我希望组合润滑间隔,以便如果它们重叠,则从内部第一个时间中取出最小值,并从内部最后一个时间中取出最大值,并汇总以创建一个跨越整个周期的新间隔。这是一个代表:
library(lubridate, warn.conflicts = FALSE)
library(dplyr, warn.conflicts = FALSE)
library(tibble)
dat <- tibble(
animal = rep(c("elk", "wolf", "moose"), each = 2),
date_interval = c(
interval(as.Date("2020-04-01"), as.Date("2020-04-05")),
interval(as.Date("2020-04-10"), as.Date("2020-04-15")),
interval(as.Date("2020-03-01"), as.Date("2020-04-01")),
interval(as.Date("2020-02-15"), as.Date("2020-03-15")),
interval(as.Date("2020-10-01"), as.Date("2020-11-01")),
interval(as.Date("2020-09-15"), as.Date("2020-10-15"))
)
)
dat
#> # A tibble: 6 x 2
#> animal date_interval
#> <chr> <Interval>
#> 1 elk 2020-04-01 UTC--2020-04-05 UTC
#> 2 elk 2020-04-10 UTC--2020-04-15 UTC
#> 3 wolf 2020-03-01 UTC--2020-04-01 UTC
#> 4 wolf 2020-02-15 UTC--2020-03-15 UTC
#> 5 moose 2020-10-01 UTC--2020-11-01 UTC
#> 6 moose 2020-09-15 UTC--2020-10-15 UTC
好的,所以在wolf
and moose
级别,我们有重叠的间隔。假设这是same狼和驼鹿之类的东西会重复计算天数:
dat %>%
group_by(animal) %>%
mutate(time = time_length(date_interval)) %>%
summarise(time_cumu = sum(time))
#> `summarise()` ungrouping output (override with `.groups` argument)
#> # A tibble: 3 x 2
#> animal time_cumu
#> <chr> <dbl>
#> 1 elk 777600
#> 2 moose 5270400
#> 3 wolf 5184000
这是我想要得到的输出类型,它总结了重叠间隔:
tibble(
animal = c("elk", "elk", "wolf", "moose"),
date_interval = c(
interval(as.Date("2020-04-01"), as.Date("2020-04-05")),
interval(as.Date("2020-04-10"), as.Date("2020-04-15")),
interval(as.Date("2020-02-15"), as.Date("2020-04-01")),
interval(as.Date("2020-09-15"), as.Date("2020-11-01"))
)
)
#> # A tibble: 4 x 2
#> animal date_interval
#> <chr> <Interval>
#> 1 elk 2020-04-01 UTC--2020-04-05 UTC
#> 2 elk 2020-04-10 UTC--2020-04-15 UTC
#> 3 wolf 2020-02-15 UTC--2020-04-01 UTC
#> 4 moose 2020-09-15 UTC--2020-11-01 UTC
Ideas?