R 根据条件跨列搜索最小日期

Question

我有以下数据框架，其中有几个日期变量。

    x <- structure(list(id = c(1, 2, 3, 4), date = structure(c(18611, 
16801, 16801, 17532), class = "Date"), s1 = c(0, 1, 1, NA), date1 = structure(c(17880, 
16450, 16416, NA), class = "Date"), s2 = c(0, 0, 1, NA), date2 = structure(c(17880, 
NA, 15869, NA), class = "Date"), DN = structure(c(18611, 15869, 
15869, NA), class = "Date")), class = c("tbl_df", "tbl", "data.frame"
), row.names = c(NA, -4L))

我想将date1和date2和生成dn根据case_when 。我目前正在使用此代码：

x <- mutate(date = as.Date(date),
           date1 =  as.Date(date1),
           date2 =  as.Date(date2),
           DN = case_when(
               s1 == 1 | s2 == 1 ~ min(date1, date2, na.rm = T),
               s1 == 0 | s2 == 0 ~ date,
               is.na(s1) & is.na(s2) ~ NA_real_
           ))

但是，我会得到一个奇怪的结果！对于id = 2 dn的值是从id = 3中获取的，我无法理解！

有什么想法吗？ 感谢向前

Answer 1

您可以使用 pmin 选择指定列的第一个日期。您可以使用以下代码：

library(dplyr)
x %>% 
  mutate(DN = case_when(
    s1 == 1 | s2 == 1 ~ pmin(date1, date2, na.rm = T),
    s1 == 0 | s2 == 0 ~ date,
    is.na(s1) & is.na(s2) ~ NA_real_
  ))

输出：

# A tibble: 4 × 7
     id date          s1 date1         s2 date2      DN        
  <dbl> <date>     <dbl> <date>     <dbl> <date>     <date>    
1     1 2020-12-15     0 2018-12-15     0 2018-12-15 2020-12-15
2     2 2016-01-01     1 2015-01-15     0 NA         2015-01-15
3     3 2016-01-01     1 2014-12-12     1 2013-06-13 2013-06-13
4     4 2018-01-01    NA NA            NA NA         NA 

Answer 2

如果您添加rowwise（）（即行分组），则将获得所需的行最小值：

x %>% 
  rowwise() %>% 
  mutate(date = as.Date(date),
         date1 =  as.Date(date1),
         date2 =  as.Date(date2),
         DN = case_when(
           s1 == 1 | s2 == 1 ~ pmin(date1, date2, na.rm = T),
           s1 == 0 | s2 == 0 ~ date,
           is.na(s1) & is.na(s2) ~ NA_real_
         ))