有条件的按组的最后日期答案

【问题标题】：Last date by group with conditions有条件的按组的最后日期
【发布时间】：2022-11-02 04:14:39
【问题描述】：

这个问题是this question 的后续问题，但每个idPerson 可以有多个decision == "d"。有多个idPerson，但一个足以解释这个问题。 idAppt 嵌套在 idPerson 中。考虑这个数据框。

   idPerson idAppt decision date      
 1 A             1 a        2021-09-10
 2 A             1 b        2021-09-11
 3 A             1 c        2021-09-12
 4 A             1 d        2021-09-13
 5 A             2 a        2021-09-20
 6 A             2 b        2021-09-21
 7 A             3 a        2021-09-10
 8 A             3 b        2021-09-11
 9 A             4 a        2021-09-21
10 A             4 b        2021-09-22
11 A             4 c        2021-09-23
12 A             4 d        2021-09-24
13 A             5 a        2021-09-10
14 A             5 b        2021-09-11
15 A             6 a        2021-10-10
16 A             6 b        2021-10-11

我想构建一个回复这些条件的date2 列：

对于给定的idAppt，如果decision == "a" 晚于同一idPerson 的decision == "d" 时的任何其他日期，则在decision == "d" 的idPerson 时报告date 的最新值（之前最接近）。例如idAppt == 2组中decision == "a"的日期晚于idAppt == 1组decision == "d"的日期，所以date2应该是2021-09-13。同样适用于idAppt == 6 组，但这里有两个更早的decision == "d"（第 4 行和第 12 行）。在这种情况下，date2 应该是2021-10-10 之前最接近的，即2021-09-23。
对于给定的idAppt，当没有decision == "d" 的date 早于decision == "a" 的date 时，取给定idPerson 中最早的一个。

这给出了以下所需的输出：

   idPerson idAppt decision date       date2       
 1 A             1 a        2021-09-10 2021-09-10
 2 A             1 b        2021-09-11 2021-09-10
 3 A             1 c        2021-09-12 2021-09-10
 4 A             1 d        2021-09-13 2021-09-10
 5 A             2 a        2021-09-20 2021-09-13 #<- correspond to value of row 4
 6 A             2 b        2021-09-21 2021-09-13  
 7 A             3 a        2021-09-10 2021-09-10 
 8 A             3 b        2021-09-11 2021-09-10
 9 A             4 a        2021-09-21 2021-09-13
10 A             4 b        2021-09-22 2021-09-13
11 A             4 c        2021-09-23 2021-09-13
12 A             4 d        2021-09-24 2021-09-13
13 A             5 a        2021-09-11 2021-09-10 #<- earliest value because 2021-09-10 is earlier than 2021-09-13
14 A             5 b        2021-09-12 2021-09-10
15 A             6 a        2021-10-10 2021-09-24 #<- correspond to value of row 12
16 A             6 b        2021-10-11 2021-09-24

数据

df <- structure(list(idPerson = c("A", "A", "A", "A", "A", "A", "A", 
"A", "A", "A", "A", "A", "A", "A", "A", "A"), idAppt = c(1L, 
1L, 1L, 1L, 2L, 2L, 3L, 3L, 4L, 4L, 4L, 4L, 5L, 5L, 6L, 6L), 
    decision = c("a", "b", "c", "d", "a", "b", "a", "b", "a", 
    "b", "c", "d", "a", "b", "a", "b"), date = structure(c(18880, 
    18881, 18882, 18883, 18890, 18891, 18880, 18881, 18891, 18892, 
    18893, 18894, 18881, 18882, 18910, 18911), class = "Date")), class = c("tbl_df", 
"tbl", "data.frame"), row.names = c(NA, -16L))

