Question

我正在尝试计算主队和球队得分的平均进球数，但是今天＆＃34;游戏。

可在此处找到数据：http://www.football-data.co.uk/mmz4281/1415/E0.csv

我的代码

pl <- pl[,2:6]
pl$Date <- as.Date(pl$Date, "%d/%m/%y")

pl$HomeTeam <- as.character(pl$HomeTeam)
pl$AwayTeam <- as.character(pl$AwayTeam)

pl.func <- function(tf){
  tf$avg.ht <- rep(NA,nrow(tf))
  tf$avg.at <- rep(NA,nrow(tf))

  for(i in 1:nrow(tf)){
     tf$avg.ht[i] <- (sum(tf$FTHG[tf$HomeTeam == tf$HomeTeam[i] & tf$Date < tf$Date[i]]) + sum(tf$FTAG[tf$AwayTeam == tf$HomeTeam[i] & tf$Date <tf$Date[i]])) / sum(tf$HomeTeam == tf$HomeTeam[i] & tf$Date < tf$Date[i] | tf$AwayTeam == tf$HomeTeam[i] & tf$Date < tf$Date[i])
     tf$avg.at[i] <- (sum(tf$FTHG[tf$HomeTeam == tf$AwayTeam[i] & tf$Date < tf$Date[i]]) + sum(tf$FTAG[tf$AwayTeam == tf$AwayTeam[i] & tf$Date <tf$Date[i]])) / sum(tf$HomeTeam == tf$AwayTeam[i] & tf$Date < tf$Date[i] | tf$AwayTeam == tf$AwayTeam[i] & tf$Date < tf$Date[i])
  }
  return(tf)
}

pl <- pl.func(pl)

我需要＆＃34;匹配＆＃34;在团队和更早的日期。上面的代码可以工作，但是因为我想计算几百个计算速度很慢。任何人都可以通过某种应用功能暗示或展示我如何做到这一点？我无法成功，因为我不知道以正确的方式从循环中替换[i]参数。

Answer 1

您实际需要的是运行条件平均值。最近，我回答了一个类似question的问题，其中OP需要每15分钟按小组运行平均值，你需要为团队所玩的每一场比赛运行平均值。

因此，请考虑以下使用示例数据并同时运行代码的sapply()方法，返回等效输出。可能会根据您的需求提高性能：

pl$runavgHT <- sapply(1:nrow(pl),
                    function(i) {
                      (sum(((pl[1:i, c("Date")] < (pl$Date[i]))
                           & (pl[1:i, c("HomeTeam")] == pl$HomeTeam[i]))
                          *  pl[1:i,]$FTHG) +
                       sum(((pl[1:i, c("Date")] < (pl$Date[i]))
                           & (pl[1:i, c("AwayTeam")] == pl$HomeTeam[i]))
                          *  pl[1:i,]$FTAG)) /

                       sum(((pl[1:i, c("Date")] < (pl$Date[i])) & 
                            (pl[1:i, c("HomeTeam")] == pl$HomeTeam[i])) 
                          |((pl[1:i, c("Date")] < (pl$Date[i])) & 
                            (pl[1:i, c("AwayTeam")] == pl$HomeTeam[i])))
                    }
             )

pl$runavgAT <- sapply(1:nrow(pl),
                    function(i) {
                      (sum(((pl[1:i, c("Date")] < (pl$Date[i]))
                            & (pl[1:i, c("HomeTeam")] == pl$AwayTeam[i]))
                           *  pl[1:i,]$FTHG) +
                       sum(((pl[1:i, c("Date")] < (pl$Date[i]))
                            & (pl[1:i, c("AwayTeam")] == pl$AwayTeam[i]))
                           *  pl[1:i,]$FTAG)) /

                       sum(((pl[1:i, c("Date")] < (pl$Date[i])) & 
                            (pl[1:i, c("HomeTeam")] == pl$AwayTeam[i])) 
                          |((pl[1:i, c("Date")] < (pl$Date[i])) & 
                            (pl[1:i, c("AwayTeam")] == pl$AwayTeam[i])))
                    }
)

Answer 2

以下是一些可能的改进（以及最终基准）：

1）这是你的函数的修改版本，只有一些改进：

pl.func2 <- function(DF){
  DF$avg.ht <- rep(NA,nrow(DF))
  DF$avg.at <- rep(NA,nrow(DF))

  for(i in 1:nrow(DF)){
     currDate <- DF$Date[i]
     currHT <- DF$HomeTeam[i]
     currAT <- DF$AwayTeam[i]

     prevHT.eq.HT <- which(DF$HomeTeam == currHT & DF$Date < currDate)
     prevHT.eq.AT <- which(DF$HomeTeam == currAT & DF$Date < currDate)
     prevAT.eq.HT <- which(DF$AwayTeam == currHT & DF$Date < currDate)
     prevAT.eq.AT <- which(DF$AwayTeam == currAT & DF$Date < currDate)

     DF$avg.ht[i] <- (sum(DF$FTHG[prevHT.eq.HT]) + sum(tf$FTAG[prevAT.eq.HT])) / (length(prevHT.eq.HT) + length(prevAT.eq.HT))
     DF$avg.at[i] <- (sum(DF$FTHG[prevHT.eq.AT]) + sum(tf$FTAG[prevAT.eq.AT])) / (length(prevHT.eq.AT) + length(prevAT.eq.AT))

  }
  return(DF)
}

2）这是你的函数的另一个修改版本，它使用累积信息来避免前几天的子集化和求和（N.B.这要求data.frame按日期排序）：

pl.func3 <- function(DF){
  DF$avg.ht <- rep(NA,nrow(DF))
  DF$avg.at <- rep(NA,nrow(DF))

  teams <- unique(c(DF$HomeTeam,DF$AwayTeam))
  cumul.info <- t(sapply(teams,FUN=function(team) c(cumulFTG=0,cumulMatches=0)))

  # store column indexes to reuse them
  cumulFTG <- 1
  cumulMatches <- 2

  for(i in 1:nrow(DF)){
     currHT <- DF$HomeTeam[i]
     currAT <- DF$AwayTeam[i]

     DF$avg.ht[i] <- cumul.info[currHT,cumulFTG] / cumul.info[currHT,cumulMatches]
     DF$avg.at[i] <- cumul.info[currAT,cumulFTG] / cumul.info[currAT,cumulMatches]

     cumul.info[currHT,cumulFTG] = cumul.info[currHT,cumulFTG] + DF$FTHG[i]
     cumul.info[currHT,cumulMatches] = cumul.info[currHT,cumulMatches] + 1

     cumul.info[currAT,cumulFTG] = cumul.info[currAT,cumulFTG] + DF$FTAG[i]
     cumul.info[currAT,cumulMatches] = cumul.info[currAT,cumulMatches] + 1

  }
  return(DF)
}

检查和基准：

# this is necessary for pl.func3
pl <- pl[order(pl$Date),] 

# are the results identical ? -> TRUE
identical(pl.func(pl),pl.func2(pl)) && identical(pl.func(pl),pl.func3(pl))

# benchmark
library(microbenchmark)
microbenchmark(pl.func(pl),pl.func2(pl),pl.func3(pl))

Unit: milliseconds
         expr       min        lq      mean    median        uq      max neval cld
  pl.func(pl) 184.36644 186.10643 188.38130 187.16322 188.80065 255.2101   100   c
 pl.func2(pl)  84.95047  85.80966  89.27945  87.41589  88.33845 159.6284   100  b 
 pl.func3(pl)  30.72683  31.05515  32.02944  31.41211  33.22858  35.8644   100 a

用R中的apply函数替换循环

2 个答案: