从部分信息生成完整的邻接矩阵
producing a full adjacency matrix from partial information
我 have
一个包含构造 5x5 邻接矩阵所需的所有信息的矩阵。每行代表一个矩阵:
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 1 1 1 1 1 1 1 0 1 0
[2,] 0 0 0 1 1 1 1 0 1 0
...
I want
从第 n 行数据创建邻接矩阵。对于 have
的第一行,want
矩阵如下所示:
[,1] [,2] [,3] [,4] [,5]
[1,] 0 1 1 1 1
[2,] 1 0 1 1 1
[3,] 1 1 0 0 1
[4,] 1 1 0 0 0
[5,] 1 1 1 0 0
如何从 have
到 want
?
这是一个使用 lower.tri
和 upper.tri
的选项
unlist(apply(mat, 1, function(x) {
m <- matrix(0, nrow = 5, ncol = 5)
m[lower.tri(m)] <- x
m[upper.tri(m)] <- x
list(m)
}), recursive = F)
#[[1]]
# [,1] [,2] [,3] [,4] [,5]
#[1,] 0 1 1 1 1
#[2,] 1 0 1 1 0
#[3,] 1 1 0 1 1
#[4,] 1 1 0 0 0
#[5,] 1 1 1 0 0
#
#[[2]]
# [,1] [,2] [,3] [,4] [,5]
#[1,] 0 0 0 1 1
#[2,] 0 0 0 1 0
#[3,] 0 1 0 1 1
#[4,] 0 1 0 0 0
#[5,] 1 1 1 0 0
unlist(..., recursive = F)
部分看起来有些笨拙,但对于防止 apply
简化结果和丢弃 dims 是必要的。另一种方法是在 data.frame
上使用 lapply
而不是 matrix
:
lapply(as.data.frame(t(mat)), function(x) {
m <- matrix(0, nrow = 5, ncol = 5)
m[lower.tri(m)] <- x
m[upper.tri(m)] <- x
return(m)
})
给出相同的结果。
示例数据
mat <- as.matrix(read.table(text =
"1 1 1 1 1 1 1 0 1 0
0 0 0 1 1 1 1 0 1 0", header = F))
colnames(mat) <- NULL
我 have
一个包含构造 5x5 邻接矩阵所需的所有信息的矩阵。每行代表一个矩阵:
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 1 1 1 1 1 1 1 0 1 0
[2,] 0 0 0 1 1 1 1 0 1 0
...
I want
从第 n 行数据创建邻接矩阵。对于 have
的第一行,want
矩阵如下所示:
[,1] [,2] [,3] [,4] [,5]
[1,] 0 1 1 1 1
[2,] 1 0 1 1 1
[3,] 1 1 0 0 1
[4,] 1 1 0 0 0
[5,] 1 1 1 0 0
如何从 have
到 want
?
这是一个使用 lower.tri
和 upper.tri
unlist(apply(mat, 1, function(x) {
m <- matrix(0, nrow = 5, ncol = 5)
m[lower.tri(m)] <- x
m[upper.tri(m)] <- x
list(m)
}), recursive = F)
#[[1]]
# [,1] [,2] [,3] [,4] [,5]
#[1,] 0 1 1 1 1
#[2,] 1 0 1 1 0
#[3,] 1 1 0 1 1
#[4,] 1 1 0 0 0
#[5,] 1 1 1 0 0
#
#[[2]]
# [,1] [,2] [,3] [,4] [,5]
#[1,] 0 0 0 1 1
#[2,] 0 0 0 1 0
#[3,] 0 1 0 1 1
#[4,] 0 1 0 0 0
#[5,] 1 1 1 0 0
unlist(..., recursive = F)
部分看起来有些笨拙,但对于防止 apply
简化结果和丢弃 dims 是必要的。另一种方法是在 data.frame
上使用 lapply
而不是 matrix
:
lapply(as.data.frame(t(mat)), function(x) {
m <- matrix(0, nrow = 5, ncol = 5)
m[lower.tri(m)] <- x
m[upper.tri(m)] <- x
return(m)
})
给出相同的结果。
示例数据
mat <- as.matrix(read.table(text =
"1 1 1 1 1 1 1 0 1 0
0 0 0 1 1 1 1 0 1 0", header = F))
colnames(mat) <- NULL