相关疑难解决方法(0)

反正有没有让grid.arrange()充当split.screen()？我想安排一张桌子直接位于图例下方.

#create histogram
my_hist<-ggplot(diamonds, aes(clarity, fill=cut)) + geom_bar()

#create inset table
my_table<- tableGrob(head(diamonds)[,1:3],gpar.coretext =gpar(fontsize=8),gpar.coltext=gpar(fontsize=8), gpar.rowtext=gpar(fontsize=8))

grid.arrange(my_hist,my_table, ncol=2)

生产:

但我希望它看起来像这样:

我尝试了split.screen()但它似乎不适用于ggplot类型的图形.有什么建议？谢谢.

我想要两个情节+他们的传奇的组合情节如下:

library(ggplot2) 
library(grid)
library(gridExtra)
dsamp <- diamonds[sample(nrow(diamonds), 1000), ]    
p1 <- qplot(price, carat, data=dsamp, colour=clarity)
p2 <- qplot(price, depth, data=dsamp, colour=clarity)
g <- ggplotGrob(p1 + theme(legend.position="bottom"))$grobs
legend <- g[[which(sapply(g, function(x) x$name) == "guide-box")]]
grid.arrange(arrangeGrob(p1+theme(legend.position="right"),p2+theme(legend.position="none"),legend,ncol=3,widths=c(3/7,3/7,1/7)))

但是我不想猜测图和图例的宽度(并指定ncol)但是从中提取p1并p2 如此处所示.

所以我希望我需要这样的东西(来自链接的改编代码):

grid_arrange_shared_legend_row <- function(...) {
  plots <- list(...)
  g <- ggplotGrob(plots[[1]] + theme(legend.position="right"))$grobs
  legend <- g[[which(sapply(g, function(x) x$name) == "guide-box")]]
  lwidth <- sum(legend$width)
  grid.arrange(
    do.call(arrangeGrob, lapply(plots, function(x)
      x + theme(legend.position="none"))),
    legend,
    ncol = length(plots)+1,
    widths = unit.c(rep(unit(1, "npc") …