我有以下 df，它根据日期总结值:
df <- data_frame(
ID = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20),
Date = c("28/01/2019", "28/01/2019", "29/01/2019", "29/01/2019",
"30/01/2019", "30/01/2019", "31/01/2019", "31/01/2019",
"01/02/2019", "01/02/2019", "04/02/2019", "04/02/2019",
"05/02/2019", "05/02/2019", "06/02/2019", "06/02/2019",
"07/02/2019", "07/02/2019", "08/02/2019", "08/02/2019"),
WeekYear = c("2019-05", "2019-05", "2019-05", "2019-05", "2019-05", "2019-
05", "2019-05", "2019-05", "2019-05", "2019-05", "2019-06",
"2019-06", "2019-06", "2019-06", "2019-06", "2019-06", "2019-
06", "2019-06", "2019-06", "2019-06"),
DayFilter = c("Public Holiday", "Public Holiday", "Work-Day", "Work-Day",
"Work-Day", "Work-Day", "Work-Day", "Work-Day", "Work-Day",
"Work-Day", "Work-Day", "Work-Day", "Work-Day", "Work-Day",
"Work-Day", "Work-Day", "Work-Day", "Work-Day", "Work-Day",
"Work-Day"),
Value = c(0, 0, 7, 3, 8, 4, 1, 0, 9, 6, 4, 5, 7, 2, 4, 9, 5, 7, 4, 1))
df
ID Date WeekYear DayFilter Value
<dbl> <chr> <chr> <chr> <dbl>
1 1 28/01/2019 2019-05 Public Holiday 0
2 2 28/01/2019 2019-05 Public Holiday 0
3 3 29/01/2019 2019-05 Work-Day 7
4 4 29/01/2019 2019-05 Work-Day 3
5 5 30/01/2019 2019-05 Work-Day 8
6 6 30/01/2019 2019-05 Work-Day 4
7 7 31/01/2019 2019-05 Work-Day 1
8 8 31/01/2019 2019-05 Work-Day 0
9 9 01/02/2019 2019-05 Work-Day 9
10 10 01/02/2019 2019-05 Work-Day 6
11 11 04/02/2019 2019-06 Work-Day 4
12 12 04/02/2019 2019-06 Work-Day 5
13 13 05/02/2019 2019-06 Work-Day 7
14 14 05/02/2019 2019-06 Work-Day 2
15 15 06/02/2019 2019-06 Work-Day 4
16 16 06/02/2019 2019-06 Work-Day 9
17 17 07/02/2019 2019-06 Work-Day 5
18 18 07/02/2019 2019-06 Work-Day 7
19 19 08/02/2019 2019-06 Work-Day 4
20 20 08/02/2019 2019-06 Work-Day 1
我的目标是创建一个新列，根据 weekyear 列，将一年中任何有公共(public)假期的周分类为不完整。这样我就可以将分析限制在有五个工作日的几周内
期望输出如下
ID Date WeekYear DayFilter WeekFilter Value
<dbl> <chr> <chr> <chr> <chr> <dbl>
1 1 28/01/2019 2019-05 Public Holiday Incomplete 0
2 2 28/01/2019 2019-05 Public Holiday Incomplete 0
3 3 29/01/2019 2019-05 Work-Day Incomplete 7
4 4 29/01/2019 2019-05 Work-Day Incomplete 3
5 5 30/01/2019 2019-05 Work-Day Incomplete 8
6 6 30/01/2019 2019-05 Work-Day Incomplete 4
7 7 31/01/2019 2019-05 Work-Day Incomplete 1
8 8 31/01/2019 2019-05 Work-Day Incomplete 0
9 9 01/02/2019 2019-05 Work-Day Incomplete 9
10 10 01/02/2019 2019-05 Work-Day Incomplete 6
11 11 04/02/2019 2019-06 Work-Day Complete 4
12 12 04/02/2019 2019-06 Work-Day Complete 5
13 13 05/02/2019 2019-06 Work-Day Complete 7
14 14 05/02/2019 2019-06 Work-Day Complete 2
15 15 06/02/2019 2019-06 Work-Day Complete 4
16 16 06/02/2019 2019-06 Work-Day Complete 9
17 17 07/02/2019 2019-06 Work-Day Complete 5
18 18 07/02/2019 2019-06 Work-Day Complete 7
19 19 08/02/2019 2019-06 Work-Day Complete 4
20 20 08/02/2019 2019-06 Work-Day Complete 1
到目前为止，我的尝试已成功创建一个以 Incomplete 作为值的 WeekFilter 列。但仅限于公共(public)假期的特定日期。我正在努力弄清楚的部分是如何将其推广到本周剩余的时间，我认为需要将 WeekYear 列合并到此语句中
df$WeekFilter = df$WeekYear
df$WeekFilter[df$DayFilter == "Public Holiday"] <- "Incomplete"
任何帮助将不胜感激
最佳答案
我们可以使用group_by和mutate:
df %>% group_by(WeekYear) %>%
mutate(WeekFilter = if("Public Holiday" %in% DayFilter) "Incomplete" else "Complete")
关于r - 根据两列中的值在 R 中创建新列，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/54412291/