R将布局对象的网格单位转换为原生

我的问题与在R中使用grid将单位从npc转换为native有关。

我试图找出某些绘图元素的位置始于ggplot2对象（轴，主绘图等）。我发现以下代码：

rm(list = ls())
library(ggplot2)
library(grid)
library(gtable)

# a dummy plot
g <- ggplot(cars, aes(x = speed, y = dist)) + 
  geom_point()
g

# a layout of each element
obj <- ggplotGrob(g)
l <- gtable:::gtable_layout(obj)
grid:::grid.show.layout(l)

我需要的所有信息都必须在名为的布局对象中l。但是，此对象的高度和宽度很奇怪。即使布局有些吸引，它们也通常为零！我进行了调整，grid:::grid.show.layout以打印其绘制的尺寸：

# aside from sprintf and cat a copy of grid:::grid.show.layout
foo <- function(l, newpage = TRUE, vp.ex = 0.8, bg = "light grey", 
                cell.border = "blue", cell.fill = "light blue", cell.label = TRUE, 
                label.col = "blue", unit.col = "red", vp = NULL, ...) {
  if (!grid:::is.layout(l)) 
    stop("'l' must be a layout")
  if (newpage) 
    grid.newpage()
  if (!is.null(vp)) 
    pushViewport(vp)
  grid.rect(gp = gpar(col = NULL, fill = bg))
  vp.mid <- viewport(0.5, 0.5, vp.ex, vp.ex, layout = l)
  pushViewport(vp.mid)
  grid.rect(gp = gpar(fill = "white"))
  gp.red <- gpar(col = unit.col)
  objs <- matrix(list(), l$nrow, l$ncol)

  unitType <- "cm"
  for (i in 1L:l$nrow) for (j in 1L:l$ncol) {

    h <- convertX(x = l$heights[i, top = FALSE], unitTo = unitType)
    w <- convertY(x = l$widths[j, top = FALSE], unitTo = unitType)
    s1 <- sprintf("s1: i = %d, j = %d, height = %s, width = %s\n", i, j, h, w)
    cat(s1)

    vp.inner <- viewport(layout.pos.row = i, layout.pos.col = j)
    pushViewport(vp.inner)

    # an attempt so save the drawn objects
    objs[[i, j]] <- grid.rect(gp = gpar(col = cell.border, fill = cell.fill))
    if (cell.label) 
      grid.text(paste0("(", i, ", ", j, ")"), gp = gpar(col = label.col))
    if (j == 1) 
      grid.text(format(l$heights[i, top = FALSE], ...), 
                gp = gp.red, just = c("right", "centre"), x = unit(-0.05, 
                                                                   "inches"), y = unit(0.5, "npc"), rot = 0)
    if (i == l$nrow) 
      grid.text(format(l$widths[j, top = FALSE], ...), 
                gp = gp.red, just = c("centre", "top"), x = unit(0.5, 
                                                                 "npc"), y = unit(-0.05, "inches"), rot = 0)
    if (j == l$ncol) 
      grid.text(format(l$heights[i, top = FALSE], ...), 
                gp = gp.red, just = c("left", "centre"), x = unit(1, 
                                                                  "npc") + unit(0.05, "inches"), y = unit(0.5, 
                                                                                                          "npc"), rot = 0)
    if (i == 1) 
      grid.text(format(l$widths[j, top = FALSE], ...), 
                gp = gp.red, just = c("centre", "bottom"), x = unit(0.5, 
                                                                    "npc"), y = unit(1, "npc") + unit(0.05, "inches"), 
                rot = 0)
    popViewport()
  }
  popViewport()
  if (!is.null(vp)) 
    popViewport()
  return(objs)
}

运行foo(l)打印：

s1: i = 1, j = 1, height = 0.193302891933029cm, width = 0.193302891933029cm
...
s1: i = 7, j = 5, height = 0cm, width = 0cm
...
s1: i = 12, j = 9, height = 0.193302891933029cm, width = 0.193302891933029cm

奇怪的是，逐步执行该函数browser显示i = 7，j = 5在中心打印最大的矩形，但大小为0cm，0cm！原始单位（分别为1null，1null）。

所以我的问题是，如何获取以npc /本机单位表示的矩形的大小/坐标？我很高兴遍历整个结构，但是我想将每个矩形的单位转换为明智的形式。理想情况下，我为每个布局元素获取以grid.rectnpc或设备的本机单位绘制的四个角的位置。

有任何想法吗？