编辑：非常感谢下面的用户做出的巨大贡献以及Gregor的基准测试。
假设我有一个填充了像这样的整数值的矩阵......
mat <- matrix(1:100, 10, 10)
我可以像这样创建每个元素的x，y坐标列表......
addresses <- expand.grid(x = 1:10, y = 1:10)
现在对于这些坐标中的每一个（即对于垫子中的每个元素），我想找到相邻的元素（包括对角线，这应该使8个邻居）。
我确信有一种简单的方法，有人可以帮忙吗？
到目前为止，我所尝试的是遍历并为每个元素记录相邻元素，如下所示;
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])
}
当它碰到矩阵的边缘时会遇到问题，特别是当索引大于矩阵的边缘时。
答案 0 :(得分：9)
这是一个很好的例子。我做了4x4所以我们可以很容易地看到它，但它都可以通过n进行调整。它也完全矢量化，所以应该有很好的速度。
n = 4
mat = matrix(1:n^2, nrow = n)
mat.pad = rbind(NA, cbind(NA, mat, NA), NA)
使用填充矩阵，邻居只有n个子矩阵，移动。使用罗盘方向作为标签：
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
所以你可以看到第一个元素mat[1,1]，从北方开始，顺时针方向，邻居是neigh的第一列。下一个元素是mat[2,1]，依此类推mat列。 （您也可以与@ mrip的答案进行比较，看看我们的列具有相同的元素，只是顺序不同。）
小矩阵
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
在一个更大的矩阵上，我不得不取出user20650的功能，因为它试图分配一个232.8 Gb的矢量，我等了大约10分钟后也拿出了Marat的答案。
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
所以看起来在任何时间都很重要的情况下，@ mrip的解决方案是迄今为止最快的。
使用的功能：
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)
}
答案 1 :(得分：4)
这将为您提供一个矩阵，其中的列对应于矩阵中每个条目的邻居：
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
答案 2 :(得分：1)
这是另一种方法：
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
答案 3 :(得分：1)
也许您可以使用矩阵元素的行和列索引来使用距离函数。
# 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
... ....
... ....
需要整理，但也许会提出一个想法