R 创建交互术语时遇到的问题

我有一个data.frame，有7000行36列。有些列是分类变量。 现在我想使用model.matrix为回归创建一个x矩阵。 在该矩阵中，应该是交互术语，如（V1*V2*V3）。 我试着这样做：

formelhas <- formula(k ~ .^3)
 test <-  model.matrix(formelhas,testahah)

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formelhas如果没有您关于数据结构的进一步信息，将很难回答这个问题

您给出的代码应该可以工作，这表明您可能犯了一些简单的印刷错误。为了确认这一点，让我们做一个快速重现的示例，看看三方交互确实是从内置数据集中的类似代码生成的：

无交互：

> str(colnames(model.matrix(mpg ~ ., mtcars)))
 chr [1:11] "(Intercept)" "cyl" "disp" "hp" "drat" "wt" "qsec" "vs" "am" "gear" "carb"

> str(colnames(model.matrix(mpg ~ .^2, mtcars)))
 chr [1:56] "(Intercept)" "cyl" "disp" "hp" "drat" "wt" "qsec" "vs" "am" "gear" "carb" "cyl:disp" ...

> str(colnames(model.matrix(mpg ~ .^3, mtcars)))
 chr [1:176] "(Intercept)" "cyl" "disp" "hp" "drat" "wt" "qsec" "vs" "am" "gear" "carb" "cyl:disp" ...

双向互动：

> str(colnames(model.matrix(mpg ~ ., mtcars)))
 chr [1:11] "(Intercept)" "cyl" "disp" "hp" "drat" "wt" "qsec" "vs" "am" "gear" "carb"

> str(colnames(model.matrix(mpg ~ .^2, mtcars)))
 chr [1:56] "(Intercept)" "cyl" "disp" "hp" "drat" "wt" "qsec" "vs" "am" "gear" "carb" "cyl:disp" ...

> str(colnames(model.matrix(mpg ~ .^3, mtcars)))
 chr [1:176] "(Intercept)" "cyl" "disp" "hp" "drat" "wt" "qsec" "vs" "am" "gear" "carb" "cyl:disp" ...

三方互动：

> str(colnames(model.matrix(mpg ~ ., mtcars)))
 chr [1:11] "(Intercept)" "cyl" "disp" "hp" "drat" "wt" "qsec" "vs" "am" "gear" "carb"

> str(colnames(model.matrix(mpg ~ .^2, mtcars)))
 chr [1:56] "(Intercept)" "cyl" "disp" "hp" "drat" "wt" "qsec" "vs" "am" "gear" "carb" "cyl:disp" ...

> str(colnames(model.matrix(mpg ~ .^3, mtcars)))
 chr [1:176] "(Intercept)" "cyl" "disp" "hp" "drat" "wt" "qsec" "vs" "am" "gear" "carb" "cyl:disp" ...

他们真的在那里吗

> tail(colnames(model.matrix(mpg ~ .^3, mtcars)))
[1] "qsec:am:carb"   "qsec:gear:carb" "vs:am:gear"     "vs:am:carb"     "vs:gear:carb"   "am:gear:carb"

是的，是的。因此，您的代码应该可以工作。也许你只是误解了什么