Rei*_*ein 6 plot r ggplot2 ggrepel
我有一个相当密集的散点图,我用R'ggplot2'构建,我想用'ggrepel'标记一个点的子集.我的问题是我想在散点图中绘制所有点,但只用ggrepel标记一个子集,当我这样做时,ggrepel在计算放置标签的位置时没有考虑图上的其他点,这导致与图上其他点重叠的标签(我不想标记).
这是一个说明问题的示例图.
# generate data:
library(data.table)
library(stringi)
set.seed(20180918)
dt = data.table(
name = stri_rand_strings(3000,length=6),
one = rnorm(n = 3000,mean = 0,sd = 1),
two = rnorm(n = 3000,mean = 0,sd = 1))
dt[, diff := one -two]
dt[, diff_cat := ifelse(one > 0 & two>0 & abs(diff)>1, "type_1",
ifelse(one<0 & two < 0 & abs(diff)>1, "type_2",
ifelse(two>0 & one<0 & abs(diff)>1, "type_3",
ifelse(two<0 & one>0 & abs(diff)>1, "type_4", "other"))))]
# make plot
ggplot(dt, aes(x=one,y=two,color=diff_cat))+
geom_point()
Run Code Online (Sandbox Code Playgroud)
如果我只绘制我想要标记的点的子集,那么ggrepel能够相对于其他点和标签以非重叠的方式放置所有标签.
ggplot(dt[abs(diff)>2 & (!diff_cat %in% c("type_3","type_4","other"))],
aes(x=one,y=two,color=diff_cat))+
geom_point()+
geom_text_repel(data = dt[abs(diff)>2 & (!diff_cat %in% c("type_3","type_4","other"))],
aes(x=one,y=two,label=name))
Run Code Online (Sandbox Code Playgroud)
但是,当我想同时绘制这个数据子集和原始数据时,我得到了带标签的重叠点:
# now add labels to a subset of points on the plot
ggplot(dt, aes(x=one,y=two,color=diff_cat))+
geom_point()+
geom_text_repel(data = dt[abs(diff)>2 & (!diff_cat %in% c("type_3","type_4","other"))],
aes(x=one,y=two,label=name))
Run Code Online (Sandbox Code Playgroud)
如何才能使点子集的标签与原始数据中的点重叠?
您可以尝试以下方法:
""为原始数据中的所有其他点分配一个空白标签(),以便geom_text_repel在彼此排斥标签时将其考虑在内;box.padding参数从默认值0.25增加到更大的值,以增加标签之间的距离;示例代码(带box.padding = 1):
ggplot(dt,
aes(x = one, y = two, color = diff_cat)) +
geom_point() +
geom_text_repel(data = . %>%
mutate(label = ifelse(diff_cat %in% c("type_1", "type_2") & abs(diff) > 2,
name, "")),
aes(label = label),
box.padding = 1,
show.legend = FALSE) + #this removes the 'a' from the legend
coord_cartesian(xlim = c(-5, 5), ylim = c(-5, 5)) +
theme_bw()
Run Code Online (Sandbox Code Playgroud)
这是另一种尝试,包括box.padding = 2:
(注意:我正在使用ggrepel 0.8.0.我不确定早期软件包版本是否存在所有功能.)