Question

我有以下数据框：

df <- data.frame(stringsAsFactors=FALSE,
              ID = c(2016070707, 2016070707, 2016070707, 2017080808, 2017080808,
                     2017080808, 2017080808),
       team_name = c("Harlequins", "Harlequins", "Harlequins", "Bristol Rugby",
                     "Bristol Rugby", "Bristol Rugby", "Bristol Rugby"),
     player_name = c("Karl Dickson", "Tim Swiel", "Alofa Alofa",
                     "Jason Woodward", "Jack Wallace", "Will Hurrell",
                     "Tusi Pisi"),
             DGC = c(7, 5, 0, 3, 0, 8, 9),
              MR = c(3, 8, 31, 25, 0, 85, 0),
              CB = c(2, 9, 1, 13, 0, 0, 0)
  )

我想转换为宽版本：

new_df <- data.frame(stringsAsFactors=FALSE,
                ID = c(2016070707, 2017080808),
         team_name = c("Harlequins", "Bristol Rugby"),
          player_1 = c("Karl Dickson", "Jason Woodward"),
          player_2 = c("Tim Swiel", "Jack Wallace"),
          player_3 = c("Alofa Alofa", "Will Hurrell"),
          player_4 = c(NA, "Tusi Pisi"),
             DGC_1 = c(7, 3),
             DGC_2 = c(5, 0),
             DGC_3 = c(0, 8),
             DGC_4 = c(NA, 9),
              MR_1 = c(3, 25),
              MR_2 = c(8, 0),
              MR_3 = c(31, 85),
              MR_4 = c(NA, 0),
              CB_1 = c(2, 13),
              CB_2 = c(9, 0),
              CB_3 = c(1, 0),
              CB_4 = c(NA, 0)
      )

但是，您可能会看到一线队有3名球员，而第二队有4名球员。我如何将 df 转换为 new_df 并还创建这些额外的列，例如 player_4 ， DGC_4 团队中的球员人数。差异将由“ NA”补偿（就像在new_df中一样）。

Answer 1

这是使用软件包data.table

的一种解决方案

df <- data.frame(stringsAsFactors=FALSE,
                 ID = c(2016070707, 2016070707, 2016070707, 2017080808, 2017080808,
                        2017080808, 2017080808),
                 team_name = c("Harlequins", "Harlequins", "Harlequins", "Bristol Rugby",
                               "Bristol Rugby", "Bristol Rugby", "Bristol Rugby"),
                 player_name = c("Karl Dickson", "Tim Swiel", "Alofa Alofa",
                                 "Jason Woodward", "Jack Wallace", "Will Hurrell",
                                 "Tusi Pisi"),
                 DGC = c(7, 5, 0, 3, 0, 8, 9),
                 MR = c(3, 8, 31, 25, 0, 85, 0),
                 CB = c(2, 9, 1, 13, 0, 0, 0)
)

library(data.table)
#convert to data.table
setDT(df)
#create a sequence from 1 to N for each ID
df[, idx := 1:.N, by = ID]
#cast wide
dcast(ID + team_name ~ idx, data = df, value.var = c("player_name", "DGC", "MR", "CB"), )
#>            ID     team_name  player_name_1 player_name_2 player_name_3
#> 1: 2016070707    Harlequins   Karl Dickson     Tim Swiel   Alofa Alofa
#> 2: 2017080808 Bristol Rugby Jason Woodward  Jack Wallace  Will Hurrell
#>    player_name_4 DGC_1 DGC_2 DGC_3 DGC_4 MR_1 MR_2 MR_3 MR_4 CB_1 CB_2
#> 1:          <NA>     7     5     0    NA    3    8   31   NA    2    9
#> 2:     Tusi Pisi     3     0     8     9   25    0   85    0   13    0
#>    CB_3 CB_4
#> 1:    1   NA
#> 2:    0    0

^{由reprex package（v0.2.1）于2019-01-27创建}

Answer 2

还有一种tidyverse可能性：

df %>%
 rowid_to_column() %>%
 gather(var, val, -c(ID, team_name, rowid)) %>%
 arrange(rowid, ID, team_name, var) %>%
 group_by(ID, team_name) %>%
 mutate(var = paste(var, gl(n_distinct(rowid), n_distinct(var)), sep = "_")) %>%
 select(-rowid) %>%
 spread(var, val, convert = TRUE) 

           ID team_name      CB_1  CB_2  CB_3  CB_4 DGC_1 DGC_2 DGC_3 DGC_4  MR_1  MR_2
        <dbl> <chr>         <int> <int> <int> <int> <int> <int> <int> <int> <int> <int>
1 2016070707. Harlequins        2     9     1    NA     7     5     0    NA     3     8
2 2017080808. Bristol Rugby    13     0     0     0     3     0     8     9    25     0
   MR_3  MR_4 player_name_1  player_name_2 player_name_3 player_name_4
  <int> <int> <chr>          <chr>         <chr>         <chr>        
1    31    NA Karl Dickson   Tim Swiel     Alofa Alofa   <NA>         
2    85     0 Jason Woodward Jack Wallace  Will Hurrell  Tusi Pisi

首先，它将数据从宽格式转换为长格式。然后，整理数据，最后创建新的变量名并将其转换为所需的宽格式。

从长到宽，每个组/ ID的行数不同

2 个答案: