Ric*_*loo 5 r ggplot2 plotly ggplotly
考虑下面的简单示例。有没有一种方法可以格式化绘图工具提示,以便长文本标签在框中可见,而不是这个截断值的荒谬矩形?
library(ggplot2); library(plotly)
df <- data.frame(x = 1:10, y = 1:10, z = rep("the longggggggggggggggggggggggggggggestttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttt labelllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllll you can imagineeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee", 10))
p <- ggplot(df, aes(x,y,label=z)) + geom_point()
ggplotly(p, tooltip = "label")
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我很确定,某个地方存在更优雅的解决方案。我只能建议你像每个角色一样休息一下n。/sf/answers/164640451/有一个很好的解决方法:
gsub('(.{1,90})(\\s|$)', '\\1\n', s)
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它将字符串“s”分成最多90个字符的行(不包括换行符“\n”,但包括字间空格),除非有一个单词本身超过90个字符,那么该单词本身将占据整个线。
所以我们只需要添加gsub()美学ggplot:
p <- ggplot(df, aes(x,y,
text = gsub('(.{1,20})(\\s|$)', '\\1\n', z))) +
geom_point()
ggplotly(p, tooltip = "text")
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更新
这是来自@Rich Pauloo 评论的更优雅的解决方案。在这种情况下,您的字符串也将大部分被左侧填充(但实际上是自动对齐的)。但是,填充取决于绘图分辨率和标签位置。
library(stringr)
p <- ggplot(df,
aes(x, y,
text = stringr::str_wrap(
string = z,
width = 20,
indent = 1, # let's add extra space from the margins
exdent = 1 # let's add extra space from the margins
))) +
geom_point()
ggplotly(p, tooltip = "text")
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