是否可以根据 R 中的时间变量过滤掉重叠的异常值？

Question

如何通过检查 ID 列和 date/time 列与 filter() 之间的重叠来删除异常值？

例如ID=1的行如下图前2行在时间上重叠，需要删除。

ID	Time start	Time end
1	2015-03-16 10:40:00	2015-03-16 11:10:00
1	2015-03-16 10:50:00	2015-03-16 10:59:00
2	2015-03-16 10:40:00	2015-03-16 10:45:00
1	2015-03-16 11:20:00	2015-03-16 11:28:56

Answer 1

试试这个来删除组内的任何时间重叠。请使用更多数据对其进行测试，看看它是否符合您的要求。我只试了下面的小样。

library(tidyverse)
library(lubridate)
#> 
#> Attaching package: 'lubridate'
#> The following objects are masked from 'package:base':
#> 
#>     date, intersect, setdiff, union
library(slider)

tribble(
  ~id, ~start, ~end,
  1, "2015-03-16 10:40:00", "2015-03-16 11:10:00",
  1, "2015-03-16 10:50:00", "2015-03-16 10:59:00",
  1, "2015-03-16 11:09:00", "2015-03-16 11:11:00",
  2, "2015-03-16 10:40:00", "2015-03-16 10:45:00",
  1, "2015-03-16 11:20:00", "2015-03-16 11:28:56",
  1, "2015-03-16 11:27:00", "2015-03-16 11:30:56",
  2, "2015-03-16 10:44:00", "2015-03-16 11:45:00"
) |>
  mutate(
    start = ymd_hms(start, tz = Sys.timezone()),
    end = ymd_hms(end, tz = Sys.timezone())
  ) |>
  arrange(id, start, end) |>
  group_by(id) |>
  mutate(
    roll_start = slide_vec(start, min, .before = Inf),
    roll_end = slide_vec(end, max, .before = Inf),
    overlap = if_else((start >= lag(roll_start) & start <= lag(roll_end)) |
      (end >= lag(roll_start) & end <= lag(roll_end)), "yes", "no")
  ) |>
  filter(overlap == "no" | is.na(overlap)) |> 
  select(- c(starts_with("roll_"), overlap))
#> # A tibble: 3 × 3
#> # Groups:   id [2]
#>      id start               end                
#>   <dbl> <dttm>              <dttm>             
#> 1     1 2015-03-16 10:40:00 2015-03-16 11:10:00
#> 2     1 2015-03-16 11:20:00 2015-03-16 11:28:56
#> 3     2 2015-03-16 10:40:00 2015-03-16 10:45:00

^{由 reprex package (v2.0.1)}

于 2022-04-30 创建

是否可以根据 R 中的时间变量过滤掉重叠的异常值？

Is it possible to filter out outliers that overlap based on time variable in R?

r

filter

outliers

data-cleaning