3
這工作調用`〜`與前綴約定
lm(mpg ~ cyl, mtcars)
那麼,這是否:
lm(`~`(mpg, cyl), mtcars)
但這並不:
lm(base::`~`(mpg, cyl), mtcars)
Error in terms.formula(formula, data = data) :
argument is not a valid model
爲什麼第三種情況會失敗?