4

假设我有一组x, y可能定义或未定义的变量。这些变量被传递到一个名为test.

y <- 10
test <- function(a,b) { ifelse(a > b, "hello", "world") }
test(x,y)

# Error in ifelse(a > b, "hello", "world") : object 'x' not found

如果我test(x,y)在 x 尚未实例化时调用,R 将抛出“未找到对象'x'”错误。

如果我添加了存在检查,则该函数在从全局环境中调用它时会起作用

y <- 10
test <- function(a,b) { 
     print(exists(as.character(substitute(a))))
     if (!exists(as.character(substitute(a)))) {a <- 0}
     ifelse(a > b, "hello", "world")  
}
test(x,y)

# [1] FALSE
# [1] "world"

x <- 11
test(x,y)

[1] TRUE
[1] "hello"

但是,如果我包装test(x,y)blah函数中。它无法找到现有的变量。

rm(list=ls())
test <- function(a,b) { 
     print(exists(as.character(substitute(a))))
     if (!exists(as.character(substitute(a)))) {a <- 0}
     ifelse(a > b, "hello", "world")  
}
blah <- function() { x <- 11; y <- 10; test(x,y)}
blah()
[1] FALSE -- expecting TRUE
[1] "world" -- expecting "hello"

我猜失败是因为它没有在正确的环境中寻找。知道我怎样才能让它正常工作吗?

4

1 回答 1

5

您可以指定首先查看的环境:

test <- function(a,b) { 
     print(exists(as.character(substitute(a)), envir=parent.frame()))
     if (!exists(as.character(substitute(a)), envir=parent.frame())) {a <- 0}
     ifelse(a > b, "hello", "world")  
}

这边走:

y <- 10
test(x,y)

# [1] FALSE
# [1] "world"

x <- 11
test(x,y)

#[1] TRUE
#[1] "hello"

rm(list=ls())

test <- function(a,b) { 
     print(exists(as.character(substitute(a)), envir=parent.frame()))
     if (!exists(as.character(substitute(a)), envir=parent.frame())) {a <- 0}
     ifelse(a > b, "hello", "world")  
}
blah <- function() { x <- 11; y <- 10; test(x,y)}
blah()

#[1] TRUE
#[1] "hello"
于 2015-01-12T21:24:35.287 回答