使用`assign()`为列表项赋值

Question

一点点背景首先......

我写了一个中缀函数,实质上取代了成语

x[[length(x) +1]] <- y

..或仅仅是x <- append(x, y)为了矢量.

这里是:

`%+=%` <- function(x, y) {
  xcall <- substitute(x)
  xobjname <- setdiff(all.names(xcall), c("[[", "[", ":", "$"))
  # if the object doesn't exist, create it
  if (!exists(xobjname, parent.frame(), mode = "list") &&
      !exists(xobjname, parent.frame(), mode = "numeric") &&
      !exists(xobjname, parent.frame(), mode = "character")) {
    xobj <- subset(y, FALSE)
  } else {
    xobj <- eval(xcall, envir = parent.frame())
  }

  if (is.atomic(xobj)) {
    if (!is.atomic(y)) {
      stop('Cannot append object of mode ', dQuote(mode(y)), 
           ' to atomic structure ', xobjname)
    }
    assign(xobjname, append(xobj, y), envir = parent.frame())
    return(invisible())
  }

  if (is.list(xobj)) {
    if (is.atomic(y)) {
      xobj[[length(xobj) + 1]] <- y
    } else {
      for (i in seq_along(y)) {
        xobj[[length(xobj) + 1]] <- y[[i]]
        names(xobj)[length(xobj)] <- names(y[i])
      }
    }
    assign(xobjname, xobj, envir = parent.frame())
    return(invisible())
  }

  stop("Can't append to an object of mode ", 
       mode(eval(xcall, envir = parent.frame())))
}

它按照预期的方式使用向量或列表,但其当前形式的限制是我不能将值附加到列表中的项目,例如:

a <- list(a = 1, b = 2)
a$b %+=% 3

到目前为止,我还没有找到如何做到这一点.我尝试了类似下面的内容,但它没有效果:

assign("b", append(a$b, 3), envir = as.environment(a))

有任何想法吗？

Answer 1

建议不要使用assign而是:

`%+=%`<- function(x, value) eval.parent(substitute(x <- append(x, value)))

x <- 3
x %+=% 5
x
## [1] 3 5

L <- list(a = 1, b = 2)
L %+=% 3
## List of 3
## $ a: num 1
## $ b: num 2
## $  : num 3

L <- list(a = 1, b = 2)
L$a %+=% 4
str(L)
## List of 2
##  $ a: num [1:2] 1 4
##  $ b: num 2

或尝试+ < - 语法,避免eval:

`+<-` <- append

# test
x <- 3
+x <- 1
x
## [1] 3 1

# test
L<- list(a = 1, b = 2)
+L <- 10
str(L)
## List of 3
##  $ a: num 1
##  $ b: num 2
##  $  : num 10

# test
L <- list(a = 1, b = 2)
+L$a <- 10
str(L)
## List of 2
##  $ a: num [1:2] 1 10
##  $ b: num 2

或者尝试这种类似的替换函数语法+<-.

`append<-` <- append
x <- 3
append(x) <- 7
## [1] 3 7

... etc ...