如何将 95% 的置信区间添加到 ggplot 中因子水平的比例图中？

Question

我想建立在我对之前提出的问题得到的很好答案的基础上：

ggplot2 中因子水平内的图表比例而不是计数

我希望以代码为基础：

var1 <- c("Left", "Right", NA, "Left", "Right", "Right", "Right", "Left", "Left", "Right", "Left", "Left","Left", "Right", "Left", "Right", "Right", "Right", "Left", "Left", "Right", NA, "Left", "Left","Left", "Right", NA, "Left", "Right", "Right", "Right", "Left", "Left", "Right", "Left", "Left","Left", "Right", "Left", "Right", "Right", "Right", "Left", "Left", "Right", NA, "Left", "Left")
var2 <- c("Higher", "Lower", NA, "Slightly higher", "Slightly higher", "Slightly higher", "Lower", "Slightly higher", "Higher", "Higher", "Higher", "Slightly higher","Higher", "Lower", "Slightly higher", "Slightly higher", "Slightly higher", "Lower", "Slightly higher", "Higher", "Higher", "Higher", NA, "Slightly lower","Higher", "Lower", NA, "Slightly higher", "Slightly higher", "Slightly higher", "Lower", "Slightly higher", "Higher", "Higher", "Higher", "Slightly higher","Higher", "Lower", "Slightly higher", "Slightly higher", "Slightly higher", "Lower", "Slightly lower", "Higher", "Higher", "Higher", NA, "Slightly lower")
df <- as.data.frame(cbind(var1, var2))

library(dplyr)
library(ggplot2)

df %>%
  na.omit() %>%
  group_by(var1, var2) %>%
  summarise(n = n()) %>%
  mutate(n = n/sum(n)) %>%
  ungroup() %>%
  ggplot() + aes(var2, n, fill = var1) + 
  geom_bar(position = "dodge", stat = "identity") + 
  labs(x="Left or Right",y="Count")+
  scale_y_continuous() +
  scale_fill_discrete(name = "Answer:")+ theme_classic()+ 
  theme(legend.position="top")  +
  scale_fill_manual(values = c("black", "red"))

以 95% 置信区间的形式向图表上的每个条添加误差条。 我试图在术语中添加

upperE=(1.96*sqrt(n/sum(n))*(1-(n/sum(n)))/n), lowerE=(-1.96*sqrt(n/sum(n))*(1-(n/sum(n)))/n).

但是，唉，我不断收到错误...

我还尝试为图表制作一个全新的 dataframe，因此：

var1 <- c("Left", "Right", NA, "Left", "Right", "Right", "Right", "Left", "Left", "Right", "Left", "Left","Left", "Right", "Left", "Right", "Right", "Right", "Left", "Left", "Right", NA, "Left", "Left","Left", "Right", NA, "Left", "Right", "Right", "Right", "Left", "Left", "Right", "Left", "Left","Left", "Right", "Left", "Right", "Right", "Right", "Left", "Left", "Right", NA, "Left", "Left")
var2 <- c("Higher", "Lower", NA, "Slightly higher", "Slightly higher", "Slightly higher", "Lower", "Slightly higher", "Higher", "Higher", "Higher", "Slightly higher","Higher", "Lower", "Slightly higher", "Slightly higher", "Slightly higher", "Lower", "Slightly higher", "Higher", "Higher", "Higher", NA, "Slightly lower","Higher", "Lower", NA, "Slightly higher", "Slightly higher", "Slightly higher", "Lower", "Slightly higher", "Higher", "Higher", "Higher", "Slightly higher","Higher", "Lower", "Slightly higher", "Slightly higher", "Slightly higher", "Lower", "Slightly lower", "Higher", "Higher", "Higher", NA, "Slightly lower")
df <- as.data.frame(cbind(var1, var2))



dat <- df %>%
  na.omit() %>%
  group_by(var1, var2) %>%
  summarise(n = n()) %>%
  mutate(prop = n/sum(n),upperE=1.96*sqrt(n/sum(n))*(1-(n/sum(n)))/n, lowerE=-1.96*sqrt(n/sum(n))*(1-(n/sum(n)))/n)

test <- ggplot(dat, aes(x=var2, y = prop, fill = var1))+ 
  geom_bar(position = "dodge", stat = "identity") + geom_errorbar(aes(ymin = lowerE, ymax = upperE),position="dodge")+
  labs(x="Answer",y="Proportion")+
  scale_fill_discrete(name = "Condition:")+ theme_classic()+ 
  theme(legend.position="top") 

这给了我错误条，但在 Y 轴上位于 0，而不是在每个条的顶部......

有没有人有什么建议？ 谢谢！

Answer 1

我现在已经弄清楚了如何让误差条位于每个条上适当的 position - 我需要将误差条的 ymin 和 ymax 规范与正在绘制的值相关联，因此：

dat <- df %>%
  na.omit() %>%
  group_by(var1, var2) %>%
  summarise(n = n()) %>%
  mutate(prop = n/sum(n),upperE=1.96*sqrt(n/sum(n))*(1-(n/sum(n)))/n, lowerE=-1.96*sqrt(n/sum(n))*(1-(n/sum(n)))/n)

test <- ggplot(dat, aes(x=var2, y = prop, fill = var1))+ 
  geom_bar(position = "dodge", stat = "identity") + geom_errorbar(aes(ymin = prop+lowerE, ymax = prop+upperE),width = .2, position=position_dodge(.9))+
  labs(x="Answer",y="Proportion")+
  scale_fill_discrete(name = "Condition:")+ theme_classic()+ 
  theme(legend.position="top") 

这给了：

Answer 2

95%CI 中 SE 的比例公式为： se = sqrt((p * (1-p))/n 。所以我认为在上面的解决方案中说明了： sqrt(n/sum(n) * 1-(n/sum(n))/n) 。但是， n只有成功的计数。完整的样本是sum(n) 。所以它实际上应该是sqrt(n/sum(n) * (1-(n/sum(n))/**sum**(n)) 。

Answer 3

超级旧线程，但以防万一有人仍然偶然发现这一点：已投票答案中的置信区间公式不正确。

它应该是：

mutate(prop = n/sum(n),
         upperE=1.96*sqrt(n/sum(n)*(1-(n/sum(n)))/sum(n)), 
         lowerE=-1.96*sqrt(n/sum(n)*(1-(n/sum(n)))/sum(n)))

. 使用用于置信区间的公式，您只需对公式的第一位求平方根。 但是，您需要对整个公式取平方根（Z 分数除外）。