以下是http://cran.r-project.org/doc/Rnews/Rnews_2001-3.pdf的代码:
defmacro <- function(..., expr){
expr <- substitute(expr)
a <- substitute(list(...))[-1]
## process the argument list
nn <- names(a)
if (is.null(nn)) nn <- rep("", length(a))
nn
for(i in seq(length=length(a))) {
if (nn[i] == "") {
nn[i] <- paste(a[[i]])
msg <- paste(a[[i]], "not supplied")
a[[i]] <- substitute(stop(foo),
list(foo = msg))
print(a)
}
}
names(a) = nn
a = as.list(a)
ff = eval(substitute(
function() {
tmp = substitute(body)
# # new environment to eval expr
# private_env = new.env()
# pf = parent.frame()
# for(arg_name in names(a)) {
# private_env[[a]] = pf[[a]]
# }
# eval(tmp, private_env)
eval(tmp, parent.frame())
},
list(body = expr)))
formals(ff) = a
mm = match.call()
mm$expr = NULL
mm[[1]] = as.name("macro")
mm_src = c(deparse(mm), deparse(expr))
attr(ff, "source") = mm_src
ff
}
setna = defmacro(a, b, values, expr = {a$b[a$b %in% values] = NA; a})
dat = data.frame(x = 1:4, y = rep(-9, 4))
setna(dat, y, -9)
dat
作者质疑读者想出一个新的defmacro,它在父框架中使用局部变量而不是eval(这可能是危险的,因为它可以修改父框架中的对象)。
我尝试创建一个新环境并从父环境中复制变量,并在那里评估函数体(代码注释掉),但结果是它根本没有评估体。
有人可以帮忙吗?
@bergant建议eval(tmp, new.env())
会这样做,而且当宏没有嵌套时它确实有效,但是我们遇到了一个问题:
#' TODO: doc
#' @export
defmacro <- function(..., expr){
expr <- substitute(expr)
a <- substitute(list(...))[-1]
## process the argument list
nn <- names(a)
if (is.null(nn)) nn <- rep("", length(a))
nn
for(i in seq(length=length(a))) {
if (nn[i] == "") {
nn[i] <- paste(a[[i]])
msg <- paste(a[[i]], "not supplied")
a[[i]] <- substitute(stop(foo),
list(foo = msg))
print(a)
}
}
names(a) = nn
a = as.list(a)
ff = eval(substitute(
function() {
tmp = substitute(body)
eval(tmp, parent.frame())
},
list(body = expr)))
formals(ff) = a
mm = match.call()
mm$expr = NULL
mm[[1]] = as.name("macro")
mm_src = c(deparse(mm), deparse(expr))
attr(ff, "source") = mm_src
ff
}
#' IfLen macro
#'
#' Check whether a object has non-zero length, and
#' eval expression accordingly.
#'
#' @param df An object which can be passed to \code{length}
#' @param body1 If \code{length(df)} is not zero, then this clause is evaluated, otherwise, body2 is evaluated.
#' @param body2 See above.
#'
#' @examples
#' ifLen(c(1, 2), { print('yes!') }, {print("no!")})
#'
#' @author kaiyin
#' @export
ifLen = defmacro(df, body1, body2 = {}, expr = {
if(length(df) != 0) {
body1
} else {
body2
}
})
#' IfLet macro
#'
#' Eval expression x, assign it to a variable, and if that is TRUE, continue
#' to eval expression1, otherwise eval expression2. Inspired by the clojure
#' \code{if-let} macro.
#'
#' @param sym_str a string that will be converted to a symbol to hold value of \code{x}
#' @param x the predicate to be evalueated, and to be assigned to a temporary variable as described in \code{sym_str}
#' @param body1 expression to be evaluated when the temporary variable is TRUE.
#' @param body2 expression to be evaluated when the temporary variable is FALSE.
