R/tidyr：转向更宽的格式并返回更长的格式以完成年份数据
R/tidyr: Pivot to wider format and back to longer format to complete year data
我生成的 table 看起来像这样：
my_data <- tibble(Year = c(rep(2020, 4), rep(2021, 12)),
Month = c(lubridate::month(1:4, label = TRUE), lubridate::month(1:12, label = TRUE)),
foo = 16:1, bar = 1:16)
Year Month foo bar
<dbl> <ord> <int> <int>
1 2020 Jan 16 1
2 2020 Feb 15 2
3 2020 Mär 14 3
4 2020 Apr 13 4
5 2021 Jan 12 5
6 2021 Feb 11 6
7 2021 Mär 10 7
8 2021 Apr 9 8
9 2021 Mai 8 9
10 2021 Jun 7 10
11 2021 Jul 6 11
12 2021 Aug 5 12
13 2021 Sep 4 13
14 2021 Okt 3 14
15 2021 Nov 2 15
16 2021 Dez 1 16
我现在想使用 pivot_wider 将我的 table 转换为宽格式，然后转回较长格式，以便在我的 table 中包含缺失的月份。 foo 和 bar 中的值应为 0.
tidyr::pivot_wider(my_data, names_from = Year, values_from = foo:bar, values_fill = 0)
Month foo_2020 foo_2021 bar_2020 bar_2021
<ord> <int> <int> <int> <int>
1 Jan 16 12 1 5
2 Feb 15 11 2 6
3 Mär 14 10 3 7
4 Apr 13 9 4 8
5 Mai 0 8 0 9
6 Jun 0 7 0 10
7 Jul 0 6 0 11
8 Aug 0 5 0 12
9 Sep 0 4 0 13
10 Okt 0 3 0 14
11 Nov 0 2 0 15
12 Dez 0 1 0 16
预期输出：
Year Month foo bar
<dbl> <ord> <dbl> <dbl>
1 2020 Jan 16 1
2 2020 Feb 15 2
3 2020 Mär 14 3
4 2020 Apr 13 4
5 2020 Mai 0 0
6 2020 Jun 0 0
7 2020 Jul 0 0
8 2020 Aug 0 0
9 2020 Sep 0 0
10 2020 Okt 0 0
11 2020 Nov 0 0
12 2020 Dez 0 0
13 2021 Jan 12 5
14 2021 Feb 11 6
15 2021 Mär 10 7
16 2021 Apr 9 8
17 2021 Mai 8 9
18 2021 Jun 7 10
19 2021 Jul 6 11
20 2021 Aug 5 12
21 2021 Sep 4 13
22 2021 Okt 3 14
23 2021 Nov 2 15
24 2021 Dez 1 16
使用 pivot_longer() 函数中的“names_sep”选项
df <- structure(list(Month = c("Jan", "Feb", "Mär", "Apr", "Mai", "Jun", "Jul", "Aug", "Sep", "Okt", "Nov", "Dez"),
foo_2020 = c(16L, 15L, 14L, 13L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L),
foo_2021 = 12:1,
bar_2020 = c(1L, 2L, 3L, 4L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L),
bar_2021 = 5:16), class = "data.frame", row.names = c(NA, -12L))
library(tidyr)
pivot_longer(df, -Month, names_sep = "_", names_to=c(".value", "Year"))
您不需要旋转两次。这就是 tidyr::complete 的用途：
complete(my_data, expand(my_data, Year, Month), fill = list(foo = 0, bar = 0))
#> Year Month foo bar
#> 1 2020 Jan 16 1
#> 2 2020 Feb 15 2
#> 3 2020 Mar 14 3
#> 4 2020 Apr 13 4
#> 5 2020 May 0 0
#> 6 2020 Jun 0 0
#> 7 2020 Jul 0 0
#> 8 2020 Aug 0 0
#> 9 2020 Sep 0 0
#> 10 2020 Oct 0 0
#> 11 2020 Nov 0 0
#> 12 2020 Dec 0 0
#> 13 2021 Jan 12 5
#> 14 2021 Feb 11 6
#> 15 2021 Mar 10 7
#> 16 2021 Apr 9 8
#> 17 2021 May 8 9
#> 18 2021 Jun 7 10
#> 19 2021 Jul 6 11
#> 20 2021 Aug 5 12
#> 21 2021 Sep 4 13
#> 22 2021 Oct 3 14
#> 23 2021 Nov 2 15
#> 24 2021 Dec 1 16
另一个可能的解决方案：
library(tidyverse)
months <- c("Jan", "Feb", "Mar", "Apr", "Mai", "Jun", "Jul", "Aug", "Sep", "Okt", "Nov", "Dez")
my_data %>%
group_by(Year) %>%
group_modify(~ add_row(.x, Month = setdiff(months, .x$Month))) %>%
ungroup %>%
mutate(across(foo:bar, replace_na, 0))
#> # A tibble: 27 × 4
#> Year Month foo bar
#> <dbl> <chr> <int> <int>
#> 1 2020 Jan 16 1
#> 2 2020 Feb 15 2
#> 3 2020 Mar 14 3
#> 4 2020 Apr 13 4
#> 5 2020 Mai 0 0
#> 6 2020 Jun 0 0
#> 7 2020 Jul 0 0
#> 8 2020 Aug 0 0
#> 9 2020 Sep 0 0
#> 10 2020 Okt 0 0
#> # … with 17 more rows
我生成的 table 看起来像这样：
my_data <- tibble(Year = c(rep(2020, 4), rep(2021, 12)),
Month = c(lubridate::month(1:4, label = TRUE), lubridate::month(1:12, label = TRUE)),
foo = 16:1, bar = 1:16)
Year Month foo bar
<dbl> <ord> <int> <int>
1 2020 Jan 16 1
2 2020 Feb 15 2
3 2020 Mär 14 3
4 2020 Apr 13 4
5 2021 Jan 12 5
6 2021 Feb 11 6
7 2021 Mär 10 7
8 2021 Apr 9 8
9 2021 Mai 8 9
10 2021 Jun 7 10
11 2021 Jul 6 11
12 2021 Aug 5 12
13 2021 Sep 4 13
14 2021 Okt 3 14
15 2021 Nov 2 15
16 2021 Dez 1 16
我现在想使用 pivot_wider 将我的 table 转换为宽格式，然后转回较长格式，以便在我的 table 中包含缺失的月份。 foo 和 bar 中的值应为 0.
