识别矩阵中的“簇”或“组”

Question

我有一个填充了离散元素的矩阵，我需要将它们聚集成完整的组。因此，例如，采用这个矩阵：

[A B B C A]
[A A B A A]
[A B B C C]
[A A A A A]

A 有两个单独的集群，C 有两个单独的集群，B 有一个集群。

我正在寻找的输出理想情况下会为每个集群分配一个唯一的 ID，如下所示：

[1 2 2 3 4]
[1 1 2 4 4]
[1 2 2 5 5]
[1 1 1 1 1]

现在我有一个 R 代码，它通过迭代检查最近邻来递归地执行此操作，但是当矩阵变大（即 100x100）时它会迅速溢出。

R 中是否有内置函数可以做到这一点？我研究了光栅和图像处理，但没有运气。我确信它一定在那里。

谢谢！

Answer 1

您可以通过构建表示矩阵的格图来解决此问题，其中仅当顶点具有相同类型时才保留边：

# Build initial matrix and lattice graph
library(igraph)
mat <- matrix(c(1, 1, 1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 3, 1, 3, 1, 1, 1, 3, 1), nrow=4)
labels <- as.vector(mat)
g <- graph.lattice(dim(mat))
lyt <- layout.auto(g)

# Remove edges between elements of different types
edgelist <- get.edgelist(g)
retain <- labels[edgelist[,1]] == labels[edgelist[,2]]
g <- delete.edges(g, E(g)[!retain])

# Take a look at what we have
plot(g, layout=lyt)

顶点按列编号。很容易看出，我们需要做的就是抓取该图的组件：

matrix(clusters(g)$membership, nrow=nrow(mat))
#      [,1] [,2] [,3] [,4] [,5]
# [1,]    1    2    2    3    4
# [2,]    1    1    2    4    4
# [3,]    1    2    2    5    5
# [4,]    1    1    1    1    1

如果您想在晶格中包含对角线，您可以从邻域大小为 2 的晶格开始，然后限制元素之间的距离不超过一行或一列。考虑以下矩阵：

[A B C B]
[B A A A]

由于包含对角链接，以下代码将捕获 4 个组，而不是 6 个：

# Build initial matrix and lattice graph (neighborhood size 2)
mat <- matrix(c(1, 2, 2, 1, 3, 1, 2, 1), nrow=2)
labels <- as.vector(mat)
rows <- (seq(length(labels)) - 1) %% nrow(mat)
cols <- ceiling(seq(length(labels)) / nrow(mat))
g <- graph.lattice(dim(mat), nei=2)

# Remove edges between elements of different types or that aren't diagonal
edgelist <- get.edgelist(g)
retain <- labels[edgelist[,1]] == labels[edgelist[,2]] &
  abs(rows[edgelist[,1]] - rows[edgelist[,2]]) <= 1 &
  abs(cols[edgelist[,1]] - cols[edgelist[,2]]) <= 1
g <- delete.edges(g, E(g)[!retain])

# Cluster to obtain final groups
matrix(clusters(g)$membership, nrow=nrow(mat))
#      [,1] [,2] [,3] [,4]
# [1,]    1    2    3    4
# [2,]    2    1    1    1

Answer 2

我不太确定这是否解决了同样的问题，但我最近编写了一些代码，它以相同的方式将墙段分组在迷宫中，即最近邻。我的是迭代的，并利用 dist() 函数。这是我使用的一些代码。

我从一个包含所有墙段的 N*4 矩阵开始（使用 Prim 的树算法生成）；列是 (x0,y0,x1,y1) 定义给定段的端点。所有段都在整数网格点上开始和结束，长度为 1。treelist 的每个元素都包含所有聚集的段。对于发布的问题，这应该更容易一些，因为每个项目只有一个坐标（行、列）而不是两个。

treelist<-list()
 treecnt<-1
 #kill edge walls, i.e. wall segments on the border of the maze.
 #  edges<- which(dowalls[,1]==dowalls[,3] | dowalls[,2]==dowalls[,4])
 vedges <- which( (dowalls[,1]==dowalls[,3]) & (dowalls[,1]==1 | dowalls[,1]==dimx+1) )
 hedges <- which( (dowalls[,2]==dowalls[,4]) & (dowalls[,2]==1 | dowalls[,1]==dimy+1) )
 dowalls<-dowalls[-c(vedges,hedges),,drop=FALSE]
 # now sort into trees 
 while(nrow(dowalls)>0 ) {
     tree <- matrix(dowalls[1,],nr=1) #force dimensions 
     dowalls<-dowalls[-1,,drop=FALSE]
     treerow <- 1 #current row of tree we're looking at
     while ( treerow <= nrow(tree) ) {
         #only examine the first 'column' of the dist() matrix 'cause those are the
         # distances from the tree[] endpoints
         touch <- c( which(dist(rbind(tree[treerow,1:2],dowalls[,1:2]) )[1:nrow(dowalls)]==0),  which(dist(rbind(tree[treerow,1:2],dowalls[,3:4]) )[1:nrow(dowalls)]==0), which(dist(rbind(tree[treerow,3:4],dowalls[,1:2]) )[1:nrow(dowalls)]==0), which(dist(rbind(tree[treerow,3:4],dowalls[,3:4]) )[1:nrow(dowalls)]==0) )
         if(length(touch) ) {
            tree <- rbind(tree,dowalls[c(touch),])
            dowalls <- dowalls[-c(touch),,drop=FALSE] 
            }
    # now be careful: want to track the row of tree[] we're working with AND
    # track how many rows there currently are in tree[]
        treerow <- treerow + 1 
    } #end of while treerow <= nrow 
    treelist[[treecnt]]<-tree
    treecnt <- treecnt + 1 
} #end ; all walls have been classified