在R中找到矩阵的相邻元素 [英] find neighbouring elements of a matrix in R
问题描述
非常感谢以下用户做出的巨大贡献，并感谢Gregor进行了基准测试.
huge thanks to the users below for great contributions and to Gregor for benchmarking.
说我有一个像这样的整数值填充的矩阵...
Say I have a matrix filled with integer values like this...
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...
I can create a list of x, y coordinates of each element like this...
addresses <- expand.grid(x = 1:10, y = 1:10)
现在，对于这些坐标中的每个坐标(即，对于垫中的每个元素)，我都希望找到相邻的元素(包括对角线，它应该使8个相邻).
Now for each of these coordinates (i.e. for each element in mat) i would like to find the neighboring elements (including diagonals this should make 8 neighbors).
我敢肯定有一种简单的方法，任何人都可以帮忙吗?
I'm sure there is an easy way, can anyone help?
到目前为止，我一直尝试遍历并为每个元素记录相邻元素，如下所示；
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])
}
当它碰到矩阵的边缘时，尤其是当索引大于矩阵的边缘时，就会遇到问题.
This runs into problems when it hits the edge of the matrix, particularly when the index is greater than the edge of the matrix.
推荐答案
这是一个很好的示例.我做了4x4，所以我们可以很容易地看到它，但是都可以通过n进行调整.它也已完全矢量化，因此应该具有良好的速度.
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)
在填充矩阵的情况下，相邻元素只是n×n个子矩阵而四处移动.使用指南针方向作为标签:
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
因此您可以看到第一个元素mat[1,1]，它从North开始并按顺时针方向旋转，相邻元素是neigh的第一列.下一个元素是mat[2,1]，依此类推，直到mat的列. (您也可以将其与@mrip的答案进行比较，以了解我们的列具有相同的元素，只是顺序不同.)
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.)
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
在更大的矩阵上，我不得不使用user20650的功能，因为它试图分配232.8 Gb向量，并且在等待大约10分钟后，我也拿出了Marat的答案.
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
因此，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.
So it looks like in any case where time matters, @mrip's solutions is by far the fastest.
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)
}
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