问使用变量名称的长到宽格式
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Stack Overflow用户

提问于 2021-12-30 08:23:01

回答 3查看 180关注 0票数 0

我有一个很宽的数据集，如下所示：

dataset <- data.frame(id = c(1, 2, 3, 4, 5),
                      basketball.time1 = c(2, 5, 4, 3, 3),
                      basketball.time2 = c(3, 4, 5, 3, 2),
                      basketball.time3 = c(1, 8, 4, 3, 1),
                      volleyball.time1 = c(2, 3, 4, 0, 1),
                      volleyball.time2 = c(3, 4, 3, 1, 3),
                      volleyball.time3 = c(1, 8, 12, 2, 3))

我想要的是长格式的数据集，将id、time、basketball和volleyball作为单独的变量。我希望使用由“”分隔的字符串创建time列，其中包含三个因素(time1、time2和time3)。在篮球和排球专栏的末尾。

非常感谢！

编辑:固定排版

tidyr

melt

dplyr

reshape

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回答 3

Stack Overflow用户

回答已采纳

发布于 2021-12-30 08:31:59

我们可以使用pivor_longer %>% pivot_wider。如果我们将适当的参数设置为separate，则不需要pivor_longer。

library(tidyr)

dataset %>%
        pivot_longer(cols = matches('time\\d+$'), names_to = c('sport', 'time'), names_pattern = '(.*)\\.(.*)') %>%
        pivot_wider(names_from = sport, values_from = value)

# A tibble: 15 × 5
      id time  basketball volleyball vollyeball
   <dbl> <chr>      <dbl>      <dbl>      <dbl>
 1     1 time1          2          2         NA
 2     1 time2          3          3         NA
 3     1 time3          1         NA          1
 4     2 time1          5          3         NA
 5     2 time2          4          4         NA
 6     2 time3          8         NA          8
 7     3 time1          4          4         NA
 8     3 time2          5          3         NA
 9     3 time3          4         NA         12
10     4 time1          3          0         NA
11     4 time2          3          1         NA
12     4 time3          3         NA          2
13     5 time1          3          1         NA
14     5 time2          2          3         NA
15     5 time3          1         NA          3

票数 1

Stack Overflow用户

发布于 2021-12-30 08:30:45

一种可能的解决办法：

library(tidyverse)

dataset <- data.frame(id = c(1, 2, 3, 4, 5),
                      basketball.time1 = c(2, 5, 4, 3, 3),
                      basketball.time2 = c(3, 4, 5, 3, 2),
                      basketball.time3 = c(1, 8, 4, 3, 1),
                      volleyball.time1 = c(2, 3, 4, 0, 1),
                      volleyball.time2 = c(3, 4, 3, 1, 3),
                      vollyeball.time3 = c(1, 8, 12, 2, 3))

dataset %>% 
  pivot_longer(cols = -id) %>% 
  separate(name,into = c("name", "time")) %>% 
  pivot_wider(id_cols = c(id, name, time))

#> # A tibble: 15 × 5
#>       id time  basketball volleyball vollyeball
#>    <dbl> <chr>      <dbl>      <dbl>      <dbl>
#>  1     1 time1          2          2         NA
#>  2     1 time2          3          3         NA
#>  3     1 time3          1         NA          1
#>  4     2 time1          5          3         NA
#>  5     2 time2          4          4         NA
#>  6     2 time3          8         NA          8
#>  7     3 time1          4          4         NA
#>  8     3 time2          5          3         NA
#>  9     3 time3          4         NA         12
#> 10     4 time1          3          0         NA
#> 11     4 time2          3          1         NA
#> 12     4 time3          3         NA          2
#> 13     5 time1          3          1         NA
#> 14     5 time2          2          3         NA
#> 15     5 time3          1         NA          3

票数 1

Stack Overflow用户

发布于 2021-12-30 08:31:35

pivot_longer
separate in sport和time column
pivot_wider sport列

library(dplyr)
library(tidyr)

dataset %>% 
  pivot_longer(
    -id
  ) %>% 
  separate(name, c("sport", "time")) %>% 
  pivot_wider(
    names_from = sport
  )

      id time  basketball volleyball vollyeball
   <dbl> <chr>      <dbl>      <dbl>      <dbl>
 1     1 time1          2          2         NA
 2     1 time2          3          3         NA
 3     1 time3          1         NA          1
 4     2 time1          5          3         NA
 5     2 time2          4          4         NA
 6     2 time3          8         NA          8
 7     3 time1          4          4         NA
 8     3 time2          5          3         NA
 9     3 time3          4         NA         12
10     4 time1          3          0         NA
11     4 time2          3          1         NA
12     4 time3          3         NA          2
13     5 time1          3          1         NA
14     5 time2          2          3         NA
15     5 time3          1         NA          3