将最近的邻居列表转换为 r 中的二进制邻接矩阵
Convert nearest neighbours list into a binary adjacency matrix in r
我有一个包含 24 个站点的列表 near_neigh，以及这些站点中所有站点的 3 个最近邻居。第一列列出了站点，其他列列出了 3 个邻居。
near_neigh <-
structure(list(row.names.near_neigh. = c("1", "2", "3", "4",
"5", "6", "7", "8", "9", "10", "11", "12", "13", "14", "15",
"16", "17", "18", "19", "20", "21", "22", "23", "24"),
nn.index.1 = c(6L,
7L, 2L, 5L, 6L, 1L, 2L, 9L, 8L, 7L, 8L, 14L, 16L, 17L, 19L, 20L,
18L, 17L, 22L, 23L, 17L, 19L, 20L, 23L),
nn.index.2 = c(2L, 1L,
7L, 8L, 4L, 7L, 6L, 4L, 5L, 6L, 9L, 13L, 15L, 18L, 20L, 13L,
21L, 21L, 20L, 19L, 18L, 23L, 22L, 16L),
nn.index.3 = c(7L, 3L,
1L, 9L, 9L, 5L, 10L, 11L, 6L, 9L, 14L, 16L, 20L, 21L, 13L, 15L,
14L, 14L, 15L, 22L, 14L, 20L, 19L, 21L)),
class = "data.frame", row.names = c(NA,
-24L))
我想将此列表转换为 24x24 的二进制邻接矩阵，描述所有站点的邻居。
1) 创建一个 nr by nr 零矩阵并使用 Reduce 从每一列连续插入 1。
nr <- nrow(near_neigh)
f <- function(m, x) replace(m, cbind(1:nr, x), 1)
Reduce(f, init = matrix(0, nr, nr), near_neigh[-1])
2) 或在索引列上使用循环：
nr <- nrow(near_neigh)
m <- matrix(0, nr, nr)
for(x in near_neigh[-1]) m[cbind(1:nr, x)] <- 1
3) 或遍历行：
nr <- nrow(near_neigh)
m <- matrix(0, nr, nr)
for(i in 1:nr) m[i, unlist(near_neigh[i, -1])] <- 1
4) 或使用 igraph
library(tidyr)
library(igraph)
set.seed(7)
edgelist <- sapply(pivot_longer(near_neigh, -1)[-2], as.numeric)
g <- graph_from_edgelist(edgelist, directed = FALSE)
mm <- as_adjacency_matrix(g) # class "dgCMatrix"
# rest is optional but converting to unnamed matrix allows comparison
# to other solutions
mm <- as.matrix(mm)
dimnames(mm) <- NULL
plot(g, vertex.shape = "none")
使用 igraph 的图形方法：
near_neigh <- sapply(near_neigh, as.integer)
library(igraph)
g <- graph_from_edgelist(rbind(near_neigh[, c(1, 2)],
near_neigh[, c(1, 3)],
near_neigh[, c(1, 4)]),
directed = TRUE)
plot(g)
as.matrix(as_adjacency_matrix(g))
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14] [,15] [,16] [,17] [,18] [,19] [,20] [,21] [,22] [,23] [,24]
[1,] 0 1 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[2,] 1 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[3,] 1 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[4,] 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[5,] 0 0 0 1 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[6,] 1 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[7,] 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[8,] 0 0 0 1 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
[9,] 0 0 0 0 1 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[10,] 0 0 0 0 0 1 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[11,] 0 0 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0
[12,] 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 1 0 0 0 0 0 0 0 0
[13,] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0
[14,] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0
[15,] 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0
[16,] 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0 0
[17,] 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 1 0 0 0
[18,] 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0
[19,] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 1 0 0
[20,] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0
[21,] 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0
[22,] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0
[23,] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 1 0 0
[24,] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 1 0
我有一个包含 24 个站点的列表 near_neigh，以及这些站点中所有站点的 3 个最近邻居。第一列列出了站点，其他列列出了 3 个邻居。
near_neigh <-
structure(list(row.names.near_neigh. = c("1", "2", "3", "4",
"5", "6", "7", "8", "9", "10", "11", "12", "13", "14", "15",
"16", "17", "18", "19", "20", "21", "22", "23", "24"),
nn.index.1 = c(6L,
7L, 2L, 5L, 6L, 1L, 2L, 9L, 8L, 7L, 8L, 14L, 16L, 17L, 19L, 20L,
18L, 17L, 22L, 23L, 17L, 19L, 20L, 23L),
nn.index.2 = c(2L, 1L,
7L, 8L, 4L, 7L, 6L, 4L, 5L, 6L, 9L, 13L, 15L, 18L, 20L, 13L,
21L, 21L, 20L, 19L, 18L, 23L, 22L, 16L),
nn.index.3 = c(7L, 3L,
1L, 9L, 9L, 5L, 10L, 11L, 6L, 9L, 14L, 16L, 20L, 21L, 13L, 15L,
14L, 14L, 15L, 22L, 14L, 20L, 19L, 21L)),
class = "data.frame", row.names = c(NA,
-24L))
我想将此列表转换为 24x24 的二进制邻接矩阵，描述所有站点的邻居。
1) 创建一个 nr by nr 零矩阵并使用 Reduce 从每一列连续插入 1。
nr <- nrow(near_neigh)
f <- function(m, x) replace(m, cbind(1:nr, x), 1)
Reduce(f, init = matrix(0, nr, nr), near_neigh[-1])
2) 或在索引列上使用循环：
nr <- nrow(near_neigh)
m <- matrix(0, nr, nr)
for(x in near_neigh[-1]) m[cbind(1:nr, x)] <- 1
3) 或遍历行：
nr <- nrow(near_neigh)
m <- matrix(0, nr, nr)
for(i in 1:nr) m[i, unlist(near_neigh[i, -1])] <- 1
4) 或使用 igraph
library(tidyr)
library(igraph)
set.seed(7)
edgelist <- sapply(pivot_longer(near_neigh, -1)[-2], as.numeric)
g <- graph_from_edgelist(edgelist, directed = FALSE)
mm <- as_adjacency_matrix(g) # class "dgCMatrix"
# rest is optional but converting to unnamed matrix allows comparison
# to other solutions
mm <- as.matrix(mm)
dimnames(mm) <- NULL
plot(g, vertex.shape = "none")
使用 igraph 的图形方法：
near_neigh <- sapply(near_neigh, as.integer)
library(igraph)
g <- graph_from_edgelist(rbind(near_neigh[, c(1, 2)],
near_neigh[, c(1, 3)],
near_neigh[, c(1, 4)]),
directed = TRUE)
plot(g)
as.matrix(as_adjacency_matrix(g))
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14] [,15] [,16] [,17] [,18] [,19] [,20] [,21] [,22] [,23] [,24]
[1,] 0 1 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[2,] 1 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[3,] 1 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[4,] 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[5,] 0 0 0 1 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[6,] 1 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[7,] 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[8,] 0 0 0 1 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
[9,] 0 0 0 0 1 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[10,] 0 0 0 0 0 1 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
[11,] 0 0 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0
[12,] 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 1 0 0 0 0 0 0 0 0
[13,] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0
[14,] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0
[15,] 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0
[16,] 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 1 0 0 0 0
[17,] 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 1 0 0 0
[18,] 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0
[19,] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 1 0 0
[20,] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0
[21,] 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0
[22,] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0
[23,] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 1 0 0
[24,] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 1 0