Question

我有一个数据库，其中包含一系列机器的错误寄存器及其对应日期。有几种错误。即：

          fechayhora        id tipo
1: 2017-03-21 11:03:00 A2_LR1_Z1  APF
2: 2017-05-03 10:34:00 A2_LR1_Z1  APF
3: 2017-05-17 08:52:00 A2_LR1_Z1  APF
4: 2017-05-17 10:46:00 A2_LR1_Z1  APF
5: 2017-05-17 14:23:00 A2_LR1_Z1  APF
6: 2017-05-17 17:29:00 A2_LR1_Z1  APF

我想添加一个包含事件总和的列，其中包括＆＃34; APF＆＃34;在之前发生过，比如说12个小时（实际上我可以改变一个参数）。

预期结果：

          fechayhora        id tipo    number_of_APF_12h
1: 2017-03-21 11:03:00 A2_LR1_Z1  APF  0
2: 2017-05-03 10:34:00 A2_LR1_Z1  APF  0
3: 2017-05-17 08:52:00 A2_LR1_Z1  APF  0
4: 2017-05-17 10:46:00 A2_LR1_Z1  APF  1
5: 2017-05-17 14:23:00 A2_LR1_Z1  APF  2
6: 2017-05-17 17:29:00 A2_LR1_Z1  APF  3

Answer 1

这是一个使用purrr::map2_dbl()的解决方案。您可以将小时数更改为您想要的任何小时数。

suppressPackageStartupMessages(library(tibble))
suppressPackageStartupMessages(library(dplyr))
suppressPackageStartupMessages(library(purrr))
suppressPackageStartupMessages(library(lubridate))

# Example data
df <- tribble(
  ~fechayhora,        ~id,       ~tipo,
  "2017-03-21 11:03:00", "A2_LR1_Z1",  "APF",
  "2017-05-03 10:34:00", "A2_LR1_Z1",  "APF",
  "2017-05-17 08:52:00", "A2_LR1_Z1",  "APF",
  "2017-05-17 10:46:00", "A2_LR1_Z1",  "APF",
  "2017-05-17 14:23:00", "A2_LR1_Z1",  "APF",
  "2017-05-17 17:29:00", "A2_LR1_Z1",  "APF"
)

# Convert fechayhora to date and add a column of the time difference
df <- df %>%
  mutate(fechayhora = as.POSIXct(fechayhora),
         minus_12   = fechayhora - hours(12))

# Map over fechayhora and minus_12
# For each (fechayhora, minus_12) pair, find all the dates between them
# and sum the logical vector that is returned
df <- df %>% mutate(
  number_of_APF_12h = map2_dbl(.x = fechayhora, 
                               .y = minus_12, 
                               .f = ~sum(between(df$fechayhora, .y, .x)) - 1))

df %>%
  select(fechayhora, number_of_APF_12h)
#> # A tibble: 6 x 2
#>            fechayhora number_of_APF_12h
#>                <dttm>             <dbl>
#> 1 2017-03-21 11:03:00                 0
#> 2 2017-05-03 10:34:00                 0
#> 3 2017-05-17 08:52:00                 0
#> 4 2017-05-17 10:46:00                 1
#> 5 2017-05-17 14:23:00                 2
#> 6 2017-05-17 17:29:00                 3

如果日期差异小于值，则求和

1 个答案: