您可以构造一个图形来表示矩阵中位置之间的有效移动：
# Construct nodes and edges from matrix(nodes <- which(m == 1 | m == 2 | m == 3, arr.ind=TRUE))# row col# [1,] 1 1# [2,] 2 1# [3,] 4 1# [4,] 2 2# [5,] 3 2# [6,] 4 2# [7,] 4 3# [8,] 2 4# [9,] 4 4edges <- which(outer(seq_len(nrow(nodes)),seq_len(nrow(nodes)), function(x, y) abs(nodes[x,"row"] - nodes[y,"row"]) + abs(nodes[x,"col"] - nodes[y,"col"]) == 1), arr.ind=T)(edges <- edges[edges[,"col"] > edges[,"row"],])# row col# [1,] 1 2# [2,] 2 4# [3,] 4 5# [4,] 3 6# [5,] 5 6# [6,] 6 7# [7,] 7 9library(igraph)g <- graph.data.frame(edges, directed=FALSE, vertices=seq_len(nrow(nodes)))
然后，您可以解决指定起点和终点之间的最短路径问题：
start.pos <- which(m == 2, arr.ind=TRUE)start.node <- which(paste(nodes[,"row"], nodes[,"col"]) == paste(start.pos[,"row"], start.pos[,"col"]))end.pos <- which(m == 3, arr.ind=TRUE)end.node <- which(paste(nodes[,"row"], nodes[,"col"]) == paste(end.pos[,"row"], end.pos[,"col"]))(sp <- nodes[get.shortest.paths(g, start.node, end.node)$vpath[[1]],])# row col# [1,] 1 1# [2,] 2 1# [3,] 2 2# [4,] 3 2# [5,] 4 2# [6,] 4 3# [7,] 4 4
最后，您可以确定方向（1：东； 2：北； 3：西； 4：南），作为对所选最终节点集的简单操纵：
dx <- diff(sp[,"col"])dy <- -diff(sp[,"row"])(dirs <- ifelse(dx == 1, 1, ifelse(dy == 1, 2, ifelse(dx == -1, 3, 4))))# [1] 4 1 4 4 1 1
此代码适用于任意大小的输入矩阵。
数据：
(m <- matrix(c(2, 1, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 0, 3), nrow=4))# [,1] [,2] [,3] [,4]# [1,] 2 0 0 0# [2,] 1 1 0 1# [3,] 0 1 0 0# [4,] 1 1 1 3
