考虑下面的简单示例。有没有办法格式化绘图工具提示,使得长文本标签在框中可见,而不是这个切断值的荒谬矩形?
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")
我很确定,某个地方存在更优雅的解决方案。我可以建议你像每个n
角色一样休息。 https://stackoverflow.com/a/2352006/9300556有一个很好的解决方法:
gsub('(.{1,90})(\\s|$)', '\\1\n', s)
它会将字符串“s”分成最多90个字符的行(不包括换行符“\ n”,但包括字间空格),除非有一个字本身超过90个字符,那么该字本身将占用一个整体线。
所以我们只需要将gsub()
添加到ggplot
美学中:
p <- ggplot(df, aes(x,y,
text = gsub('(.{1,20})(\\s|$)', '\\1\n', z))) +
geom_point()
ggplotly(p, tooltip = "text")
UPDATE
这是@Rich Pauloo评论中更优雅的解决方案。在这种情况下,你的字符串也将大部分被填充(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")