tidyr::pivot_wider(my_data, names_from = Year, values_from = foo:bar, values_fill = 0)
Month foo_2020 foo_2021 bar_2020 bar_2021
<ord> <int> <int> <int> <int>
1 Jan 16 12 1 5
2 Feb 15 11 2 6
3 Mär 14 10 3 7
4 Apr 13 9 4 8
5 Mai 0 8 0 9
6 Jun 0 7 0 10
7 Jul 0 6 0 11
8 Aug 0 5 0 12
9 Sep 0 4 0 13
10 Okt 0 3 0 14
11 Nov 0 2 0 15
12 Dez 0 1 0 16
预期输出：
Year Month foo bar
<dbl> <ord> <dbl> <dbl>
1 2020 Jan 16 1
2 2020 Feb 15 2
3 2020 Mär 14 3
4 2020 Apr 13 4
5 2020 Mai 0 0
6 2020 Jun 0 0
7 2020 Jul 0 0
8 2020 Aug 0 0
9 2020 Sep 0 0
10 2020 Okt 0 0
11 2020 Nov 0 0
12 2020 Dez 0 0
13 2021 Jan 12 5
14 2021 Feb 11 6
15 2021 Mär 10 7
16 2021 Apr 9 8
17 2021 Mai 8 9
18 2021 Jun 7 10
19 2021 Jul 6 11
20 2021 Aug 5 12
21 2021 Sep 4 13
22 2021 Okt 3 14
23 2021 Nov 2 15
24 2021 Dez 1 16
使用 pivot_longer() 函数中的“names_sep”选项
df <- structure(list(Month = c("Jan", "Feb", "Mär", "Apr", "Mai", "Jun", "Jul", "Aug", "Sep", "Okt", "Nov", "Dez"),
foo_2020 = c(16L, 15L, 14L, 13L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L),
foo_2021 = 12:1,
bar_2020 = c(1L, 2L, 3L, 4L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L),
bar_2021 = 5:16), class = "data.frame", row.names = c(NA, -12L))
library(tidyr)
pivot_longer(df, -Month, names_sep = "_", names_to=c(".value", "Year"))
您不需要旋转两次。这就是 tidyr::complete 的用途：
complete(my_data, expand(my_data, Year, Month), fill = list(foo = 0, bar = 0))
#> Year Month foo bar
#> 1 2020 Jan 16 1
#> 2 2020 Feb 15 2
#> 3 2020 Mar 14 3
#> 4 2020 Apr 13 4
#> 5 2020 May 0 0
#> 6 2020 Jun 0 0
#> 7 2020 Jul 0 0
#> 8 2020 Aug 0 0
#> 9 2020 Sep 0 0
#> 10 2020 Oct 0 0
#> 11 2020 Nov 0 0
#> 12 2020 Dec 0 0
#> 13 2021 Jan 12 5
#> 14 2021 Feb 11 6
#> 15 2021 Mar 10 7
#> 16 2021 Apr 9 8
#> 17 2021 May 8 9
#> 18 2021 Jun 7 10
#> 19 2021 Jul 6 11
#> 20 2021 Aug 5 12
#> 21 2021 Sep 4 13
#> 22 2021 Oct 3 14
#> 23 2021 Nov 2 15
#> 24 2021 Dec 1 16
另一个可能的解决方案：
library(tidyverse)
months <- c("Jan", "Feb", "Mar", "Apr", "Mai", "Jun", "Jul", "Aug", "Sep", "Okt", "Nov", "Dez")
my_data %>%
group_by(Year) %>%
group_modify(~ add_row(.x, Month = setdiff(months, .x$Month))) %>%
ungroup %>%
mutate(across(foo:bar, replace_na, 0))
#> # A tibble: 27 × 4
#> Year Month foo bar
#> <dbl> <chr> <int> <int>
#> 1 2020 Jan 16 1
#> 2 2020 Feb 15 2
#> 3 2020 Mar 14 3
#> 4 2020 Apr 13 4
#> 5 2020 Mai 0 0
#> 6 2020 Jun 0 0
#> 7 2020 Jul 0 0
#> 8 2020 Aug 0 0
#> 9 2020 Sep 0 0
#> 10 2020 Okt 0 0
#> # … with 17 more rows