首页
学习
活动
专区
圈层
工具
发布
社区首页 >问答首页 >在ggplot2中插入图例下的表

在ggplot2中插入图例下的表
EN

Stack Overflow用户
提问于 2013-12-08 17:02:14
回答 1查看 652关注 0票数 2

我发布了这个问题earlier,但事实证明我确认的答案有问题,所以我再次发布这个问题,希望能解决这个问题。

我有下面的data.frame,我想用ggplot2绘制成一个文件:

代码语言:javascript
复制
df = data.frame(mean=c(1.96535,2.133604,1.99303,1.865004,2.181713,1.909511,2.047971,1.676599,2.143763,1.939875,1.816028,1.95465,2.153445,1.802517,2.141799,1.722428),
sd=c(0.0595173,0.03884202,0.0570006,0.04934336,0.04008221,0.05108064,0.0463556,0.06272475,0.04321496,0.05283728,0.05894342,0.05160038,0.04679423,0.05305525,0.04626291,0.0573123),
par=as.factor(c("p","p","m","m","p","p","m","m","m","m","p","p","m","m","p","p")),
group=as.factor(c("iF","iF","iF","iF","iM","iM","iM","iM","RF","RF","RF","RF","RM","RM","RM","RM")),
rep=c(1,2,1,2,1,2,1,2,1,2,1,2,1,2,1,2)) 

这是我的代码,用于生成ggplot2对象:

代码语言:javascript
复制
p <- ggplot(df,aes(factor(rep),y=mean,ymin=mean-2*sd,ymax=mean+2*sd,color=factor(par)))
p <- p + geom_pointrange()+facet_wrap(~group, ncol = 4)+scale_color_manual(values = c("blue","red"),labels = c("p","m"),name = "par id")
p <- p + ggtitle("test")
p <- p + labs(y="log(y)",x="rep")

另外,我要做的是将这个data.frame作为一个表添加到图例中:

代码语言:javascript
复制
leg.df = data.frame(statistic = c("pp(par)","pp(g)","pp(s)","fc(p/m)"), value = c(0.96,0.94,0.78,1.5))

我得到了这样的解决方案:

代码语言:javascript
复制
leg.df.grob <-  tableGrob(leg.df, gpar.coretext =gpar(fontsize=8),
         par.coltext=gpar(fontsize=8), 
         gpar.rowtext=gpar(fontsize=8))

### final result
library(gridExtra)
pp <- arrangeGrob(p + theme(legend.position = "none"), 
                  arrangeGrob(leg.df.grob, legend), ncol = 2)

然后,理论上,我可以将pp保存到一个文件中,例如,使用:

代码语言:javascript
复制
ggsave('plot.png',pp)

不幸的是,命令:

代码语言:javascript
复制
pp <- arrangeGrob(p + theme(legend.position = "none"), 
                      arrangeGrob(leg.df.grob, legend), ncol = 2) 

导致绘图设备打开(这正是我试图避免的),此外,它还抛出了以下错误:

代码语言:javascript
复制
Error in arrangeGrob(leg.df.grob, legend) : input must be grobs!

知道怎么解决这个问题吗?

EN

回答 1

Stack Overflow用户

回答已采纳

发布于 2013-12-08 17:22:36

另一个解决方案是正确的。我想您得到了一个错误,因为您没有设置legend变量。因此,以R函数legend作为参数调用legend。您应该将legend定义为:

代码语言:javascript
复制
g_legend <- function(a.gplot){
    tmp <- ggplot_gtable(ggplot_build(a.gplot))
    leg <- which(sapply(tmp$grobs, function(x) x$name) == "guide-box")
    legend <- tmp$grobs[[leg]]
    return(legend)
}
legend <- g_legend(p)

我稍微修改了另一个答案,通过设置widths参数来更好地重新排列grobs:

代码语言:javascript
复制
pp <- arrangeGrob(p + theme(legend.position = "none"), 
                  widths=c(3/4, 1/4),
                  arrangeGrob( legend,leg.df.grob), ncol = 2)

票数 4
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/20456163

复制
相关文章

相似问题

领券
问题归档专栏文章快讯文章归档关键词归档开发者手册归档开发者手册 Section 归档