我认为这是游程编码(base::rle、dplyr::consecutive_id或data.table::rleid之一)和一些更简单的过滤的组合.

您将看到2*365+2作为一个判别器的用法:用于时差的POSIXt方法没有"year"作为选项，所以我们需要以天为单位，并且在2002-2007有一个闰年.

dplyr

library(dplyr)
dat %>%
  group_by(grp = consecutive_id(value <= -50)) %>%
  filter(any(value <= -50), row_number() == n()) %>%
  ungroup() %>%
  filter(row_number() == 1L | difftime(time, time[1], units="day") >= (2*365+2))
# # A tibble: 2 × 3
#   time       value   grp
#   <date>     <dbl> <int>
# 1 2002-12-31 -52.4     1
# 2 2007-12-31 -52.4     5

data.table

library(data.table)
as.data.table(dat)[, .SD[any(value <= -50), .(time, value)][.N,], by = .(grp = rleid(value <= -50))
  ][(seq(.N) == 1 | difftime(time, time[1], units="day") >= (2*365+2)),]
#      grp       time  value
#    <int>     <Date>  <num>
# 1:     1 2002-12-31 -52.44
# 2:     5 2007-12-31 -52.40

base R

(还有一点工作要做.)

# a home-grown base-R version of `rleid` and `consecutive_id` above
my_rleid <- function(...) {
  r <- rle(do.call(paste, c(list(...), sep = "_")))$lengths
  rep(seq_along(r), times = r)
}
dat |>
  transform(grp = my_rleid(value <= -50)) |>
  subset(ave(value <= -50, grp, FUN = function(z) any(z) & seq_along(z) == length(z))) |>
  subset(seq_along(time) == 1L | difftime(time, time[1], units="day") >= (2*365+2))
#         time  value grp
# 2 2002-12-31 -52.44   1
# 7 2007-12-31 -52.40   5