Question

我有一个遵循以下格式的数据框。

 match  team1 team2 winningTeam
 1      A     D     A
 2      B     E     E
 3      C     F     C
 4      D     C     C
 5      E     B     B
 6      F     A     A
 7      A     D     D
 8      D     A     A

我想要做的是创建变量，用于计算最后x次匹配的第1组和第2组的形式。例如，我想创建一个名为team1_form_last3_matches的变量，其中匹配8将为0.33（因为他们赢得了他们最后3场比赛中的1场），并且还会有一个名为team2_form_last3_matches的变量，在比赛8中为0.66（因为他们赢了）他们最后3场比赛中的2场）。理想情况下，我希望能够指定在计算团队 x _form_last y 变量时要考虑的先前匹配数以及要自动创建的变量。我尝试了一堆使用dplyr，动物园滚动平均函数和嵌套for / if语句的方法。但是，我还没有完全破解它，当然也不是优雅的方式。我觉得我错过了这个通用问题的简单解决方案。任何帮助将不胜感激！

干杯，

杰克

Answer 1

如下：

dat <- data.frame(match = c(1:8), team1 = c("A","B","C","D","E","F","A","D"), team2 = c("D","E","F","C","B","A","D","A"), winningTeam = c("A","E","C","C","B","A","D","A"))
  match team1 team2 winningTeam
1     1     A     D           A
2     2     B     E           E
3     3     C     F           C
4     4     D     C           C
5     5     E     B           B
6     6     F     A           A
7     7     A     D           D
8     8     D     A           A

Allteams <- c("A","B","C","D","E","F")

# A vectorized function for you to use to do as you ask:
teamX_form_lastY <- function(teams, games, dat){
  sapply(teams, function(x) {
    games_info <- rowSums(dat[,c("team1","team2")] == x) + (dat[,"winningTeam"] == x)
    lookup <- ifelse(rev(games_info[games_info != 0])==2,1,0)
    games_won <- sum(lookup[1:games])
    if(length(lookup) < games) warning(paste("maximum games for team",x,"should be",length(lookup)))
    games_won/games
  })
}

teamX_form_lastY("A", 4, dat)
A 
0.75 

# Has a warning for the number of games you should be using
teamX_form_lastY("A", 5, dat)
A 
NA 
Warning message:
  In FUN(X[[i]], ...) : maximum games for team A should be 4

# vectorized input
teamX_form_lastY(teams = c("A","B"), games = 2, dat = dat)
A   B 
0.5 0.5 

# so you ca do all teams
teamX_form_lastY(teams = Allteams, 2, dat)
A   B   C   D   E   F 
0.5 0.5 1.0 0.5 0.5 0.0

Answer 2

这适用于t1l3，您需要将其复制为t2。

dat <- data.frame(match = c(1:8), team1 = c("A","B","C","D","E","F","A","D"), team2 = c("D","E","F","C","B","A","D","A"), winningTeam = c("A","E","C","C","B","A","D","A"),stringsAsFactors = FALSE)

dat$t1l3 <- c(NA,sapply(2:nrow(dat),function(i) {
  df <- dat[1:(i-1),] #just previous games, i.e. excludes current game
  df <- df[df$team1==dat$team1[i] | df$team2==dat$team1[i],] #just those containing T1
  df <- tail(df,3) #just the last three (or fewer if there aren't three previous games)
  return(sum(df$winningTeam==dat$team1[i])/nrow(df)) #total wins/total games (up to three)
}))

R

2 个答案: