使用 dplyr 从每组的另一列中识别另一个日期之前的最近日期？答案

【问题标题】：Identify the most recent date before another date from another column per group using dplyr?使用 dplyr 从每组的另一列中识别另一个日期之前的最近日期？
【发布时间】：2020-08-17 21:58:58
【问题描述】：

我有一个如下结构的数据框：

example <- data.frame(id = c(1,1,1,2,2,2,3,3,3),
                         delivereddate = c("7/20/2019","7/24/2019","7/28/2019","3/24/2019","4/13/2019","4/25/2019","11/13/2019","11/20/2019","11/27/2019"),
                         applieddate = c("7/22/2019","7/22/2019","7/22/2019",NA,NA,NA,"11/21/2019","11/21/2019","11/21/2019"))

我正在尝试添加一个列，用于标识每个 ID 的应用日期之前的最近交付日期。我试图获得的最终结果的一个示例如下：

desiredresult <- data.frame(id = c(1,1,1,2,2,2,3,3,3),
                            delivereddate = c("7/20/2019","7/24/2019","7/28/2019","3/24/2019","4/13/2019","4/25/2019","11/13/2019","11/20/2019","11/27/2019"),
                            applieddate = c("7/22/2019","7/22/2019","7/22/2019",NA,NA,NA,"11/21/2019","11/21/2019","11/21/2019"),
                            applied = c(1,0,0,0,0,0,0,1,0))

我需要应用的列是二进制的（0 或 1），并且每个 id 只能有 1 行带有 1 标志。如果一个 id 没有应用日期，那么所有行的应用标志都是 0。

【问题讨论】：

标签： r dplyr

【解决方案1】：

我们可以使用findInterval

library(dplyr)
library(lubridate)
example %>% 
      dplyr::group_by(id) %>% 
      dplyr::mutate(applied = +(row_number() %in% 
              findInterval(lubridate::mdy(first(applieddate)), 
                          lubridate::mdy(delivereddate))))
# A tibble: 9 x 4
# Groups:   id [3]
#     id delivereddate applieddate applied
#  <dbl> <chr>         <chr>         <int>
#1     1 7/20/2019     7/22/2019         1
#2     1 7/24/2019     7/22/2019         0
#3     1 7/28/2019     7/22/2019         0
#4     2 3/24/2019     <NA>              0
#5     2 4/13/2019     <NA>              0
#6     2 4/25/2019     <NA>              0
#7     3 11/13/2019    11/21/2019        0
#8     3 11/20/2019    11/21/2019        1
#9     3 11/27/2019    11/21/2019        0

【讨论】：

【解决方案2】：

您可以将列转换为日期类，从delivereddate 中减去applieddate 并取绝对值。然后，对于每个id，我们将 1 分配给观察到最小差异的索引。

library(dplyr)

example %>%
  mutate(across(ends_with('date'), lubridate::mdy), 
         applied = abs(delivereddate - applieddate)) %>%
  group_by(id) %>%
  mutate(applied = +(row_number() %in% which.min(applied)))

#     id delivereddate applieddate applied
#  <dbl> <date>        <date>        <int>
#1     1 2019-07-20    2019-07-22        1
#2     1 2019-07-24    2019-07-22        0
#3     1 2019-07-28    2019-07-22        0
#4     2 2019-03-24    NA                0
#5     2 2019-04-13    NA                0
#6     2 2019-04-25    NA                0
#7     3 2019-11-13    2019-11-21        0
#8     3 2019-11-20    2019-11-21        1
#9     3 2019-11-27    2019-11-21        0

【讨论】：