Question

我对R有些新之处,而且我有这样一部法典,它会产生一个不了解哪类的变量。 R是否有任何预测设施可以告诉我这种变数属于哪一类?

以下是这一变量的财产:

我正在研究线性模型选择,而我拥有的资源是另一个模型产生的<代码>lm。现在我想检索<代码>lm。指挥摘要(模拟)呼吁,以便我无需硬化示范结构。然而,由于我不得不改变数据集,我需要就“扼杀”作一些修改,但显然这并非简单明了。我想知道是否有任何指挥类似于扼杀。取而代之的是,我能够从可变的召唤中操纵这一变量。

> str<-summary(rdnM)$call
> str
lm(formula = y ~ x1, data = rdndat)
> str[1]
lm()
> str[2]
y ~ x1()
> str[3]
rdndat()
> str[3] <- data
Warning message:
In str[3] <- data :
  number of items to replace is not a multiple of replacement length
> str
lm(formula = y ~ x1, data = c(10, 20, 30, 40))
> str<-summary(rdnM)$call
> str
lm(formula = y ~ x1, data = rdndat)
> str[3] <-  data 
> str
lm(formula = y ~ x1, data = "data")
> str<-summary(rdnM)$call
> type str
Error: unexpected symbol in "type str"
>

Answer 1

In terms of introspection: R allows you to easily examine and operate on language objects.
For more details, see R Language Definition, particularly sections 2 and 6. For instance, in your case, summary(rdnM)$call is a "call" object. You can retrieve pieces of it by indexing, but you can t construct another call object by assigning to indices like you are trying to do. You d have to construct a new call.

在你的情况下,你正在从现有呼吁中发出一条最新的呼吁。如果您希望重新使用关于不同数据的公式,请通过<代码>formula(foo$ calls)从呼吁标的中提取公式。

 foo <- lm(formula = y ~ x1, data = data.frame(y=rnorm(10),x1=rnorm(10)))
 bar <- lm(formula(foo$call), data = data.frame(y=rnorm(10),x1=rnorm(10)))

另一方面,如果你试图更新该公式,你可使用<编码>更新日期():

baz <- update(bar, . ~ . - 1)
baz$call
##>lm(formula = y ~ x1 - 1, data = data.frame(y = rnorm(10), x1 = rnorm(10)))

友情链接