将 R 表达式传递给函数，然后更改其输入参数

发布于 2025-01-13 15:16:42 字数 1367 浏览 1 评论 0原文

我想将 R 表达式传递给自己编写的函数，而不需要首先评估它，然后操纵输入参数该函数中的 R 表达式。

我想，用一个例子更容易解释。

在下面的示例函数 fun1 中，应采用 lm(y ~ x, data=datatrain) 作为参数，不带一开始评价，我想更换 lm(y ~ x, data=datatrain) 中的 datatrain 通过函数内的 newdatatrain （更准确地说， newdatatrain 将是带有一些列的 datatrain 在函数内排列）。

我需要提供 lm(y ~ x, data=datatrain) 的原因作为函数参数，函数应该也能够采用其他类型的模型以及可能的其他模型输入参数（例如， randomForest::randomForest(y ~ x, data=datatrain, Replace=FALSE))。

这是我尝试过的：

# Does not run through:

fun1 <- function(mycall) {
  
  obj <- quote(mycall)
  
  newdatatrain <- data.frame(y=c(0.22,  1.09,  0.93, -0.02, -0.27), 
                             x=c(1.50,  0.53,  0.04, -2.43, -0.54))
  
  obj$data <- newdatatrain
  
  res <- eval(obj)
  
  return(res)
  
}

fun1(lm(y ~ x, data=datatrain))



# Does run through, but is not
# practical, because lm(y ~ x, data=datatrain)
# should be a function argument:


fun2 <- function(mycall) {
  
  obj <- quote(lm(y ~ x, data=datatrain))
  
  newdatatrain <- data.frame(y=c(0.22,  1.09,  0.93, -0.02, -0.27), 
                             x=c(1.50,  0.53,  0.04, -2.43, -0.54))
  
  obj$data <- newdatatrain
  
  res <- eval(obj)
  
  return(res)
  
}

fun2(lm(y ~ x, data=datatrain))

如何解决这个问题？

原文

I would like to pass an R expression to a self-written function without
evaluating it at first and then manipulate an input parameter of
the R expression in that function.

I guess, it is easier to explain with an example.

In the below example function fun1, which should take
lm(y ~ x, data=datatrain) as an argument without
evaluating it at first, I would like to replace
datatrain in lm(y ~ x, data=datatrain) by
newdatatrain within the function (to be more precise,
newdatatrain will be datatrain with some columns
permuted within the function).

The reason why I need to provide lm(y ~ x, data=datatrain)
as a function argument is that the function should
also be able to take other types of models with possibly other
input parameters (e.g.,
randomForest::randomForest(y ~ x, data=datatrain, replace=FALSE)).

Here is what I tried:

# Does not run through:

fun1 <- function(mycall) {
  
  obj <- quote(mycall)
  
  newdatatrain <- data.frame(y=c(0.22,  1.09,  0.93, -0.02, -0.27), 
                             x=c(1.50,  0.53,  0.04, -2.43, -0.54))
  
  obj$data <- newdatatrain
  
  res <- eval(obj)
  
  return(res)
  
}

fun1(lm(y ~ x, data=datatrain))



# Does run through, but is not
# practical, because lm(y ~ x, data=datatrain)
# should be a function argument:


fun2 <- function(mycall) {
  
  obj <- quote(lm(y ~ x, data=datatrain))
  
  newdatatrain <- data.frame(y=c(0.22,  1.09,  0.93, -0.02, -0.27), 
                             x=c(1.50,  0.53,  0.04, -2.43, -0.54))
  
  obj$data <- newdatatrain
  
  res <- eval(obj)
  
  return(res)
  
}

fun2(lm(y ~ x, data=datatrain))

How could this be tackled?

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℡寂寞咖啡 2025-01-20 15:16:42

如果我误解了，我很抱歉，但是你能不能做这样的事情：

f <- function(estimator, formula, data) {
  newdata = data[sample(nrow(data),2),]
  predict(estimator(formula,data=data), newdata=newdata)
}

f(lm, y ~ x, datatrain)
f(glm, y ~ x, data=datatrain)

I'm sorry if I misunderstood, but can you not do something like this:

f <- function(estimator, formula, data) {
  newdata = data[sample(nrow(data),2),]
  predict(estimator(formula,data=data), newdata=newdata)
}

f(lm, y ~ x, datatrain)
f(glm, y ~ x, data=datatrain)

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