在条形图中添加每组的观察数（ggplot2）答案

【问题标题】：Add number of observations per group in barplot (ggplot2)在条形图中添加每组的观察数（ggplot2）
【发布时间】：2020-02-28 14:06:24
【问题描述】：

我正在尝试将观察次数添加到我的情节中，但我不知道我是否接近这个权利。我使用 ggplot2 和 tidyverse 包绘制了我的情节。

我想要的是在图表的每一列顶部添加观察数（在数据的 Obs 列中）。但是，看起来我做得不对，因为它需要列的每个部分的值。

有谁知道如何解决这个问题？非常感谢！

图表代码

Euk_one %>% 
  drop_na(`Length Stretch`) %>%
  mutate_at(c("GGA","GGG","GGT","GGC"),~.x/`Length Stretch`) %>% 
  group_by(interval) %>% select(c(interval, "GGA","GGG","GGT","GGC", Obs))%>%
  summarize_all(mean) %>%pivot_longer(-c(interval, Obs)) %>%
  ggplot(aes(x=interval, y=value, fill=name)) + geom_col() +
  ggtitle("Eukarya 1 Codon Stretches - Distribution of codons ")+
  labs(fill= "Codons") +
  theme(axis.text.x = element_text(angle = 45)) + geom_text(aes(label=Obs), vjust = -.25)

示例数据

structure(list(interval = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 
4L, 4L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("[6,7]", 
"(7,8]", "(8,9]", "(9,10]"), class = "factor"), `Length Stretch` = c(8, 
8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 10, 10, 10, 
10, 10, 10, 6, 6, 6, 6, 6, 7, 6, 6, 6, 6, 6, 6, 7, 6, 6, 6, 7, 
7, 6, 6, 7, 6, 6, 7, 7, 7, 6), GGA = c(0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 7, 0, 6, 0, 0, 0, 0, 0, 7, 6, 6, 0, 0, 0, 
0), GGG = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), GGT = c(8, 8, 8, 8, 8, 
8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 
6, 6, 6, 6, 6, 7, 6, 6, 6, 6, 6, 6, 0, 6, 0, 6, 7, 7, 6, 6, 0, 
0, 0, 7, 7, 7, 6), GGC = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), `Length Gene` = c(5250, 
5250, 3021, 1905, 2910, 2973, 2367, 3021, 2931, 2754, 2424, 2862, 
2436, 5061, 4935, 23949, 23988, 24024, 2733, 2733, 2358, 2358, 
2733, 2733, 2838, 2079, 2100, 1998, 2835, 1197, 1197, 2838, 1998, 
3795, 3795, 3795, 1320, 1827, 2088, 663, 10395, 10545, 2340, 
2340, 1323, 1977, 2085, 4773, 10557, 10554, 1920), NumGens = c(19297, 
19297, 19297, 19297, 19297, 19297, 19297, 19297, 19297, 19297, 
19297, 19297, 19297, 19297, 19297, 19297, 19297, 19297, 19297, 
19297, 19297, 19297, 19297, 19297, 19297, 19297, 19297, 19297, 
19297, 19297, 19297, 19297, 19297, 19297, 19297, 19297, 19297, 
19297, 19297, 19297, 19297, 19297, 19297, 19297, 19297, 19297, 
19297, 19297, 19297, 19297, 19297), Count = c(1, 1, 1, 1, 1, 
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 
1, 1, 1, 1), Obs = c(13L, 13L, 13L, 13L, 13L, 13L, 13L, 13L, 
13L, 13L, 13L, 13L, 13L, 5L, 5L, 5L, 5L, 5L, 6L, 6L, 6L, 6L, 
6L, 6L, 27L, 27L, 27L, 27L, 27L, 27L, 27L, 27L, 27L, 27L, 27L, 
27L, 27L, 27L, 27L, 27L, 27L, 27L, 27L, 27L, 27L, 27L, 27L, 27L, 
27L, 27L, 27L)), class = "data.frame", row.names = c(NA, -51L
))

情节示例

【问题讨论】：

标签： r ggplot2 plot dplyr tidyverse

【解决方案1】：

由于每个间隔只有一个值作为标签绘制，因此您可以使用 dplyr 中的 distinct 函数过滤数据框并设置单个 y 值来绘制它：

library(tidyverse)
Euk_one %>% 
  drop_na(`Length Stretch`) %>%
  mutate_at(c("GGA","GGG","GGT","GGC"),~.x/`Length Stretch`) %>% 
  group_by(interval) %>% dplyr::select(c(interval, "GGA","GGG","GGT","GGC", Obs))%>%
  summarize_all(mean) %>%pivot_longer(-c(interval, Obs)) %>%
  ggplot(aes(x=interval, y=value, fill=name)) + geom_col() +
  ggtitle("Eukarya 1 Codon Stretches - Distribution of codons ")+
  labs(fill= "Codons") +
  theme(axis.text.x = element_text(angle = 45)) + 
  geom_text(inherit.aes = FALSE, data = . %>% distinct(interval, Obs), 
            aes(x = interval, y = 0.25, label=Obs))

它回答了你的问题吗？

【讨论】：