[英]variable-scope in R functions
I'd like to specify functions in a flexible way. 我想以灵活的方式指定功能。 How can I make sure that the environment of a given function does not change when I create another function just after it. 当我在其后创建另一个函数时,如何确保给定函数的环境不会改变。
To illustrate, this works properly: 为了说明,这适用于:
make.fn2 <- function(a, b) {
fn2 <- function(x) {
return( x + a + b )
}
return( fn2 )
}
a <- 2; b <- 3
fn2.1 <- make.fn2(a, b)
fn2.1(3) # 8
fn2.1(4) # 9
a <- 4
fn2.2 <- make.fn2(a, b)
fn2.2(3) # 10
fn2.1(3) # 8
This does not 事实并非如此
make.fn2 <- function(a, b) {
fn2 <- function(x) {
return( x + a + b )
}
return( fn2 )
}
a <- 2; b <- 3
fn2.1 <- make.fn2(a, b)
a <- 4
fn2.2 <- make.fn2(a, b)
fn2.1(3) # 10
fn2.1(4) # 11
fn2.2(3) # 10
fn2.1(3) # 10
This is due to lazy evaluation. 这是由于懒惰的评估。 The function is not actually constructed until it is called. 在调用函数之前,该函数实际上并未构造。 So, in the second case, both times the new version of a
is picked up. 因此,在第二种情况下,两次都会获取新版本的a
。 See also this other question . 另见其他问题 。
You can solve this issue by using force
: 您可以使用force
来解决此问题:
make.fn2 <- function(a, b) {
force(a)
force(b)
fn2 <- function(x) {
return( x + a + b )
}
return( fn2 )
}
This forces the function to pick up the values of a
and b
when the function is created, not when the function is called. 这会强制函数在创建函数时获取a
和b
的值,而不是在调用函数时。 It produces the correct output: 它产生正确的输出:
> a <- 2; b <- 3
> fn2.1 <- make.fn2(a, b)
>
> a <- 4
> fn2.2 <- make.fn2(a, b)
>
> fn2.1(3) # 10
[1] 8
> fn2.1(4) # 11
[1] 9
> fn2.2(3) # 10
[1] 10
> fn2.1(3) # 10
[1] 8
>
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.