碱R

data[
  ave(as.logical(data$Departure),data$ID,FUN=function(x){
    t=cumsum(!x)
    t==max(t)
  }),
]

  ID Departure  Date
2  A     FALSE Jan 2
3  A      TRUE Jan 3
6  B     FALSE Jan 3
7  B      TRUE Jan 4

使用data.table的另一种方法

注意我准备了你的数据，使用逻辑TRUE/FALSE，我使用日期而不是月日字符串。
数据

ID <- rep(c("A","B"),times=c(3,4))
Departure <- as.logical(c("TRUE","FALSE","TRUE","TRUE","FALSE","FALSE","TRUE"))
Date <- lubridate::parse_date_time(c("Jan 1","Jan 2","Jan 3","Jan 1","Jan 2","Jan 3","Jan 4"), "md")

data <- data.frame(ID, Departure, Date)

解决方案

首先，我们为每个ID提取您最近的FALSE记录的子集，然后将这些结果与最近日期的所有记录连接起来。

library(data.table)
setDT(data)

results <- data[!Departure, .SD[.N], ID][data, .(ID, Departure = i.Departure, Date), on = .(ID, Date <= Date), nomatch = 0]

结果

results

   ID Departure       Date
1:  A     FALSE 2023-01-02
2:  A      TRUE 2023-01-03
3:  B     FALSE 2023-01-03
4:  B      TRUE 2023-01-04

扩展

如果您真的想获得原来的日期格式，只需在之后再次格式化它们

results[, Date := format(Date, "%b %d")]

results

   ID Departure   Date
1:  A     FALSE Jan 02
2:  A      TRUE Jan 03
3:  B     FALSE Jan 03
4:  B      TRUE Jan 04

这是一个有趣的挑战！我相信你可以想出更聪明的方法，但这是我的方法（使用dplyr）：

group_by(data, ID) %>% mutate(row=1:n(), 
                              rowF=ifelse(Departure==F, row, NA), 
                              maxF=max(rowF, na.rm=T)) %>% 
  filter(row >= maxF) %>% 
  select(-c(row:maxF)) %>%  
  # remove the above line if you want to see how the sausage was made
  ungroup()

输出：

# A tibble: 4 x 3
  ID    Departure Date 
  <chr> <chr>     <chr>
1 A     FALSE     Jan 2
2 A     TRUE      Jan 3
3 B     FALSE     Jan 3
4 B     TRUE      Jan 4

您可以检查取反的Departure列的总和是否等于取反的Departure列的cumsum。请注意，示例数据中的此列是字符列，需要更改为逻辑列：

library(dplyr)

data |>
  mutate(Departure = as.logical(Departure)) |>
  filter(sum(!Departure) == cumsum(!Departure) & any(!Departure), .by = ID)

  ID Departure  Date
1  A     FALSE Jan 2
2  A      TRUE Jan 3
3  B     FALSE Jan 3
4  B      TRUE Jan 4

我不确定是否需要在过滤条件中使用& any(!Departure)，但这将确保只保留包含FALSE值的组。

这适用于您提供的数据，尽管我不确定它是否适用于其他示例（我不清楚ID周围的规则-所以您可能想检查一下，但它应该可以工作：

df %>% 
mutate(Departure = ifelse(Departure == "TRUE", TRUE, FALSE)) %>% # change the Departure column to logical, not character
filter((Departure == TRUE & lag(Departure) == FALSE & ID == lag(ID)) | (Departure == FALSE & lead(Departure) == TRUE & lead(Departure) != FALSE) & ID == lead(ID))
# if the departure is TRUE and the previous departure is FALSE and the ID is the same as the previous ID, 
# or 
# if the departure is FALSE and the next departure is TRUE and the next departure is not FALSE and the ID is the same as the next ID, #then keep the row

# A tibble: 4 × 3
  ID    Departure Date 
  <chr> <lgl>     <chr>
1 A     FALSE     Jan 2
2 A     TRUE      Jan 3
3 B     FALSE     Jan 3
4 B     TRUE      Jan 4

使用rev重置rleby ID。

by(data, data$ID, \(x) tail(x, with(rle(rev(x$Departure)), which.max(values == 'FALSE')))) |>
  do.call(what='rbind')
#     ID Departure  Date
# A.2  A     FALSE Jan 2
# A.3  A      TRUE Jan 3
# B.6  B     FALSE Jan 3
# B.7  B      TRUE Jan 4

注意，如果Departure列为真布尔值，

data$Departure <- as.logical(data$Departure)

它简化为

by(data, data$ID, \(x) tail(x, with(rle(rev(x$Departure)), which.max(!values)))) |>
  do.call(what='rbind')