问题

y ~ x + x:x 似乎等同于 y ~ x，而 y ~ x + I(x^2) 正确地包括了二次项在模型中。

为什么我不能将二次项写为一个变量与其自身的交互？

答案1

得分: 1

我理解，以下是翻译好的部分：

1. "A start at an answer is that while :, in practice, typically multiplies the numeric columns associated with the variable (e.g. if x and y are both numeric, x:y creates an interaction column that is the product of x and y), the root meaning of :` in R's formula syntax is not "multiply columns" but "form an interaction". The interaction of a variable with itself is just itself."

   • "一个初步的答案是，虽然:在实践中通常将与变量相关的数值列相乘（例如，如果x和y都是数字，x:y会创建一个交互列，该列是x和y的乘积），但:在R的公式语法中的根本含义并不是“相乘列”，而是“形成一个交互作用”。变量与自身的交互作用就是它自己。"

2. "I would love to have a complete formal description of R's version of Wilkinson-Rogers syntax (which is what this is), but I don't know that one exists. The original framing of the formula language is in Wilkinson and Rogers (1973) [where . rather than : was used for the "interaction" operator]; I believe there's a description in the "White Book" (Chambers and Hastie 1992); but other than that, I think the only full definition is the source code of model.matrix() itself (which is not all that nice to look at ...)"

   • "我很想拥有R版本的Wilkinson-Rogers语法的完整正式描述（这就是它的名称），但我不知道是否存在这样的描述。公式语言的最初构架出现在Wilkinson和Rogers（1973）中[在那里，用.而不是:来表示“交互作用”操作符]；我相信在“白皮书”（Chambers和Hastie 1992）中有一份描述；但除此之外，我认为唯一的完整定义就是model.matrix()的源代码本身（虽然不太容易阅读...）。"

3. 参考文献：

   • Chambers, J. M., and T. Hastie, eds. Statistical Models in S. Wadsworth & Brooks/Cole, 1992.
   • Wilkinson, G. N., and C. E. Rogers. "Symbolic Description of Factorial Models for Analysis of Variance." Journal of the Royal Statistical Society. Series C (Applied Statistics) 22, no. 3 (1973): 392–99. https://doi.org/10.2307/2346786.

答案2

得分: 0

为了轻松地构造多项式，我们可以使用 poly。

lm(Petal.Length ~ Sepal.Length + I(Sepal.Length^2) + I(Sepal.Length^3), iris)$coe
# (Intercept)      Sepal.Length I(Sepal.Length^2) I(Sepal.Length^3) 
#  19.8028068       -13.5808046         2.8767502        -0.1742277

lm(Petal.Length ~ poly(Sepal.Length, 3, raw=TRUE), iris)$coe
# (Intercept) poly(Sepal.Length, 3, raw = TRUE)1 
#  19.8028068                        -13.5808046 
# poly(Sepal.Length, 3, raw = TRUE)2 poly(Sepal.Length, 3, raw = TRUE)3 
#                          2.8767502                         -0.1742277

lm(Petal.Length ~ Sepal.Length + I(Sepal.Length^2) + I(Sepal.Length^3), iris)$coe
# (Intercept)      Sepal.Length I(Sepal.Length^2) I(Sepal.Length^3) 
#  19.8028068       -13.5808046         2.8767502        -0.1742277

lm(Petal.Length ~ poly(Sepal.Length, 3, raw=TRUE), iris)$coe
# (Intercept) poly(Sepal.Length, 3, raw = TRUE)1 
#  19.8028068                        -13.5808046 
# poly(Sepal.Length, 3, raw = TRUE)2 poly(Sepal.Length, 3, raw = TRUE)3 
#                          2.8767502                         -0.1742277