r - 识别矩阵中的重复岛并更改它们的值

Question

我有一个矩阵:

m <- matrix(c(
  1,    1,    1,    0,    0,    0,
  0,    0,    0,    0,    0,    0,
  3,    0,    0,    0,    0,    0,
  3,    0,    0,    0,    0,    2,
  3,    0,    0,    0,    0,    0,
  3,    0,    0,    0,    2,    2),
  ncol = 6, byrow = TRUE)

     [,1] [,2] [,3] [,4] [,5] [,6]
[1,]    1    1    1    0    0    0
[2,]    0    0    0    0    0    0
[3,]    3    0    0    0    0    0
[4,]    3    0    0    0    0    2 # <- island 3, value 2
[5,]    3    0    0    0    0    0
[6,]    3    0    0    0    2    2 # <- island  4, also value 2

在这个矩阵中，有四个“孤岛”，即由零分隔的非零值:

(1) 由三个 1、(2) 四个 3、(3) 一个 2 和 (4) 两个 2 组成的岛屿。

因此，两个岛由值 2 组成。 .我想识别这样的“重复”岛屿，并将“岛屿”之一(要么都可以)的值更改为下一个可用数字(在本例中为 4):

     [,1] [,2] [,3] [,4] [,5] [,6]
[1,]    1    1    1    0    0    0
[2,]    0    0    0    0    0    0
[3,]    3    0    0    0    0    0
[4,]    3    0    0    0    0    2
[5,]    3    0    0    0    0    0
[6,]    3    0    0    0    4    4

Answer 1

有趣的问题!让我们来看一个更复杂的案例

(M <- matrix(c(1, 0, 3, 3, 3, 3, 1, 0, 0, 0, 0, 0, 1, 0, 3, 0, 2, 
               0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 2, 1, 0, 0, 2, 0, 2), 6, 6))
#      [,1] [,2] [,3] [,4] [,5] [,6]
# [1,]    1    1    1    0    0    1
# [2,]    0    0    0    0    0    0
# [3,]    3    0    3    3    0    0
# [4,]    3    0    0    0    0    2
# [5,]    3    0    2    0    0    0
# [6,]    3    0    0    0    2    2

这是一个基于图形的解决方案。

library(igraph)
# Indices of nonzero matrix elements
idx <- which(M != 0, arr.ind = TRUE)
# Adjacency matrix for matrix entries
# Two entries are adjacent if their column or row number differs by one
# Also, due to idx, an implicit condition is also that the two entries are the same
adj <- 1 * (as.matrix(dist(idx, method = "manhattan")) == 1)
# Creating loops as to take into account singleton islands
diag(adj) <- 1
# A corresponding graphs
g <- graph_from_adjacency_matrix(adj, mode = "undirected")
# Connected components of this graph
cmps <- clusters(g)
# Going over unique values of M
for(i in 1:max(M)) {
  # Islands of value i
  un <- unique(cmps$membership[M[idx] == i])
  # More than one island?
  if(length(un) > 1)
    # If so, let's go over islands 2, 3, ...
    for(cmp in un[-1])
      # ... and replace corresponding matrix entries by max(M) + 1
      M[idx[cmps$membership == cmp, , drop = FALSE]] <- max(M) + 1
}

M
#      [,1] [,2] [,3] [,4] [,5] [,6]
# [1,]    1    1    1    0    0    4
# [2,]    0    0    0    0    0    0
# [3,]    3    0    7    7    0    0
# [4,]    3    0    0    0    0    6
# [5,]    3    0    2    0    0    0
# [6,]    3    0    0    0    5    5

另请注意，使用 adj如果我们能找到导致具有最大块数的块对角矩阵的排列，我们就可以单独找到所有的岛。那么每个块将对应一个岛。但是，我找不到相关程序的 R 实现。