r - `pivot_longer` 操作 - 实现预期输出的更简单方法？

Question

我有一个 df 的形式:

df <- tibble(
  id = c(1,2,3),
  val02 = c(0,1,0),
  val03 = c(1,0,0),
  val04 = c(0,1,1),
  age02 = c(1,2,3),
  age03 = c(2,3,4),
  age04 = c(3,4,5)
)

我想把它整理成像这样的格式:

# A tibble: 9 x 4
     id year    val   age
  <dbl> <chr> <dbl> <dbl>
1     1 02        0     1
2     1 03        1     2
3     1 04        0     3
4     2 02        1     2
5     2 03        0     3
6     2 04        1     4
7     3 02        0     3
8     3 03        0     4
9     3 04        1     5

---

使用两个单独的 pivot_longer 操作和最后的 left_join 我实现了我想要的:

library(tidyverse)
df1 <- df %>%
  pivot_longer(cols = starts_with("val"), names_to = "year", values_to = "val", names_prefix = "val")
df2 <- df %>%
  pivot_longer(cols = starts_with("age"), names_to = "year", values_to = "age", names_prefix = "age")

left_join(df1, df2) %>%
  select(id, year, val, age)

然而，这似乎非常复杂。

如何简化此操作？有没有办法一次性执行此操作？ (在一根 pipe 里..？)

---

Answer 1

这取决于你的字符串(列名)的复杂性，但给出一个想法:

library(tidyverse)

df %>%
  pivot_longer(-id,
               names_to = c('.value', 'year'),
               names_pattern = '([a-z]+)(\\d+)'
  )

输出:

# A tibble: 9 x 4
     id year    val   age
  <dbl> <chr> <dbl> <dbl>
1     1 02        0     1
2     1 03        1     2
3     1 04        0     3
4     2 02        1     2
5     2 03        0     3
6     2 04        1     4
7     3 02        0     3
8     3 03        0     4
9     3 04        1     5