Ggplot2中的有条件着色条

Question

下面是一些简单示例数据框和绘图的代码。我想知道如何有条件地为条形着色。我熟悉 scale_fill_manual 来手动给条形上色，但是如果我希望 2016 构面中的 "Satisfied" 比 2015 中的 "Satisfied" 的百分比低，那该怎么办。可能是红色警告边框或其他颜色，如橙色（仅作为示例）。

这不是最好的例子，但如果我有一个年度顶盒得分图，如果条形图下降到某个百分比以下，这将有助于使条形图改变颜色。我尝试使用"colour = ifelse(Perc < 60, "orange", "green" 组合，但无法使其正常工作。我不确定如何构造 ifelse 语句或将其放置在 ggplot 代码中的什么位置。

Year<-c(2015, 2015, 2015, 2015, 2015, 2016, 2016, 2016, 2016, 2016)

Service<-c("Satisfied", "Satisfied", "Satisfied", "Dissatisfied", "Dissatisfied",
           "Satisfied", "Satisfied", "Dissatisfied", "Dissatisfied", "Dissatisfied")

df <- data.frame(Year, Service)

library(dplyr)
df.prop <- df %>%
            count(Year, Service) %>%
            mutate(Perc = prop.table(n))

library(ggplot2)
ggplot(df.prop, aes(x = Service, y = Perc, fill = Service)) +
       geom_bar(stat = "identity", position = "dodge") +
       geom_text(aes(label = percent(Perc)), position = position_dodge(width = 1),
                 vjust = 1.5, size = 3) +
       scale_y_continuous(labels = percent) +
       facet_grid( ~ Year)

Answer 1

向数据框添加新变量可能是最简单的方法：

df.prop$colour <- ifelse(df.prop$Service == "Satisfied" & df.prop$Perc < 0.6, "orange", NA)

那么你可以这样做：

ggplot(df.prop, aes(x = Service, y = Perc, fill = Service, colour=colour)) +
  geom_bar(stat = "identity", position = "dodge") +
  geom_text(aes(label = percent(Perc)), position = position_dodge(width = 1),
            vjust = 1.5, size = 3, colour="black") +
  scale_y_continuous(labels = percent) +
  facet_grid( ~ Year) +
  scale_colour_identity()

如果您想根据条件更改填充，您可以这样做：

df.prop$fill <- ifelse(df.prop$Service == "Satisfied" & df.prop$Perc < 0.6, "orange", ifelse(df.prop$Service == "Satisfied" & df.prop$Perc >= 0.6, "#00BFC4", "#F8766D"))

ggplot(df.prop, aes(x = Service, y = Perc, fill = fill)) +
  geom_bar(stat = "identity", position = "dodge") +
  geom_text(aes(label = percent(Perc)), position = position_dodge(width = 1),
            vjust = 1.5, size = 3) +
  scale_y_continuous(labels = percent) +
  facet_grid( ~ Year) +
  scale_fill_identity()