EO <- structure(list(idPerson = c("A", "A", "A", "A", "A", "A", "A", 
"A", "A", "A", "A", "A", "A", "A", "A", "A"), idAppt = c(1L, 
1L, 1L, 1L, 2L, 2L, 3L, 3L, 4L, 4L, 4L, 4L, 5L, 5L, 6L, 6L), 
    decision = c("a", "b", "c", "d", "a", "b", "a", "b", "a", 
    "b", "c", "d", "a", "b", "a", "b"), date = structure(c(18880, 
    18881, 18882, 18883, 18890, 18891, 18880, 18881, 18891, 18892, 
    18893, 18894, 18881, 18882, 18910, 18911), class = "Date"), 
    date2 = c("2021-09-10", "2021-09-10", "2021-09-10", "2021-09-10", 
    "2021-09-13", "2021-09-13", "2021-09-10", "2021-09-10", "2021-09-13", 
    "2021-09-13", "2021-09-13", "2021-09-13", "2021-09-10", "2021-09-10", 
    "2021-09-24", "2021-09-24")), row.names = c(NA, -16L), class = c("tbl_df", 
"tbl", "data.frame"))

【问题讨论】：

标签： r date grouping

【解决方案1】：

这是我解决问题的方法，虽然看起来有点复杂：

library(dplyr)
df %>%
  group_by(idPerson) %>%
  mutate(d_date = list(date[decision == "d"]), min_date_person = min(date)) %>% 
  group_by(idPerson, idAppt) %>%
  mutate(date3 = unlist(map(d_date, (x){
          dates <- date[decision == "a"] - x
          w <- which.min(dates[dates > 0])
          ifelse(is.null(w), NA, w)
          })),
         date2 = if_else(is.na(date3), min_date_person, do.call("c", map(d_date, ~ unique(.x[date3]))))) %>% 
  ungroup() %>% 
  select(1:4, date2)

# A tibble: 16 × 5
   idPerson idAppt decision date       date2     
   <chr>     <int> <chr>    <date>     <date>    
 1 A             1 a        2021-09-10 2021-09-10
 2 A             1 b        2021-09-11 2021-09-10
 3 A             1 c        2021-09-12 2021-09-10
 4 A             1 d        2021-09-13 2021-09-10
 5 A             2 a        2021-09-20 2021-09-13
 6 A             2 b        2021-09-21 2021-09-13
 7 A             3 a        2021-09-10 2021-09-10
 8 A             3 b        2021-09-11 2021-09-10
 9 A             4 a        2021-09-21 2021-09-13
10 A             4 b        2021-09-22 2021-09-13
11 A             4 c        2021-09-23 2021-09-13
12 A             4 d        2021-09-24 2021-09-13
13 A             5 a        2021-09-11 2021-09-10
14 A             5 b        2021-09-12 2021-09-10
15 A             6 a        2021-10-10 2021-09-24
16 A             6 b        2021-10-11 2021-09-24

【讨论】：

我昨天试了一下，但对我来说太复杂了。感谢分享。

【解决方案2】：

使用 data.table 滚动连接：

library(data.table)
setDT(df)

# rolling join between decision "d" and "a"
df[decision == "a", date2 := df[decision == "d"][.SD, on = .(idPerson, date), x.date, roll = Inf]]

# set non-matching rows for decision "a" to min(date)
df[decision == "a" & is.na(date2), date2 := min(date), by = idPerson]

# replace other NA by last observation carried forward
setnafill(df, type = "locf", cols = "date2")

    idPerson idAppt decision       date      date2
 1:        A      1        a 2021-09-10 2021-09-10
 2:        A      1        b 2021-09-11 2021-09-10
 3:        A      1        c 2021-09-12 2021-09-10
 4:        A      1        d 2021-09-13 2021-09-10
 5:        A      2        a 2021-09-20 2021-09-13
 6:        A      2        b 2021-09-21 2021-09-13
 7:        A      3        a 2021-09-10 2021-09-10
 8:        A      3        b 2021-09-11 2021-09-10
 9:        A      4        a 2021-09-21 2021-09-13
10:        A      4        b 2021-09-22 2021-09-13
11:        A      4        c 2021-09-23 2021-09-13
12:        A      4        d 2021-09-24 2021-09-13
13:        A      5        a 2021-09-11 2021-09-10
14:        A      5        b 2021-09-12 2021-09-10
15:        A      6        a 2021-10-10 2021-09-24
16:        A      6        b 2021-10-11 2021-09-24

'idAppt' 的相关性并不完全清楚，因为似乎日期的比较是在身份证.

【讨论】：