將數(shù)據(jù)幀在R軸上旋轉兩次，然后返回到原始形狀

比方說，我想知道四名籃球運動員中哪一名是最好的，我設立了一個小比賽，兩名球員1對1比賽，我記錄了一組數(shù)據(jù)

#rm(list=ls())

set.seed(1234)

# some made up scores from my tournament
df <- data.frame(
  player1 = c("a", "a", "b", "c", "d", "d"),
  player2 = c("b", "c", "d", "b", "a", "c"),
  date = c("2021-01-01", "2021-01-02", "2021-01-04", "2021-01-05", "2021-01-06", "2021-01-08"),
  p1_dunks = sample(c(4:11), 6, replace = TRUE),
  p2_dunks = sample(c(3:12), 6, replace = TRUE),
  p1_blocks = sample(c(8:10), 6, replace = TRUE),
  p2_blocks = sample(c(10:12), 6, replace = TRUE),
  p1_threepointers = sample(c(2:7), 6, replace = TRUE),
  p2_threepointers = sample(c(1:5), 6, replace = TRUE)
)

為了計算一個球員在比賽的任何一點上表現(xiàn)得有多好，我可以將其旋轉兩次，并用每個統(tǒng)計的累積和替換每個統(tǒng)計的計數(shù)

# cast to long and get cumulative stats per player
melted_df <- df %>%
  pivot_longer(cols = starts_with(c("p1", "p2")), names_to = "stat", values_to = "number") %>%
  pivot_longer(cols = starts_with("player"), names_to = "player", values_to = "name") %>%
  filter(
    (player == "player1" & grepl("^p1", stat)) |
      (player == "player2" & grepl("^p2", stat))
  ) %>%
  arrange(date) %>%
  group_by(player, stat) %>%
  mutate(number = cumsum(number))

那么我就可以很容易地對此提出質疑了

melted_df %>%
    filter(date < "2021-01-05") %>%
    filter(!duplicated(name)) %>%
    filter(grepl("dunks$", stat))

但是假設在我的用例中，我需要將這個長格式的數(shù)據(jù)強制恢復到其原始形式（使用player1、player2，然后是每個player1和player2的統(tǒng)計數(shù)據(jù)）。我可以試試

# try to cast back to original format
back_to_wider_df <- melted_df %>%
  pivot_wider(names_from = "player", values_from = "name") %>%
  pivot_wider(names_from = "stat", values_from = "number")

但這反而會給出一個數(shù)據(jù)幀，該數(shù)據(jù)幀是每匹配一行的“偏移量”，其中有一半的NA值：

> head(back_to_wider_df)
# A tibble: 6 × 9
  date       player1 player2 p1_dunks p1_blocks p1_threepointers p2_dunks p2_blocks p2_threepointers
  <chr>      <chr>   <chr>      <int>     <int>            <int>    <int>     <int>            <int>
1 2021-01-01 a       NA             7         9                6       NA        NA               NA
2 2021-01-01 NA      b             NA        NA               NA       11        11                4
3 2021-01-02 a       NA            18        18                9       NA        NA               NA
4 2021-01-02 NA      c             NA        NA               NA       18        22                8
5 2021-01-04 b       NA            23        27               15       NA        NA               NA
6 2021-01-04 NA      d             NA        NA               NA       26        32               11

是否有一種簡單的方法將其修復回原始形狀，以便前三行應為：

> df
        date player1 player2 p1_dunks p1_blocks p1_threepointers p2_dunks p2_blocks p2_three_pointers
1 2021-01-01       a       b        7         9                6       11        11                 4
2 2021-01-02       a       c       18        18                9       18        22                 8
3 2021-01-04       b       d       23        27               15       26        32                11

thanks,