r - 强制 model.matrix 遵循 R 公式中各项的顺序

Question

让我们用假数据创建一个矩阵:

data_ex <- data.frame(y = runif(5,0,1), a1 = runif(5,0,1), b2 = runif(5,0,1), 
                      c3 = runif(5,0,1), d4 = runif(5,0,1))
> data_ex
      y    a1    b2    c3    d4
1 0.162 0.221 0.483 0.989 0.558
2 0.445 0.854 0.732 0.723 0.259
3 0.884 0.041 0.893 0.985 0.947
4 0.944 0.718 0.338 0.238 0.592
5 0.094 0.867 0.026 0.334 0.314

模型的公式如下:

forml <- as.formula("y ~ a1 + b2 + a1:c3:d4 + a1:c3  + a1:b2 + a1:b2:c3")
> forml
y ~ a1 + b2 + a1:c3:d4 + a1:c3 + a1:b2 + a1:b2:c3

生成的model.matrix是:

> as.matrix(model.matrix(forml, data_ex))
  (Intercept)    a1    b2 a1:c3 a1:b2 a1:c3:d4 a1:b2:c3
1           1 0.221 0.483 0.218 0.107    0.122    0.105
2           1 0.854 0.732 0.617 0.625    0.160    0.452
3           1 0.041 0.893 0.040 0.036    0.038    0.036
4           1 0.718 0.338 0.171 0.243    0.101    0.058
5           1 0.867 0.026 0.290 0.022    0.091    0.008

正如您所看到的，各列已从最低交互等级到最高交互等级重新排序。我正在寻找一种方法，强制 model.matrix 函数遵循公式中各项的确切顺序。生成的矩阵应如下所示:

> Correct_matrix
  (Intercept)    a1    b2 a1:c3:d4 a1:c3 a1:b2  a1:b2:c3
1           1 0.221 0.107    0.483 0.218  0.122    0.105
2           1 0.854 0.625    0.732 0.617  0.160    0.452
3           1 0.041 0.036    0.893 0.040  0.038    0.036
4           1 0.718 0.243    0.338 0.171  0.101    0.058
5           1 0.867 0.022    0.026 0.290  0.091    0.008

Answer 1

您可以创建术语并使用keep.order = TRUE保持术语的顺序。生成的对象可以与 model.matrix 一起使用。

model.matrix(terms(forml, keep.order = TRUE), data_ex)

结果:

  (Intercept)        a1         b2    a1:c3:d4     a1:c3      a1:b2   a1:b2:c3
1           1 0.4604044 0.10968326 0.198301034 0.3015807 0.05049866 0.03307836
2           1 0.4795555 0.61339588 0.018934135 0.2205621 0.29415737 0.13529189
3           1 0.7560366 0.67036486 0.001418541 0.4465991 0.50682035 0.29938436
4           1 0.4490247 0.69179890 0.135388984 0.1376586 0.31063480 0.09523209
5           1 0.7198557 0.08595737 0.131564438 0.2918157 0.06187690 0.02508371
attr(,"assign")
[1] 0 1 2 3 4 5 6