我正在创建一个要在其中使用ggrepel
显示标签的图。我在下面显示一个最小的示例,该示例说明标签如何具有两个用逗号分隔的成分-首先与鸢尾花种类的类型有关,其次与该组的样本量有关。
# needed libraries
set.seed(123)
library(ggrepel)
# creating a dataframe with label column
(df <- iris %>%
dplyr::group_by(Species) %>%
dplyr::summarise(n = n(), mean = mean(Sepal.Length)) %>%
purrrlyr::by_row(
.d = .,
..f = ~ paste("list(~",
.$Species,
",",
.$n,
")",
sep = ""),
.collate = "rows",
.to = "label",
.labels = TRUE
))
#> # A tibble: 3 x 4
#> Species n mean label
#> <fct> <int> <dbl> <chr>
#> 1 setosa 50 5.01 list(~setosa,50)
#> 2 versicolor 50 5.94 list(~versicolor,50)
#> 3 virginica 50 6.59 list(~virginica,50)
# displaying labels
ggplot(iris, aes(Species, Sepal.Length)) +
geom_point() +
ggrepel::geom_label_repel(data = df,
aes(x = Species, y = mean, label = label),
parse = TRUE)
由reprex package(v0.2.1)于2018-11-17创建
我的问题是如何摆脱这两个组件之间的空间。尽管我在sep = ""
函数中指定了paste()
,但是在两个我不想要的组件之间仍然有多余的空间(例如setosa, 50
,versicolor, 50
,{{1} }标签应改为virginica, 50
,setosa,50
,versicolor,50
)。
答案 0 :(得分:2)
下面是代码的更新版本,该代码实现了一种在种类名称和样本数量之间放置逗号(无后续空格)的方法。例如,您标记的标签看起来像"~setosa*\",\"*50"
而不是list(~setosa,50)
(df <- iris %>%
dplyr::group_by(Species) %>%
dplyr::summarise(n = n(), mean = mean(Sepal.Length)) %>%
purrrlyr::by_row(
.d = .,
..f = ~ paste("~",
.$Species,
"*\",\"*",
.$n,
"",
sep = ""),
.collate = "rows",
.to = "label",
.labels = TRUE
))
#> # A tibble: 3 x 4
#> Species n mean label
#> <fct> <int> <dbl> <chr>
#> 1 setosa 50 5.01 "~setosa*\",\"*50"
#> 2 versicolor 50 5.94 "~versicolor*\",\"*50"
#> 3 virginica 50 6.59 "~virginica*\",\"*50"
# displaying labels
ggplot(iris, aes(Species, Sepal.Length)) +
geom_point() +
stat_smooth(method="lm",size=0.6,se=FALSE,colour="black")+
ggrepel::geom_label_repel(data = df,
aes(x = Species, y = mean, label = label),
parse = TRUE)
会产生以下情节:
希望有帮助。