在 R 中，根据行中的值将行转换为列

Question

这是我正在使用的table。我想将 reservation 转换成单独的列。

我想让它变成这样的。我一直在尝试使用 reshape2 和 dplyr 的 separate 来做到这一点，但我没有找到解决方案。

Answer 1

你可以试试

library(tidyverse)

df %>%
  rowwise %>%
  mutate(reservation_main = str_split(reservation,'_' ,simplify = T)[1],
         reservation_no = paste0('_',str_split(reservation,'_' ,simplify = T)[2])) %>%
  select(id, response_id, reservation_main, reservation_no) %>%
  pivot_wider(names_from = reservation_no, values_from = response_id)


     id reservation_main  `_1`  `_2`  `_3`
  <dbl> <chr>            <dbl> <dbl> <dbl>
1 31100 A                    1     1     0
2 31100 B                    1     1     0
3 31100 C                    1     0     0

Answer 2

由于reservation格式一致，我们可以用_分隔成两列。然后，我们可以将 response 转换为 0 和 1。然后，我删除 response_id。最后，我转向宽幅面。我假设您不希望在列中的数字之前使用 _。

library(tidyverse)

df %>%
  separate(reservation, c("reservation", "number"), sep = "_") %>%
  mutate(response = ifelse(response == "yes", 1, 0)) %>%
  select(-response_id) %>%
  pivot_wider(names_from = "number", values_from = "response")

输出

# A tibble: 2 × 5
     id reservation   `1`   `2`   `3`
  <dbl> <chr>       <dbl> <dbl> <dbl>
1 31100 A               1     1     0
2 31100 B               1     1     0

如果您确实希望将 _ 保留在列的数字前面，那么我们可以调整 separate.

中的正则表达式

df %>%
  separate(reservation, c("reservation", "number"), sep = "(?=\_)") %>%
  mutate(response = ifelse(response == "yes", 1, 0)) %>%
  select(-response_id) %>%
  pivot_wider(names_from = "number", values_from = "response")

# A tibble: 2 × 5
     id reservation  `_1`  `_2`  `_3`
  <dbl> <chr>       <dbl> <dbl> <dbl>
1 31100 A               1     1     0
2 31100 B               1     1     0

数据

df <- structure(
  list(
    id = c(31100, 31100, 31100, 31100, 31100, 31100),
    reservation = c("A_1", "A_2", "A_3", "B_1", "B_2", "B_3"),
    response = c("yes", "yes", "no", "yes", "yes", "no"),
    response_id = c(1,
                    1, 0, 1, 1, 0)
  ),
  class = "data.frame",
  row.names = c(NA,-6L)
)

在 R 中，根据行中的值将行转换为列

Converting rows into columns based on the values in the rows, in R

r

dplyr

tidyr

reshape2