gridGraphics::grid.echo error: EXPR must be a length 1 vector

Question

我正在尝试根据 to this link 使用 gridGraphics::grid.echo 因此我可以使用 [=16= 将 Gviz plotTracks 绘图与 ggplot 组合]的plot_grid.

继 Gviz 之后的 vignette and this link，这是我所做的：

require(Gviz)
data(geneModels)
gtrack <- GenomeAxisTrack()
itrack <- IdeogramTrack(genome = "hg19", chromosome = as.character(geneModels$chromosome[1]))
grtrack <- GeneRegionTrack(geneModels, genome = "hg19",chromosome = as.character(geneModels$chromosome[1]), name = "Gene Model")

require(gridGraphics)

gwrap_plot <- function(x) {
  tree <- grid::grid.grabExpr(gridGraphics::grid.echo(x))
  u <- grid::unit(1, 'null')
  gtable::gtable_col(NULL, list(tree), u, u)
}

graphics.off()
plotTracks(list(itrack, gtrack, grtrack))
track.plot <- recordPlot()
gwrap_plot(track.plot)

我得到这个错误：

 Error in switch(x[[2]][[1]]$name, C_abline = C_abline(x[[2]]), C_plot_new = C_plot_new(x[[2]]),  : 
  EXPR must be a length 1 vector

知道 gridGraphics 的 grid.echo 和 Gviz 的 plotTracks 情节有什么问题吗？

Answer 1

grid.echo 用于基本图形，Gviz 似乎使用 grid 图形，

p1 = grid::grid.grabExpr(plotTracks(list(itrack, gtrack, grtrack), add = TRUE))
p2 = ggplot2::qplot(1:10, 1:10)

gridExtra::grid.arrange(p1, p2, ncol=2)

gridGraphics::grid.echo error: EXPR must be a length 1 vector

gridGraphics::grid.echo error: EXPR must be a length 1 vector

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