根据每组不同的列填写 NA 直到特定日期答案

【问题标题】：Fill NA until certain date based on different column per group根据每组不同的列填写 NA 直到特定日期
【发布时间】：2023-01-25 20:59:13
【问题描述】：

我有以下数据框 df（dput 下面）：

   group      date1      date2 value
1      A 2022-01-01 2022-01-07    NA
2      A 2022-01-02 2022-01-07     1
3      A 2022-01-04 2022-01-07    NA
4      A 2022-01-10 2022-01-07    NA
5      B 2022-01-01 2022-01-06     3
6      B 2022-01-03 2022-01-06    NA
7      B 2022-01-04 2022-01-06    NA
8      B 2022-01-06 2022-01-06    NA
9      C 2022-01-01 2022-01-09    NA
10     C 2022-01-03 2022-01-09    NA
11     C 2022-01-04 2022-01-09     2
12     C 2022-01-11 2022-01-09    NA

我想在 date2 之前填写每组的 NA 值。因此，如果 date1 较小，则应填充第一行（值为 1 的 A 组）之后的所有 NA 值，直到 date2。这是所需的输出：

   group      date1      date2 value
1      A 2022-01-01 2022-01-07    NA
2      A 2022-01-02 2022-01-07     1
3      A 2022-01-04 2022-01-07     1
4      A 2022-01-10 2022-01-07    NA
5      B 2022-01-01 2022-01-06     3
6      B 2022-01-03 2022-01-06     3
7      B 2022-01-04 2022-01-06     3
8      B 2022-01-06 2022-01-06     3
9      C 2022-01-01 2022-01-09    NA
10     C 2022-01-03 2022-01-09    NA
11     C 2022-01-04 2022-01-09     2
12     C 2022-01-11 2022-01-09    NA

正如您在 A 组中看到的那样，只有下一行填充了 1，因为最后一个日期 1 在日期 2 之后。所以我想知道是否有人知道如何根据每个组的特定日期填写 NA？

dputdf:

df<-structure(list(group = c("A", "A", "A", "A", "B", "B", "B", "B", 
"C", "C", "C", "C"), date1 = c("2022-01-01", "2022-01-02", "2022-01-04", 
"2022-01-10", "2022-01-01", "2022-01-03", "2022-01-04", "2022-01-06", 
"2022-01-01", "2022-01-03", "2022-01-04", "2022-01-11"), date2 = c("2022-01-07", 
"2022-01-07", "2022-01-07", "2022-01-07", "2022-01-06", "2022-01-06", 
"2022-01-06", "2022-01-06", "2022-01-09", "2022-01-09", "2022-01-09", 
"2022-01-09"), value = c(NA, 1, NA, NA, 3, NA, NA, NA, NA, NA, 
2, NA)), class = "data.frame", row.names = c(NA, -12L))

【问题讨论】：

标签： r dataframe fill

【解决方案1】：

创建一个额外的组并使用fill

library(dplyr)
library(tidyr)
df %>% 
  group_by(group, after = date1 > date2) %>% 
  fill(value) %>% 
  ungroup() %>% 
  select(-after)

# A tibble: 12 × 4
   group date1      date2      value
   <chr> <chr>      <chr>      <dbl>
 1 A     2022-01-01 2022-01-07    NA
 2 A     2022-01-02 2022-01-07     1
 3 A     2022-01-04 2022-01-07     1
 4 A     2022-01-10 2022-01-07    NA
 5 B     2022-01-01 2022-01-06     3
 6 B     2022-01-03 2022-01-06     3
 7 B     2022-01-04 2022-01-06     3
 8 B     2022-01-06 2022-01-06     3
 9 C     2022-01-01 2022-01-09    NA
10 C     2022-01-03 2022-01-09    NA
11 C     2022-01-04 2022-01-09     2
12 C     2022-01-11 2022-01-09    NA

【讨论】：

【解决方案2】：

使用数据表

library(data.table)                                                                                                                                                    

setDT(df)[as.Date(date1) <= as.Date(date2), value := nafill(value, type = "locf"), group]

结果

    group      date1      date2 value
 1:     A 2022-01-01 2022-01-07    NA
 2:     A 2022-01-02 2022-01-07     1
 3:     A 2022-01-04 2022-01-07     1
 4:     A 2022-01-10 2022-01-07    NA
 5:     B 2022-01-01 2022-01-06     3
 6:     B 2022-01-03 2022-01-06     3
 7:     B 2022-01-04 2022-01-06     3
 8:     B 2022-01-06 2022-01-06     3
 9:     C 2022-01-01 2022-01-09    NA
10:     C 2022-01-03 2022-01-09    NA
11:     C 2022-01-04 2022-01-09     2
12:     C 2022-01-11 2022-01-09    NA

【讨论】：