ggplotGrob 的倒数? [英] Inverse of ggplotGrob?

问题描述

我有一个操作 ggplot 对象的函数，方法是将其转换为 grob，然后修改图层.我希望该函数返回一个 ggplot 对象而不是 grob.有没有一种简单的方法可以将 grob 转换回 gg?

I have a function which manipulates a ggplot object, by converting it to a grob and then modifying the layers. I would like the function to return a ggplot object not a grob. Is there a simple way to convert a grob back to gg?

ggplotGrob 上的文档非常稀少.
简单例子:

The documentation on ggplotGrob is awfully sparse.
Simple example:

P <- ggplot(iris) + geom_bar(aes(x=Species, y=Petal.Width), stat="identity")

G <- ggplotGrob(P)
... some manipulation to G ...

## DESIRED: 
P2 <- inverse_of_ggplotGrob(G)

such that, we can continue to use basic ggplot syntax, ie
`P2 + ylab ("The Width of the Petal")`

<小时>

更新:

为了回答评论中的问题，这里的动机是根据每个方面中标签名称的值，以编程方式修改方面标签的颜色.下面的函数运行良好(基于上一个问题中 baptise 的输入).

To answer the question in the comment, the motivation here is to modify the colors of facet labels programmatically, based on the value of label name in each facet. The functions below work nicely (based on input from baptise in a previous question).

我希望 colorByGroup 的返回值是一个 ggplot 对象，而不仅仅是一个 grob.

I would like for the return value from colorByGroup to be a ggplot object, not simply a grob.

这是代码，有兴趣的人

get_grob_strips <- function(G, strips=grep(pattern="strip.*", G$layout$name)) {

  if (inherits(G, "gg"))
    G <- ggplotGrob(G)
  if (!inherits(G, "gtable"))
    stop ("G must be a gtable object or a gg object")

  strip.type <- G$layout[strips, "name"]
  ## I know this works for a simple 
  strip.nms <- sapply(strips, function(i) {
     attributes(G$grobs[[i]]$width$arg1)$data[[1]][["label"]]
  })

  data.table(grob_index=strips, type=strip.type, group=strip.nms)
}


refill <- function(strip, colour){
  strip[["children"]][[1]][["gp"]][["fill"]] <- colour
  return(strip)
}

colorByGroup <- function(P, colors, showWarnings=TRUE) {
## The names of colors should match to the groups in facet
  G <- ggplotGrob(P)
  DT.strips <- get_grob_strips(G)

  groups <- names(colors)
  if (is.null(groups) || !is.character(groups)) {
    groups <- unique(DT.strips$group)
    if (length(colors) < length(groups))
      stop ("not enough colors specified")
    colors <- colors[seq(groups)]
    names(colors) <- groups
  }


  ## 'groups' should match the 'group' in DT.strips, which came from the facet_name
  matched_groups <- intersect(groups, DT.strips$group)
  if (!length(matched_groups))
    stop ("no groups match")
  if (showWarnings) {
      if (length(wh <- setdiff(groups, DT.strips$group)))
        warning ("values in 'groups' but not a facet label: 
", paste(wh, colapse=", "))
      if (length(wh <- setdiff(DT.strips$group, groups)))
        warning ("values in facet label but not in 'groups': 
", paste(wh, colapse=", "))
  }

  ## identify the indecies to the grob and the appropriate color
  DT.strips[, color := colors[group]]
  inds <- DT.strips[!is.na(color), grob_index]
  cols <- DT.strips[!is.na(color), color]

  ## Fill in the appropriate colors, using refill()
  G$grobs[inds] <- mapply(refill, strip = G$grobs[inds], colour = cols, SIMPLIFY = FALSE)

  G
}

推荐答案

我会说不.ggplotGrob 是一条单行道.grob 对象是由 grid 定义的绘图基元.您可以从头开始创建任意 grob.没有通用的方法可以将 grobs 的随机集合转换回生成它们的函数(它不可逆，因为它不是 1:1).一旦你去grob，你就永远不会回去.

I would say no. ggplotGrob is a one-way street. grob objects are drawing primitives defined by grid. You can create arbitrary grobs from scratch. There's no general way to turn a random collection of grobs back into a function that would generate them (it's not invertible because it's not 1:1). Once you go grob, you never go back.

您可以将 ggplot 对象包装在自定义类中并重载绘图/打印命令以执行一些自定义的 grob 操作，但这可能更加黑客化.

You could wrap a ggplot object in a custom class and overload the plot/print commands to do some custom grob manipulation, but that's probably even more hack-ish.

这篇关于ggplotGrob 的倒数?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！