`pivot_longer`操作-实现预期产出的更简单方法?
我有一个`pivot_longer`操作-实现预期产出的更简单方法?,r,dplyr,tidyverse,tidyr,R,Dplyr,Tidyverse,Tidyr,我有一个df格式: 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) ) 然而,这似乎非常复杂 如何简化此操作?是否有一种一次性执行此操作的方法?(在一根管子里?) 这取决于字符串(列名)的复杂程度,但请给出一个想法: library(tid
df
格式:
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)
)
然而,这似乎非常复杂
如何简化此操作?是否有一种一次性执行此操作的方法?(在一根管子里?)
这取决于字符串(列名)的复杂程度,但请给出一个想法:
library(tidyverse)
df %>%
pivot_longer(-id,
names_to = c('.value', 'year'),
names_pattern = '([a-z]+)(\\d+)'
)
输出:
#一个tible:9 x 4
id年份val年龄
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
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)
library(tidyverse)
df %>%
pivot_longer(-id,
names_to = c('.value', 'year'),
names_pattern = '([a-z]+)(\\d+)'
)
# 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