将矢量转换为逻辑矩阵

时间:2012-08-15 13:57:33

标签: r

是否存在一个原生R函数,它将获取一个输入向量并返回相应的二进制矩阵,其中矩阵的列数与输入向量中的唯一值相同?

例如,给定x <- 1:3,我想返回以下矩阵:

     [,1] [,2] [,3]
[1,]    1    0    0
[2,]    0    1    0
[3,]    0    0    1

函数contrasts接近,但我似乎无法绕过返回的n-1列:

> contrasts(as.factor(x))
  2 3
1 0 0
2 1 0
3 0 1

2 个答案:

答案 0 :(得分:5)

model.matrix()可能会有所帮助,但你需要抑制拦截:

> model.matrix(~ factor(1:3) - 1)
  factor(1:3)1 factor(1:3)2 factor(1:3)3
1            1            0            0
2            0            1            0
3            0            0            1
attr(,"assign")
[1] 1 1 1
attr(,"contrasts")
attr(,"contrasts")$`factor(1:3)`
[1] "contr.treatment"

稍微复杂一些:

> set.seed(1)
> fac <- factor(sample(1:3, 10, replace = TRUE))
> model.matrix(~ fac - 1)
   fac1 fac2 fac3
1     1    0    0
2     0    1    0
3     0    1    0
4     0    0    1
5     1    0    0
6     0    0    1
7     0    0    1
8     0    1    0
9     0    1    0
10    1    0    0
attr(,"assign")
[1] 1 1 1
attr(,"contrasts")
attr(,"contrasts")$fac
[1] "contr.treatment"

答案 1 :(得分:5)

实际上,contrasts就是你想要的。

contrasts(as.factor(1:3), contrasts=FALSE)

  1 2 3
1 1 0 0
2 0 1 0
3 0 0 1