r - 如何将每小时降雨数据数据聚合成24小时系列

Question

我的数据框如下

df <- tibble::tribble(
              ~date, ~pcp,
   "9/27/2017 9:00",    0,
  "9/27/2017 10:00",    0,
  "9/27/2017 11:00",    0,
  "9/27/2017 12:00",    0,
  "9/27/2017 13:00",    0,
  "9/27/2017 14:00",    0,
  "9/27/2017 15:00",    0,
  "9/27/2017 16:00",    0,
  "9/27/2017 17:00",    0,
  "9/27/2017 18:00",    0,
  "9/27/2017 19:00",    0,
  "9/27/2017 20:00",    0,
  "9/27/2017 21:00",    0,
  "9/27/2017 22:00",    0,
  "9/27/2017 23:00",    0,
   "9/28/2017 0:00",    0,
   "9/28/2017 1:00",    0,
   "9/28/2017 2:00",    0,
   "9/28/2017 3:00",    0,
   "9/28/2017 4:00",    0,
   "9/28/2017 5:00",    0,
   "9/28/2017 6:00",    0,
   "9/28/2017 7:00", 0.15,
   "9/28/2017 8:00", 8.76,
   "9/28/2017 9:00", 0.02,
  "9/28/2017 10:00",    0,
  "9/28/2017 11:00",    0,
  "9/28/2017 12:00",    0,
  "9/28/2017 13:00",    0,
  "9/28/2017 14:00",    0,
  "9/28/2017 15:00",    0,
  "9/28/2017 16:00",    0,
  "9/28/2017 17:00",    0,
  "9/28/2017 18:00",    0,
  "9/28/2017 19:00",    0,
  "9/28/2017 20:00",    0,
  "9/28/2017 21:00",    0,
  "9/28/2017 22:00",    0,
  "9/28/2017 23:00",    0,
   "9/29/2017 0:00",    0,
   "9/29/2017 1:00",    0,
   "9/29/2017 2:00",    0,
   "9/29/2017 3:00",    0,
   "9/29/2017 4:00",    0,
   "9/29/2017 5:00",    0,
   "9/29/2017 6:00",    0,
   "9/29/2017 7:00",    0,
   "9/29/2017 8:00", 0.31
  )

我想每天汇总数据(总和)。而不是从同一天的 00:00 到 23:59 聚合，我希望它的初始时间从当天的 09:00 开始 i并在当天的 08:59 结束 i + 1 (24 小时后)。
输出是欲望如下

9/28/2017,8.91
9/29/2017,0.33

我是在 Excel 中手动完成的，我不确定要使用什么代码来解决这个问题。
提供的示例是长数据帧的摘录。
谢谢...

Answer 1

如果你想要每个日期的数据

library(tidyverse)
library(lubridate)
df %>% 
    mutate(datetime = parse_date_time(date, "mdy H:M"),
           date = date(datetime)) %>%
    group_by(date) %>%
    summarise(sum_pcp = sum(pcp))

会产生

# A tibble: 3 x 2
  date       sum_pcp
  <date>          <dbl>
1 2017-09-27       0   
2 2017-09-28       8.93
3 2017-09-29       0.31

如果你想从 9:00 到第二天的 9:00 计算，你可以引入一个 subjective_day通过从原始 datetime 中减去 9 小时目的。

df %>% 
  mutate(datetime = parse_date_time(date, "mdy H:M"),
         date = date(datetime),
         initial_day = date(first(date)),
         time = hour(datetime),
         subjective_day = datetime - hours(9)) %>%
  group_by(subjective_day = floor_date(subjective_day, "1 day")) %>% 
  summarise(sum_pcp = sum(pcp))

会产生

  subjective_day      sum_pcp
  <dttm>                <dbl>
1 2017-09-27 00:00:00    8.91
2 2017-09-28 00:00:00    0.33

您的主观一天将始终落后 1 天，因此您可以对此进行调整，或者记住世界上有 9 小时时差的地方实际上将其作为正确的日期时间:)