Question

我有一个填充了离散元素的矩阵，我需要将它们聚类成完整的组。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有两个独立的集群，C有两个独立的集群，B有一个集群。

我正在寻找的输出理想情况下会为每个clister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，通过迭代检查最近邻居来递归地执行此操作，但是当矩阵变大（即100x100）时它会快速溢出。

R中是否有内置函数可以执行此操作？我查看了光栅和图像处理，但没有运气。我确信它必须在那里。

谢谢！

Answer 1

您可以通过构建表示矩阵的点阵图来处理此问题，其中只有顶点具有相同类型时才会保留边：

# 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)

enter image description here

顶点按列向下编号。很容易看出我们需要做的就是抓住这个图的组成部分：

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

如果要在网格中包含对角线，可以从邻域大小为2的网格开始，然后限制为不超过一行或一列的元素。考虑以下矩阵：

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

由于包含对角链接，这里的代码将捕获4组，而不是6组：

# 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

我不太确定这是否能解决同样的问题，但我最近写了一些代码，它们以相同的方式将迷宫中的墙段分组，即最近邻居。我是迭代的，并使用dist（）函数。这是我使用过的一些代码。

我从包含所有墙段的N * 4矩阵开始（使用Prim＆＃39树Alg生成）;列（x0，y0，x1，y1）定义给定段的端点。所有段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簇段。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

识别矩阵中的“簇”或“组”

2 个答案: