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]]
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
})
and
var <- 0
and
f1 <- function() {
sum_var <- 0
sum_var2 = 0
sum_var3 <- 0
}
and
f2 <- (function() {
sum_var <- 0
sum_var2 = 0
sum_var3 <- 0
})
and
f3 = function(x) {
sum_var <- 0
sum_var2 = 0
sum_var3 <- 0
}
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?
can I be sure that I always get "<-" operator in syntax tree
Let's see …
> quote(b -> a)
a <- b
> identical(quote(b -> a), quote(a <- b))
[1] TRUE
So yes, the ->
assignment is always parsed as <-
(the same is not true when invoking ->
as a function name! 1 ).
Your first display is the other way round because of parse
's keep.source
argument :
> parse(text = 'b -> a')
expression(b -> a)
> parse(text = 'b -> a', keep.source = FALSE)
expression(a <- b)
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
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.
This is slightly off topic, but I think that a left-to-right assignment operator can be used in cases beyond fast command-line typing. So, here's one way to define such an operator:
"%op%"<- function(a,b) eval(substitute(b<-a), parent.frame())
I am using this to avoid duplicating R-code when storing/loading a set of matrices and vectors to/from a vector ( Error: object '->' not found in R ).
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