如何使用 R 有条件地将行插入数据框中？答案

【问题标题】：How to conditionally insert rows into a data frame with R?如何使用 R 有条件地将行插入数据框中？
【发布时间】：2022-01-18 15:35:48
【问题描述】：

我正在尝试根据变异列 (Day) 的 Sys.Date() 是否为 Tue 有条件地插入行。如果是这样，我想插入前两天在MaxDate 中列出的行。如果Day 列不是Tue，那么我只想让数据框保持原样。我认为您不能在数据框上使用 if_else() 并且不确定如何去做。也许以某种方式使用add_row()？

这就是我所拥有的：

ID	Product	MaxDate	Day
100	candy	2022-01-18	Tue
100	chips	2022-01-18	Tue
101	candy	2022-01-18	Tue
101	chips	2022-01-18	Tue
102	candy	2022-01-18	Tue
103	candy	2022-01-13	Tue
103	chips	2022-01-13	Tue

如果是星期二，这就是我想要的：

ID	Product	MaxDate	Day
100	candy	2022-01-16	Tue
100	chips	2022-01-16	Tue
100	candy	2022-01-17	Tue
100	chips	2022-01-17	Tue
100	candy	2022-01-18	Tue
100	chips	2022-01-18	Tue
101	candy	2022-01-16	Tue
101	chips	2022-01-16	Tue
101	candy	2022-01-17	Tue
101	chips	2022-01-17	Tue
101	candy	2022-01-18	Tue
101	chips	2022-01-18	Tue
102	candy	2022-01-16	Tue
102	candy	2022-01-17	Tue
102	candy	2022-01-18	Tue
103	candy	2022-01-16	Tue
103	chips	2022-01-16	Tue
103	candy	2022-01-17	Tue
103	chips	2022-01-17	Tue
103	candy	2022-01-13	Tue
103	chips	2022-01-13	Tue

如果不是Tue，我希望数据框不变：

ID	Product	MaxDate	Day
100	candy	2022-01-17	Mon
100	chips	2022-01-17	Mon
101	candy	2022-01-17	Mon
101	chips	2022-01-17	Mon
102	candy	2022-01-17	Mon
103	candy	2022-01-13	Mon
103	chips	2022-01-13	Mon

谢谢。

【问题讨论】：

我听到你的声音 - 刚刚做到了！谢谢

标签： r date dplyr lubridate

【解决方案1】：

library(dplyr, warn.conflicts = FALSE)

df = read.table(text = 'ID  Product MaxDate Day
100 candy   2022-01-18  Tue
100 chips   2022-01-18  Tue
101 candy   2022-01-18  Tue
101 chips   2022-01-18  Tue
102 candy   2022-01-18  Tue
103 candy   2022-01-13  Wed
103 chips   2022-01-13  Tue', header = T) %>%
  mutate(MaxDate = as.Date(MaxDate))


df %>% 
  left_join(tibble(Day = 'Tue', lagged_days = 2:0)) %>% 
  mutate(MaxDate = MaxDate - coalesce(lagged_days, 0),
         lagged_days = NULL)
#> Joining, by = "Day"
#>     ID Product    MaxDate Day
#> 1  100   candy 2022-01-16 Tue
#> 2  100   candy 2022-01-17 Tue
#> 3  100   candy 2022-01-18 Tue
#> 4  100   chips 2022-01-16 Tue
#> 5  100   chips 2022-01-17 Tue
#> 6  100   chips 2022-01-18 Tue
#> 7  101   candy 2022-01-16 Tue
#> 8  101   candy 2022-01-17 Tue
#> 9  101   candy 2022-01-18 Tue
#> 10 101   chips 2022-01-16 Tue
#> 11 101   chips 2022-01-17 Tue
#> 12 101   chips 2022-01-18 Tue
#> 13 102   candy 2022-01-16 Tue
#> 14 102   candy 2022-01-17 Tue
#> 15 102   candy 2022-01-18 Tue
#> 16 103   candy 2022-01-13 Wed
#> 17 103   chips 2022-01-11 Tue
#> 18 103   chips 2022-01-12 Tue
#> 19 103   chips 2022-01-13 Tue

^{由reprex package 创建于 2022-01-18 (v2.0.1)}

【讨论】：

有一个很好的通用解决方案！比我的好多了。
我要感谢您的评论让我走上了这条道路。 “可能有一种更优雅的方式”。 “有？嗯……”

【解决方案2】：

如果您需要概括这一点，可能有一种更优雅的方式，但这很快并且可以完成工作：

bind_rows(
    df,
    df %>% filter(Day == "Tue") %>% mutate(MaxDate = MaxDate - 1),
    df %>% filter(Day == "Tue") %>% mutate(MaxDate = MaxDate - 2)
  ) %>%
  arrange(ID, MaxDate, Product)
#     ID Product    MaxDate Day
# 1  100   candy 2022-01-16 Tue
# 2  100   chips 2022-01-16 Tue
# 3  100   candy 2022-01-17 Tue
# 4  100   chips 2022-01-17 Tue
# 5  100   candy 2022-01-18 Tue
# 6  100   chips 2022-01-18 Tue
# 7  101   candy 2022-01-16 Tue
# 8  101   chips 2022-01-16 Tue
# 9  101   candy 2022-01-17 Tue
# 10 101   chips 2022-01-17 Tue
# 11 101   candy 2022-01-18 Tue
# 12 101   chips 2022-01-18 Tue
# 13 102   candy 2022-01-16 Tue
# 14 102   candy 2022-01-17 Tue
# 15 102   candy 2022-01-18 Tue
# 16 103   candy 2022-01-11 Tue
# 17 103   chips 2022-01-11 Tue
# 18 103   candy 2022-01-12 Tue
# 19 103   chips 2022-01-12 Tue
# 20 103   candy 2022-01-13 Tue
# 21 103   chips 2022-01-13 Tue

使用这个可重现的数据：

df = read.table(text = 'ID  Product MaxDate Day
100 candy   2022-01-18  Tue
100 chips   2022-01-18  Tue
101 candy   2022-01-18  Tue
101 chips   2022-01-18  Tue
102 candy   2022-01-18  Tue
103 candy   2022-01-13  Tue
103 chips   2022-01-13  Tue', header = T) %>%
  mutate(MaxDate = as.Date(MaxDate))

【讨论】：