英文:
Create a specific selection matrix
问题
如果我有n个主题，每个主题重复t次。如果我想创建一个选择矩阵，它将如下所示：
n = 5t = 3Select_M = diag(n) %x% matrix(1, t)Select_M[,1] [,2] [,3] [,4] [,5][1,] 1 0 0 0 0[2,] 1 0 0 0 0[3,] 1 0 0 0 0[4,] 0 1 0 0 0[5,] 0 1 0 0 0[6,] 0 1 0 0 0[7,] 0 0 1 0 0[8,] 0 0 1 0 0[9,] 0 0 1 0 0[10,] 0 0 0 1 0[11,] 0 0 0 1 0[12,] 0 0 0 1 0[13,] 0 0 0 0 1[14,] 0 0 0 0 1[15,] 0 0 0 0 1
我希望为每个主题创建不同的时间段。换句话说，第一个主题重复7次，第二个主题重复11次，依此类推。
我如何高效地为这些特定的重复创建选择矩阵？
英文:
Say, I have n subjects and each is repeated t times. If I want to create a selection matrix, it would be as follows
n = 5t = 3Select_M = diag(n) %x% matrix(1, t)Select_M[,1] [,2] [,3] [,4] [,5][1,] 1 0 0 0 0[2,] 1 0 0 0 0[3,] 1 0 0 0 0[4,] 0 1 0 0 0[5,] 0 1 0 0 0[6,] 0 1 0 0 0[7,] 0 0 1 0 0[8,] 0 0 1 0 0[9,] 0 0 1 0 0[10,] 0 0 0 1 0[11,] 0 0 0 1 0[12,] 0 0 0 1 0[13,] 0 0 0 0 1[14,] 0 0 0 0 1[15,] 0 0 0 0 1
My interest is to have different time periods for each subject. In other words, the first subject is repeated 7 times, the second subject is repeated 11 times, and so on.
How can I efficiently create a selection matrix for these specific repeats?
答案1
得分: 2
你可以定义一个表示不同时间段的因子列，然后使用 model.matrix() 来创建一个设计矩阵，不包括截距项。
例如，第一个主题重复了2次，第二个主题重复了3次，第三个主题重复了4次，你可以尝试如下：
t = c(2, 3, 4)data.frame(sub = factor(rep(seq_along(t), t))) |>model.matrix(~ sub - 1, data = _)# sub1 sub2 sub3# 1 1 0 0# 2 1 0 0# 3 0 1 0# 4 0 1 0# 5 0 1 0# 6 0 0 1# 7 0 0 1# 8 0 0 1# 9 0 0 1
注意：我只提供翻译，请不要再问我需要翻译的内容。
英文:
You can define a factor column indicating different time periods, and then use model.matrix() to create a design matrix from it without the intercept term.
For example, the first subject is repeated 2 times, the second subject 3 times, and the third subject 4 times, you can try as follows:
t = c(2, 3, 4)data.frame(sub = factor(rep(seq_along(t), t))) |>model.matrix(~ sub - 1, data = _)# sub1 sub2 sub3# 1 1 0 0# 2 1 0 0# 3 0 1 0# 4 0 1 0# 5 0 1 0# 6 0 0 1# 7 0 0 1# 8 0 0 1# 9 0 0 1
答案2
得分: 2
使用 bdiag 函数
library(Matrix)t <- 2:4as.matrix(bdiag(lapply(as.list(t), \(x) rep(1, x))))
输出
[,1] [,2] [,3][1,] 1 0 0[2,] 1 0 0[3,] 0 1 0[4,] 0 1 0[5,] 0 1 0[6,] 0 0 1[7,] 0 0 1[8,] 0 0 1[9,] 0 0 1
英文:
Using bdiag
library(Matrix)t <- 2:4as.matrix(bdiag(lapply(as.list(t), \(x) rep(1, x))))
-output
[,1] [,2] [,3][1,] 1 0 0[2,] 1 0 0[3,] 0 1 0[4,] 0 1 0[5,] 0 1 0[6,] 0 0 1[7,] 0 0 1[8,] 0 0 1[9,] 0 0 1
答案3
得分: 2
可以使用以下基本的R代码：
> v <- 2:4> +(col(m <- replicate(length(v), rep(seq_along(v), v))) == m)[,1] [,2] [,3][1,] 1 0 0[2,] 1 0 0[3,] 0 1 0[4,] 0 1 0[5,] 0 1 0[6,] 0 0 1[7,] 0 0 1[8,] 0 0 1[9,] 0 0 1
英文:
We can use the following base R code
> v <- 2:4> +(col(m <- replicate(length(v), rep(seq_along(v), v))) == m)[,1] [,2] [,3][1,] 1 0 0[2,] 1 0 0[3,] 0 1 0[4,] 0 1 0[5,] 0 1 0[6,] 0 0 1[7,] 0 0 1[8,] 0 0 1[9,] 0 0 1
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