#'
#' @examples
#' ifLet(..temp.., TRUE, {print(paste("true.", as.character(..temp..)))},
#' {print(paste("false.", as.character(..temp..)))})
#' ifLet("..temp..", TRUE, {print(paste("true.", as.character(..temp..)))},
#' {print(paste("false.", as.character(..temp..)))})
#'
#' @author kaiyin
#' @export
ifLet = defmacro(sym_str, x, body1, body2={}, expr = {
stopifnot(is.character(sym_str))
stopifnot(length(sym_str) == 1)
assign(sym_str, x)
if(eval(as.symbol(sym_str))) {
body1
} else {
body2
}
})
#
#setMethod("ifLet",
# signature(sym = "character", x = "ANY", body1 = "ANY", body2 = "ANY"),
# function(sym, x, body1, body2 = {}) {
# e = new.env()
# sym_str = deparse(substitute(sym))
# ifLet(sym_str, x, body1, body2)
# })
#
##' TODO: doc
##' @export
#setMethod("ifLet",
# signature(sym = "character", x = "ANY", body1 = "ANY", body2 = "ANY"),
# function(sym, x, body1, body2 = {}) {
# stopifnot(length(sym) == 1)
# e = new.env()
# assign(sym, x, envir = e)
# if(e[[sym]]) {
# eval(substitute(body1), e, parent.frame())
# } else {
# eval(substitute(body2), e, parent.frame())
# }
# })
#' IfLetLen macro
#'
#' Similar to ifLet, but conditioned on whether the length of
#' the result of \code{eval(x)} is 0.
#'
#'
#' @param x the predicate to be evalueated, and to be assigned to a temporary var called \code{..temp..}
#' @param body1 expression to be evaluated when \code{..temp..} is TRUE.
#' @param body2 expression to be evaluated when \code{..temp..} is FALSE.
#'
#' @examples
#' ifLetLen("..temp..", 1:3, {print(paste("true.", as.character(..temp..)))},
#' {print(paste("false.", as.character(..temp..)))})
#'
#' @author kaiyin
#' @export
ifLetLen = defmacro(sym_str, x, body1, body2={}, expr = {
stopifnot(is.character(sym_str))
stopifnot(length(sym_str) == 1)
assign(sym_str, x)
ifLen(eval(as.symbol(sym_str)), {
body1
}, {
body2
})
})
如果您运行此测试:
ifLetLen("..temp..", 1:3, {print(paste("true.", as.character(..temp..)))},
{print(paste("false.", as.character(..temp..)))})
您将获得object not found error
。
答案 0 :(得分:2)
您可以将环境作为属性添加到defmacro
:
defmacro <- function(..., expr, env = parent.frame()){
expr <- substitute(expr)
a <- substitute(list(...))[-1]
## process the argument list
nn <- names(a)
if (is.null(nn)) nn <- rep("", length(a))
nn
for(i in seq(length=length(a))) {
if (nn[i] == "") {
nn[i] <- paste(a[[i]])
msg <- paste(a[[i]], "not supplied")
a[[i]] <- substitute(stop(foo),
list(foo = msg))
print(a)
}
}
names(a) = nn
a = as.list(a)
ff = eval(substitute(
function() {
tmp = substitute(body)
eval(tmp, env)
},
list(body = expr)))
formals(ff) = a
mm = match.call()
mm$expr = NULL
mm[[1]] = as.name("macro")
mm_src = c(deparse(mm), deparse(expr))
attr(ff, "source") = mm_src
ff
}
我们在这里使用new.env
:
ifLen = defmacro(df, body1, body2 = {}, expr = {
if(length(df) != 0) {
body1
} else {
body2
}
}, env = new.env())
但在这里我们不是:
ifLetLen = defmacro(sym_str, x, body1, body2={}, expr = {
stopifnot(is.character(sym_str))
stopifnot(length(sym_str) == 1)
assign(sym_str, x)
ifLen(eval(as.symbol(sym_str)), {
body1
}, {
body2
})
})
ifLetLen("..temp..", 1:3, {print(paste("true.", as.character(..temp..)))},
{print(paste("false.", as.character(..temp..))); xxx <- 69})
# [1] "true. 1" "true. 2" "true. 3"
第一个例子:
setna = defmacro(a, b, values, expr = {a$b[a$b %in% values] = NA; a}, env = new.env())
dat = data.frame(x = 1:4, y = rep(-9, 4))
> setna(dat, y, -9)
# x y
# 1 1 NA
# 2 2 NA
# 3 3 NA
# 4 4 NA
> dat
# x y
# 1 1 -9
# 2 2 -9
# 3 3 -9
# 4 4 -9
提出的解决方案的问题在于您必须关注环境(什么功能和表达式评估的位置可见)。我认为它不是一个非常透明的编程工具。
注意:它没有解决局部变量的问题(来自原始论文) - 它只是把所有东西放在不同的环境中(就像典型的R函数一样)。