R中用于因子实验的无约束设计矩阵

Question

我希望在R中为因子实验创建一个不受限制的设计矩阵，以下代码为我提供了所需的矩阵。但是该代码要求对每个因子以及拦截项分别使用model.matrix命令。我想知道是否可以通过一行代码获得相同的结果。谢谢。

y <- c(55, 56, 57, 53, 54, 55, 51, 52, 53, 61, 62, 63)
N <- gl(n = 2, k = 6, length = 2 * 6
        , labels = c("Low", "High")
        , ordered = FALSE)
P <- gl(n = 2, k = 3, length = 2 * 6
        , labels = c("Low", "High")
        , ordered = FALSE)
Data <- data.frame(y, N, P)

X <-
  cbind(
      model.matrix(object = y ~ 1,        data = Data)
    , model.matrix(object = y ~ -1 + N,   data = Data)
    , model.matrix(object = y ~ -1 + P,   data = Data)
    , model.matrix(object = y ~ -1 + N:P, data = Data)
    )

print(x = X)

- MYaseen208

Answer 1

我认为关键是将所有的对比度都设置为FALSE。从技术上讲，这可能只需要一行代码...只是这会变成一行很长的代码。

model.matrix(y ~ N +P + N:P, data=Data, 
      contrasts.arg = lapply(Data[,sapply(Data, is.factor)], 
                             contrasts, contrasts=FALSE))


   (Intercept) NLow NHigh PLow PHigh NLow:PLow NHigh:PLow NLow:PHigh NHigh:PHigh
1            1    1     0    1     0         1          0          0           0
2            1    1     0    1     0         1          0          0           0
3            1    1     0    1     0         1          0          0           0
4            1    1     0    0     1         0          0          1           0
5            1    1     0    0     1         0          0          1           0
6            1    1     0    0     1         0          0          1           0
7            1    0     1    1     0         0          1          0           0
8            1    0     1    1     0         0          1          0           0
9            1    0     1    1     0         0          1          0           0
10           1    0     1    0     1         0          0          0           1
11           1    0     1    0     1         0          0          0           1
12           1    0     1    0     1         0          0          0           1
attr(,"assign")
[1] 0 1 1 2 2 3 3 3 3
attr(,"contrasts")
attr(,"contrasts")$N
     Low High
Low    1    0
High   0    1

attr(,"contrasts")$P
     Low High
Low    1    0
High   0    1

- SchaunW

Answer 2

虽不是一行代码，但也许更简明易懂:

contrasts(N, nlevels(N)) <- diag(nlevels(N))
contrasts(P, nlevels(P)) <- diag(nlevels(P))
Data2 <- data.frame(y, N, P)
X2 <- model.matrix(y ~ 1 + N + P + N:P, data=Data2)

- Jan van der Laan