rbind(N  = as.vector(mat.pad[ind - 1, ind    ]),
                  NE = as.vector(mat.pad[ind - 1, ind + 1]),
                  E  = as.vector(mat.pad[ind    , ind + 1]),
                  SE = as.vector(mat.pad[ind + 1, ind + 1]),
                  S  = as.vector(mat.pad[ind + 1, ind    ]),
                  SW = as.vector(mat.pad[ind + 1, ind - 1]),
                  W  = as.vector(mat.pad[ind    , ind - 1]),
                  NW = as.vector(mat.pad[ind - 1, ind - 1]))
    return(neigh)
}

mrip = function(mat) {
    m2<-cbind(NA,rbind(NA,mat,NA),NA)
    addresses <- expand.grid(x = 1:4, y = 1:4)
    ret <- c()
    for(i in 1:-1)
        for(j in 1:-1)
            if(i!=0 || j !=0)
                ret <- rbind(ret,m2[addresses$x+i+1+nrow(m2)*(addresses$y+j)]) 
    return(ret)
}

get.neighbors <- function(rw, z, mat) {
    # Convert to absolute addresses 
    z2 <- t(z + unlist(rw))
    # Choose those with indices within mat 
    b.good <- rowSums(z2 > 0)==2  &  z2[,1] <= nrow(mat)  &  z2[,2] <= ncol(mat)
    mat[z2[b.good,]]
}

marat = function(mat) {
    n.row = n.col = nrow(mat)
    addresses <- expand.grid(x = 1:n.row, y = 1:n.col)
    # Relative addresses
    z <- rbind(c(-1,0,1,-1,1,-1,0,1), c(-1,-1,-1,0,0,1,1,1))
    apply(addresses, 1,
          get.neighbors, z = z, mat = mat) # Returns a list with neighbors
}

u20650 = function(mat) {
    w <-  which(mat==mat, arr.ind=TRUE)
    d <- as.matrix(dist(w, "maximum", diag=TRUE, upper=TRUE))
    # extract neighbouring values for each element
    # extract where max distance is one
    a <- apply(d, 1, function(i) mat[i == 1] )
    names(a)  <- mat
    return(a)
}

Answer 2

This will get you a matrix with columns corresponding to neighbors of each entry in the matrix:

mat <- matrix(1:16, 4, 4)
m2<-cbind(NA,rbind(NA,mat,NA),NA)
addresses <- expand.grid(x = 1:4, y = 1:4)

ret<-c()
for(i in 1:-1)
  for(j in 1:-1)
    if(i!=0 || j !=0)
      ret<-rbind(ret,m2[addresses$x+i+1+nrow(m2)*(addresses$y+j)]) 


> ret
     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14]
[1,]    6    7    8   NA   10   11   12   NA   14    15    16    NA    NA    NA
[2,]    2    3    4   NA    6    7    8   NA   10    11    12    NA    14    15
[3,]   NA   NA   NA   NA    2    3    4   NA    6     7     8    NA    10    11
[4,]    5    6    7    8    9   10   11   12   13    14    15    16    NA    NA
[5,]   NA   NA   NA   NA    1    2    3    4    5     6     7     8     9    10
[6,]   NA    5    6    7   NA    9   10   11   NA    13    14    15    NA    NA
[7,]   NA    1    2    3   NA    5    6    7   NA     9    10    11    NA    13
[8,]   NA   NA   NA   NA   NA    1    2    3   NA     5     6     7    NA     9
     [,15] [,16]
[1,]    NA    NA
[2,]    16    NA
[3,]    12    NA
[4,]    NA    NA
[5,]    11    12
[6,]    NA    NA
[7,]    14    15
[8,]    10    11

Answer 3

Here is another approach:

n.col <- 5
n.row <- 10
mat <- matrix(seq(n.col * n.row), n.row, n.col)

addresses <- expand.grid(x = 1:n.row, y = 1:n.col)

# Relative addresses
z <- rbind(c(-1,0,1,-1,1,-1,0,1),c(-1,-1,-1,0,0,1,1,1))

get.neighbors <- function(rw) {
  # Convert to absolute addresses 
  z2 <- t(z + unlist(rw))
  # Choose those with indices within mat 
  b.good <- rowSums(z2 > 0)==2  &  z2[,1] <= nrow(mat)  &  z2[,2] <=ncol(mat)
  mat[z2[b.good,]]
}

apply(addresses,1, get.neighbors) # Returns a list with neighbors

Answer 4

Perhaps you may be able to use a distance function here using the row and column indices of the matrix elements.

# data
(mat <- matrix(16:31, 4, 4))
     [,1] [,2] [,3] [,4]
[1,]   16   20   24   28
[2,]   17   21   25   29
[3,]   18   22   26   30
[4,]   19   23   27   31

# find distances between row and column indexes
# interested in values where the distance is one
w <-  which(mat==mat, arr.ind=TRUE)
d <- as.matrix(dist(w, "maximum", diag=TRUE, upper=TRUE))

# extract neighbouring values for each element
# extract where max distance is one
a <- apply(d, 1, function(i) mat[i == 1] )
names(a)  <- mat
a

$`16`
[1] "17" "20" "21"

$`17`
[1] "16" "18" "20" "21" "22"

$`18`
[1] "17" "19" "21" "22" "23
... ....
... ....

Needs tidied, but maybe gives an idea

Answer 5

There was also an much faster solution posted here: List of n first Neighbors from a 3d Array R

for large matrices this solution is much faster, see my benchmark here:

For small matrices I want to show another very interesting solution, which is fast for small matrices, but slow for bigger ones:

Using complex numbers and the abs function.

get_neighbor <- function(matrix, x=1,y=1){

    z <- complex(real = rep(1:nrow(matrix), ncol(matrix)),
                 imaginary = rep(1:ncol(matrix), each = nrow(matrix)))

    lookup <- lapply(seq_along(z), function(x){
      ## calculate distance
      dist <- which(abs(z - z[x]) < 2)
      ## remove those with dist == 0 -> it´s the number itself
      dist[which(dist != x)]
    })
    index <- (y-1)*(nrow(matrix))+x
    matrix[lookup[[index]]]

}

Answer 6

We can try the following base R code

f <- Vectorize(function(mat, x, y) {
  X <- pmin(pmax(x + c(-1:1), 1), nrow(mat))
  Y <- pmin(pmax(y + c(-1:1), 1), ncol(mat))
  mat[as.matrix(subset(
    unique(expand.grid(X = X, Y = Y)),
    !(X == x & Y == y)
  ))]
},vectorize.args = c("x","y"))

neighbors <- with(addresses, f(mat,x,y))

and we will see

> head(neighbors)
[[1]]
[1]  2 11 12

[[2]]
[1]  1  3 11 12 13

[[3]]
[1]  2  4 12 13 14

[[4]]
[1]  3  5 13 14 15

[[5]]
[1]  4  6 14 15 16

[[6]]
[1]  5  7 15 16 17