3

我有许多表的数据，例如：

event_id   player   finish
1          a        1
1          b        2
1          c        3
1          d        4
2          b        1
2          e        2
2          f        3
2          a        3
2          g        5

许多 event_id 的，每个从 5 到 20 名玩家，完成可能并列。

为了在R中使用 PlayerRatings 包，我想将表格重新格式化为：

event_id   player1   player2 result
1          a         b       1
1          a         c       1
1          a         d       1
1          b         c       1
1          b         d       1
1          c         d       1
2          b         e       1
2          b         f       1
2          b         a       1
2          b         g       1
2          e         f       1
2          e         a       1
2          e         g       1
2          f         a       0.5
2          f         g       1
2          a         g       1

4 名玩家的 event_id 将在新表中有 4*3/2 = 6 条记录，5 名玩家将有 5*4/2 = 10 条记录，依此类推。如果玩家“a”的“完成”小于玩家“b”，则“结果”为 1。如果“完成”相等，则“结果”为 0.5。如果玩家“a”的完成次数大于玩家“b”，则“结果”将为 0。

任何帮助表示赞赏！

4

2 回答 2

3

Here a data.table solution. I am using it for the grouping and syntax features. The code is a little bit complicated so I give here the idea.

  1. group per event_id
  2. for each event, create a combinations of player , suing combn
  3. for each combinations of player computer the finish score using a nested ifelse

Here the whole code:

library(data.table)
DT <- as.data.table(dat)
DT[,{ids <- do.call(rbind,combn(seq_along(player),2,simplify=FALSE))
      z <- mapply(function(x,y){
            z <- ifelse(finish[x]>finish[y],0,
                   ifelse(finish[x]<finish[y],1,0.5))
            data.frame(player[x],player[y],z)
            },
            ids[,1],
            ids[,2])
      data.frame(t(z))

     },event_id]

    event_id player.x. player.y.   z
 1:        1         a         b   1
 2:        1         a         c   1
 3:        1         a         d   1
 4:        1         b         c   1
 5:        1         b         d   1
 6:        1         c         d   1
 7:        2         b         e   1
 8:        2         b         f   1
 9:        2         b         a   1
10:        2         b         g   1
11:        2         e         f   1
12:        2         e         a   1
13:        2         e         g   1
14:        2         f         a 0.5
15:        2         f         g   1
16:        2         a         g   1
于 2013-06-28T05:13:32.613 回答
3

这是一个合并解决方案：第二行就是它的全部内容。

a<-data.frame(event_id=c(1,1,1,1,1,2,2,2,2,2,2),player=letters[c(1:5,3:8)],finish=c(1,1,3:5,1:6))
b<-merge(a,a,by.x="event_id",by.y="event_id",suffixes = c(".x",".y"))
b$score<-b$finish.x<b$finish.y
b$score[b$finish.x==b$finish.y]<-0.5
c<-b[b$player.x!=b$player.y & as.character(b$player.x)<as.character(b$player.y),c("event_id","player.x","player.y","score")]
于 2013-06-28T05:34:12.063 回答