按 r 中的时间间隔计算

Question

我正在尝试建立 windows 包含条件 "T_mean between 18-23 and rh>=70%" 的天数，自指定时间向后（24 年 9 月 24 日，蓝色）[=5 天间隔]

理想的列是 5day_int，这是我手工计算的。我的尝试包括 dplyr::lag 但到目前为止没有成功...

任何帮助都会非常有帮助！

数据：

tibble::tribble(
        ~date, ~t_mean, ~rh,
  "15-sep-14",      20,  65,
  "16-sep-14",      19,  68,
  "17-sep-14",      21,  68,
  "18-sep-14",      20,  76,
  "19-sep-14",      18,  68,
  "20-sep-14",      19,  78,
  "21-sep-14",      17,  82,
  "22-sep-14",      20,  54,
  "23-sep-14",      21,  57,
  "24-sep-14",      20,  75
  )

Answer 1

使用zoo和dplyr一种可能的解决方案：

library(dplyr)
library(zoo)

df %>%
  mutate(condition = ifelse(rh >= 70 & between(t_mean, 18, 23), 1, 0),
         fiveday_int = rollapplyr(condition, width = 5, FUN = sum, partial = TRUE))

# A tibble: 10 x 5
   date      t_mean    rh condition fiveday_int
   <chr>      <dbl> <dbl>     <dbl>       <dbl>
 1 15-sep-14     20    65         0           0
 2 16-sep-14     19    68         0           0
 3 17-sep-14     21    68         0           0
 4 18-sep-14     20    76         1           1
 5 19-sep-14     18    68         0           1
 6 20-sep-14     19    78         1           2
 7 21-sep-14     17    82         0           2
 8 22-sep-14     20    54         0           2
 9 23-sep-14     21    57         0           1
10 24-sep-14     20    75         1           2

数据：

df <- tibble::tribble(
  ~date, ~t_mean, ~rh,
  "15-sep-14",      20,  65,
  "16-sep-14",      19,  68,
  "17-sep-14",      21,  68,
  "18-sep-14",      20,  76,
  "19-sep-14",      18,  68,
  "20-sep-14",      19,  78,
  "21-sep-14",      17,  82,
  "22-sep-14",      20,  54,
  "23-sep-14",      21,  57,
  "24-sep-14",      20,  75
)

按 r 中的时间间隔计算

count if by time interval in r

r

lubridate

dplyr