试试这个：

library(dplyr)
quux %>%
  mutate(Status.Date = as.Date(Status.Date, format = "%m/%d/%Y")) %>%
  group_by(ID) %>%
  mutate(
    time_to_5000 =
      if (5000 %in% Status) Status.Date[Status==5000][1] - Status.Date else NA
  ) %>%
  ungroup()
# # A tibble: 5 × 6
#   ID     Status Status.Date Model Ship.to.Location time_to_5000
#   <chr>   <dbl> <date>      <chr> <chr>            <drtn>      
# 1 ABC123   3000 2023-01-01  M     X                14 days     
# 2 ABC123   3100 2023-01-10  M     X                 5 days     
# 3 ABC124   2000 2023-11-01  N     Y                NA days     
# 4 ABC124   2500 2023-12-10  N     Y                NA days     
# 5 ABC123   5000 2023-01-15  M     X                 0 days

从技术上讲，我们不需要if语句。也就是说，这也起作用：

... %>%
  mutate(time_to_5000 = Status.Date[Status==5000][1] - Status.Date) %>%
  ...

我把它包括在上面，因为我不确定它是否有意义有多个条件（e。例如，both 3000/5000必须存在），以便返回非NA值。有很多方法可以实现这个逻辑。
数据（增加到至少包括一个5000）：

quux <- structure(list(ID = c("ABC123", "ABC123", "ABC124", "ABC124", "ABC123"), Status = c(3000, 3100, 2000, 2500, 5000), Status.Date = c("1/1/2023", "1/10/2023", "11/1/2023", "12/10/2023", "1/15/2023"), Model = c("M", "M", "N", "N", "M"), Ship.to.Location = c("X", "X", "Y", "Y", "X")), row.names = c(NA, -5L), class = "data.frame")