如何在R中使用pivot_longer()按类别将列分成多行？

Question

以下是一些虚构的数据：

tibble(fruit = rep(c("apple", "pear", "orange"), each = 3),
       size = rep(c("big", "medium", "small"), times = 3),
       # summer stock
       shopA_summer_wk1 = abs(round(rnorm(9, 10, 5), 0)),
       shopA_summer_wk2 = abs(round(rnorm(9, 10, 5), 0)),
       shopB_summer_wk1 = abs(round(rnorm(9, 10, 5), 0)),
       shopB_summer_wk2 = abs(round(rnorm(9, 10, 5), 0)),
       shopC_summer_wk1 = abs(round(rnorm(9, 10, 5), 0)),
       shopC_summer_wk2 = abs(round(rnorm(9, 10, 5), 0)),
       # winter stock
       shopA_winter_wk1 = abs(round(rnorm(9, 8, 4), 0)),
       shopA_winter_wk2 = abs(round(rnorm(9, 8, 4), 0)),
       shopA_winter_wk3 = abs(round(rnorm(9, 8, 4), 0)),
       shopB_winter_wk1 = abs(round(rnorm(9, 8, 4), 0)),
       shopB_winter_wk2 = abs(round(rnorm(9, 8, 4), 0)),
       shopB_winter_wk3 = abs(round(rnorm(9, 8, 4), 0)),
       shopC_winter_wk1 = abs(round(rnorm(9, 8, 4), 0)),
       shopC_winter_wk2 = abs(round(rnorm(9, 8, 4), 0)),
       shopC_winter_wk3 = abs(round(rnorm(9, 8, 4), 0)))

一些数据是在夏季 2 周和冬季 3 周内收集 3 个商店（A、B、C）的数据。收集的数据是商店在该特定周内每种尺寸（大、中、小）的水果（苹果、梨、橙子）库存数量。

以下是数据集的前 6 行：

# fruit  size   shopA_summer_wk1   shopA_summer_wk2 shopB_summer_wk1 shopB_summer_wk2 shopC_summer_wk1 shopC_summer_wk2 shopA_winter_wk1 shopA_winter_wk2 shopA_winter_wk3
#   <chr>  <chr>             <dbl>            <dbl>            <dbl>            <dbl>            <dbl>            <dbl>            <dbl>            <dbl>            <dbl>
# 1 apple  big                   9               12               12               16               15                5               14                4                0
# 2 apple  medium               21               16               16                1               12               11                8                8                9
# 3 apple  small                10                6               18               18               22               12                4                2                0
# 4 pear   big                  13                7                4               12               13                6               10                6                2
# 5 pear   medium               13               12                8                0                8                5               11                7                3
# 6 pear   small                16               18                4                3               13                8                7                5                0

我想使用pivot_longer()R 中的函数来重构这个数据集。鉴于有相当多的组类别，我很难为此编写代码。

我希望它看起来像下面这样：

我将非常感谢任何意见:)

Answer 1

使用该names_pattern参数，我们可以这样做：

pivot_longer(df, c(-fruit, -size), names_pattern = '(^.*)_wk(.*$)',
              names_to = c('Shop_season', 'week'))
#> # A tibble: 135 x 5
#>    fruit size  Shop_season  week  value
#>    <chr> <chr> <chr>        <chr> <dbl>
#>  1 apple big   shopA_summer 1        11
#>  2 apple big   shopA_summer 2         8
#>  3 apple big   shopB_summer 1         4
#>  4 apple big   shopB_summer 2        24
#>  5 apple big   shopC_summer 1         9
#>  6 apple big   shopC_summer 2        10
#>  7 apple big   shopA_winter 1         9
#>  8 apple big   shopA_winter 2        12
#>  9 apple big   shopA_winter 3         5
#> 10 apple big   shopB_winter 1         5
#> # ... with 125 more rows

您可能还想separate购物和季节，因为这实际上是两个不同的变量：

pivot_longer(df, c(-fruit, -size), names_pattern = '(^.*)_wk(.*$)',
              names_to = c('Shop_season', 'week')) %>%
   separate(Shop_season, into = c('shop', 'season'))
#> # A tibble: 135 x 6
#>    fruit size  shop  season week  value
#>    <chr> <chr> <chr> <chr>  <chr> <dbl>
#>  1 apple big   shopA summer 1        11
#>  2 apple big   shopA summer 2         8
#>  3 apple big   shopB summer 1         4
#>  4 apple big   shopB summer 2        24
#>  5 apple big   shopC summer 1         9
#>  6 apple big   shopC summer 2        10
#>  7 apple big   shopA winter 1         9
#>  8 apple big   shopA winter 2        12
#>  9 apple big   shopA winter 3         5
#> 10 apple big   shopB winter 1         5
#> #... with 125 more rows