解析“-" R中的赋值运算符 [英] Parsing "->" assignment operator in R

问题描述

我的问题是关于用R语言解析表达式.让我跳入一个例子:

My question is about parsing expressions in R language. Let me jump right into an example:

fun_text <- c("
0 -> var
f1 <- function()
{
    0 -> sum_var
    sum_var2 = 0
    sum_var3 <- 0
}

(function()
{
    0 -> sum_var
    sum_var2 = 0
    sum_var3 <- 0
})->f2

f3 = function(x)
{
  0 -> sum_var
  sum_var2 = 0
  sum_var3 <- 0
}

")

fun_tree <- parse(text=fun_text)
fun_tree 
fun_tree[[1]]
fun_tree[[2]]
fun_tree[[3]]
fun_tree[[4]]

之后，我们获得这些结果:

After that, we obtain those results:

expression(0 -> var, f1 <- function()
{
    0 -> sum_var
    sum_var2 = 0
    sum_var3 <- 0
}, (function()
{
    0 -> sum_var
    sum_var2 = 0
    sum_var3 <- 0
})->f2, f3 = function(x)
{
    0 -> sum_var
    sum_var2 = 0
    sum_var3 <- 0
})

和

var <- 0

和

f1 <- function() {
    sum_var <- 0
    sum_var2 = 0
    sum_var3 <- 0
}

和

f2 <- (function() {
    sum_var <- 0
    sum_var2 = 0
    sum_var3 <- 0
})

和

f3 = function(x) {
    sum_var <- 0
    sum_var2 = 0
    sum_var3 <- 0
}

如您所见，所有->"赋值运算符均更改为<--"，但在第一个示例中没有更改(仅"fun_tree").我的问题是:为什么呢?并且可以确定我总是在语法树中得到<-"运算符，这样我就不会在实现->"大小写时费心吗?

As you can see, all "->" assignment operators are changed to "<-", but not in the first example ("fun_tree" only). My question is: why is that? and can I be sure that I always get "<-" operator in syntax tree, so I can do not bother myself in implementing "->" case?

推荐答案

我可以确定我总是在语法树中得到<-"运算符

can I be sure that I always get "<-" operator in syntax tree

让我们看看...

> quote(b -> a)
a <- b
> identical(quote(b -> a), quote(a <- b))
[1] TRUE

所以是的，->分配始终被解析为<-(当调用->作为函数名称时，不为true！ ¹ )

So yes, the -> assignment is always parsed as <- (the same is not true when invoking -> as a function name!¹).

由于 parse的keep.source参数:

> parse(text = 'b -> a')
expression(b -> a)
> parse(text = 'b -> a', keep.source = FALSE)
expression(a <- b)

---

¹ 将<-作为函数调用与将其用作运算符相同:

---

¹ Invoking <- as a function is the same as using it as an operator:

> quote(`<-`(a, b))
a <- b
> identical(quote(a <- b), quote(`<-`(a, b)))
[1] TRUE

但是，没有->函数(尽管您可以定义一个函数)，并且编写b -> a从不调用->函数，它总是被解析为a <- b，进而调用<-函数或基元.

However, there is no -> function (although you can define one), and writing b -> a never calls a -> function, it always gets parsed as a <- b, which, in turn, invokes the <- function or primitive.

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