r - 在 geom_tile() 上将 x 轴标签更改为小时(时间)

Question

这是一个geom_tile显示小时数和星期几，如何显示每个小时(即 x 轴上的 00:00 到 23:00)？

library(tidyverse)
df %>% 
  ggplot(aes(hour, day, fill = value)) +
  geom_tile(colour = "ivory") 

目前它每五个小时显示一次:

我尝试了很多不同的方法，并且更喜欢“最佳实践”方式(即无需手动生成标签)，但如果需要标签，这里有一种生成方式 hour_labs <- 0:23 %>% { ifelse(nchar(.) == 1, paste0("0", .), .) } %>% paste0(., ":00")

可重现示例的数据

df <- structure(list(day = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 
4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 5L, 5L, 
5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 
5L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 
6L, 6L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 7L), .Label = c("Sunday", 
"Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"
), class = c("ordered", "factor")), hour = c(0L, 2L, 3L, 5L, 
6L, 7L, 8L, 10L, 11L, 12L, 13L, 18L, 21L, 22L, 23L, 0L, 1L, 2L, 
3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 20L, 21L, 22L, 
23L, 0L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 
20L, 21L, 22L, 23L, 0L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 
11L, 13L, 14L, 20L, 21L, 22L, 23L, 0L, 1L, 2L, 3L, 4L, 5L, 6L, 
7L, 8L, 9L, 10L, 11L, 12L, 13L, 15L, 20L, 21L, 22L, 23L, 0L, 
1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 11L, 13L, 14L, 15L, 16L, 
19L, 21L, 0L, 1L, 2L, 3L, 7L, 8L, 10L, 13L, 14L, 22L, 23L), value = c(1L, 
1L, 1L, 2L, 1L, 3L, 1L, 1L, 2L, 1L, 3L, 1L, 2L, 13L, 13L, 24L, 
39L, 21L, 17L, 25L, 22L, 27L, 28L, 19L, 6L, 2L, 2L, 1L, 2L, 2L, 
7L, 23L, 38L, 18L, 26L, 21L, 20L, 31L, 40L, 35L, 22L, 5L, 3L, 
2L, 7L, 4L, 3L, 3L, 3L, 17L, 13L, 23L, 24L, 19L, 31L, 13L, 35L, 
50L, 22L, 13L, 7L, 2L, 1L, 1L, 1L, 1L, 3L, 14L, 17L, 33L, 32L, 
32L, 25L, 29L, 27L, 38L, 26L, 11L, 8L, 4L, 5L, 5L, 3L, 1L, 1L, 
3L, 14L, 21L, 24L, 22L, 25L, 26L, 23L, 58L, 36L, 26L, 6L, 3L, 
1L, 5L, 3L, 1L, 1L, 3L, 1L, 2L, 2L, 1L, 1L, 1L, 2L, 1L, 1L, 2L, 
1L, 1L)), row.names = c(NA, -116L), groups = structure(list(day = structure(1:7, .Label = c("Sunday", 
"Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"
), class = c("ordered", "factor")), .rows = structure(list(1:15, 
    16:33, 34:51, 52:69, 70:88, 89:105, 106:116), ptype = integer(0), class = c("vctrs_list_of", 
"vctrs_vctr"))), row.names = c(NA, 7L), class = c("tbl_df", "tbl", 
"data.frame"), .drop = TRUE), class = c("grouped_df", "tbl_df", 
"tbl", "data.frame"))

Answer 1

这是使用 sprintf 构造标签的一种方法。

library(dplyr)
library(ggplot2)

df %>%
  mutate(lab = sprintf('%02d:00', hour)) %>%
  ggplot() + aes(lab, day, fill = value) +
  geom_tile(colour = "ivory") + 
  theme(axis.text.x = element_text(angle = 90, hjust = 1))

---

除了@Eric Watt 的建议，我们还可以使用complete 来完成缺失的时间。

df %>%
  mutate(lab = sprintf('%02d:00', hour)) %>% 
  tidyr::complete(lab = sprintf('%02d:00', 0:23)) %>%
  ggplot() + aes(lab, day, fill = value) +
  geom_tile(colour = "ivory") + 
  theme(axis.text.x = element_text(angle = 90, hjust = 1))