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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

该模型的公式如下：

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

结果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

如您所见，列从最低交互等级重新排序到最高。我正在寻找一种强制model.matrix函数遵循公式中术语的确切顺序的方法。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
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1 回答 1

6

您可以使用 来创建terms并保持条款的顺序keep.order = TRUE。生成的对象可以与model.matrix.

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")
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于 2017-08-28T15:29:40.313 回答