R: (un)reduce 数据框

Question

我有以下假数据集。在每天 ( dates ) 的一段时间内，所有元素 ( status ) 的状态 ( id ) 都会被记录下来。

df <- data.frame( id = c(1, 1, 1, 1, 1,  2, 2, 2, 2, 2,  3, 3, 3, 3, 3,  4, 4, 4, 4, 4),
                  dates = c("2021-01-01",
                           "2021-01-02",
                           "2021-01-03",
                           "2021-01-04",
                           "2021-01-05",
                           
                           "2021-01-01",
                           "2021-01-02",
                           "2021-01-03",
                           "2021-01-04",
                           "2021-01-05",
                           
                           "2021-01-01",
                           "2021-01-02",
                           "2021-01-03",
                           "2021-01-04",
                           "2021-01-05",
                           
                           "2021-01-01",
                           "2021-01-02",
                           "2021-01-03",
                           "2021-01-04",
                           "2021-01-05"),
                 
                 status = c("A", "A", "A", "B", "C",
                            "A", "A", "B", "C", "C",
                            "A", "B", "C", "D", "E",
                            "A", "B", "B", "B", "B")
                 ) 

> df
   id      dates status
1   1 2021-01-01      A
2   1 2021-01-02      A
3   1 2021-01-03      A
4   1 2021-01-04      B
5   1 2021-01-05      C
6   2 2021-01-01      A
7   2 2021-01-02      A
8   2 2021-01-03      B
9   2 2021-01-04      C
10  2 2021-01-05      C
11  3 2021-01-01      A
12  3 2021-01-02      B
13  3 2021-01-03      C
14  3 2021-01-04      D
15  3 2021-01-05      E
16  4 2021-01-01      A
17  4 2021-01-02      B
18  4 2021-01-03      B
19  4 2021-01-04      B
20  4 2021-01-05      B

不幸的是，为了节省空间，减少了数据帧:如果在随后的两天中状态相同，则删除第二个条目。假设状态保持不变，直到再次更改，因此实际数据集如下所示:

> df %>% group_by(id) %>%
+   mutate(dupl = duplicated(status, 2)) %>%
+   ungroup() %>%
+   filter(dupl == FALSE) %>%
+   select(-dupl)
# A tibble: 13 x 3
      id dates      status
   <dbl> <chr>      <chr> 
 1     1 2021-01-01 A     
 2     1 2021-01-04 B     
 3     1 2021-01-05 C     
 4     2 2021-01-01 A     
 5     2 2021-01-03 B     
 6     2 2021-01-04 C     
 7     3 2021-01-01 A     
 8     3 2021-01-02 B     
 9     3 2021-01-03 C     
10     3 2021-01-04 D     
11     3 2021-01-05 E     
12     4 2021-01-01 A     
13     4 2021-01-02 B 

我现在的问题是:我怎样才能再次回到数据集的第一个(完整)版本？所有 id 的时间段始终相同s (2021-01-01 至 2021-01-05)

Answer 1

library(tidyverse)

# the reduced version can be created like this instead
df_reduced <- df %>% 
  mutate(dates = lubridate::ymd(dates)) %>% 
  distinct(id, status, .keep_all = TRUE)

对于这样的问题，我会查看 tidyr 中的函数相关 missing values .我们可以用expand生成完整的id/dates组合序列，然后填写 NA值与 fill(status, .direction = "down") .

df_reduced %>% 
  expand(id, dates = full_seq(dates, 1)) %>% 
  left_join(df_reduced) %>% 
  group_by(id) %>% 
  fill(status, .direction = "down")

#> Joining, by = c("id", "dates")
#> # A tibble: 20 x 3
#> # Groups:   id [4]
#>       id dates      status
#>    <dbl> <chr>      <chr> 
#>  1     1 2021-01-01 A     
#>  2     1 2021-01-02 A     
#>  3     1 2021-01-03 A     
#>  4     1 2021-01-04 B     
#>  5     1 2021-01-05 C     
#>  6     2 2021-01-01 A     
#>  7     2 2021-01-02 A     
#>  8     2 2021-01-03 B     
#>  9     2 2021-01-04 C     
#> 10     2 2021-01-05 C     
#> 11     3 2021-01-01 A     
#> 12     3 2021-01-02 B     
#> 13     3 2021-01-03 C     
#> 14     3 2021-01-04 D     
#> 15     3 2021-01-05 E     
#> 16     4 2021-01-01 A     
#> 17     4 2021-01-02 B     
#> 18     4 2021-01-03 B     
#> 19     4 2021-01-04 B     
#> 20     4 2021-01-05 B

创建于 2021-07-06 由 reprex package (v1.0.0)