ELO 评级算法中的 for 循环错误

Question

我正在尝试从这里运行 ELO 评级代码

https://github.com/xsankar/hairy-octo-hipster/blob/master/ELO-538.R

我使用的数据集在这里

https://www.mediafire.com/file/0a5v393ki775vw9/sport2.xlsx/file

我正在运行的代码略有修改，我只使用了 17 周而不是原始代码中的 22 周，但我无法让它在 22 周内工作

##########################################################

library(dplyr)

data<- read_xlsx("C:/User/sport2.xlsx")
 
data$PtsW <- as.numeric(data$PtsW)
data$PtsL <- as.numeric(data$PtsL)
data$YdsW <- as.numeric(data$YdsW)
data$YdsL <- as.numeric(data$YdsL)
data$TOW <- as.numeric(data$TOW)
data$TOL <- as.numeric(data$TOL)

rankings <- data_frame()
teams <- data %>% distinct(Winner.tie) %>% select(Winner.tie)
teams
rankings <- bind_rows(rankings,teams)
#print(data)
#class(rankings)
#print(rankings)
rankings

for (i in 2:19) {
  rankings[,i] <- 0
}
colnames(rankings) <- c("Team","Week.0","Week.1","Week.2","Week.3","Week.4","Week.5","Week.6",
                        "Week.7","Week.8","Week.9","Week.10","Week.11","Week.12","Week.13",
                        "Week.14","Week.15","Week.16","Week.17")
rankings$Week.0 <- 1500

# Iterate for each week of play
week.no <- 1
k_factor <- 20.0
week.data <- data[data$Week == week.no,]


for (j in 1:17) {
  week.no <- j
  k_factor <- 20.0
  week.data <- data[data$Week == week.no,]
  for (i in 1:nrow(week.data)) {
    winner <- week.data[i,"Winner.tie"]
 
    loser <- week.data[i,"Loser.tie"]
    old.rank.w <- rankings[rankings$Team == winner,week.no+1]
    old.rank.w <- old.rank.w[[1]]
    old.rank.l <- rankings[rankings$Team == loser,week.no+1]
    old.rank.l <- old.rank.l[[1]]
    
    # Calculate Margin of Victory Multiplier 
    # mv_mult = LN(ABS(PD)+1) * (2.2/((ELOW-ELOL)*.001+2.2))
    pd <- week.data$PtsW[i] - week.data$PtsL[i]
    mv_mult <- 1 #Margin For Victory Multiplier
    mv_mult <- log(pd +1) * (2.2/((old.rank.w - old.rank.l)*.001+2.2))
    #
    
    # Use old ELO Algorithm
    #
    w_w <- 1.0
    w_l <- 0.0
    if (pd == 0) {
      w_w <- 0.5
      w_l <- 0.5
    }
    #
    
    #
    d_ij_w <- old.rank.w - old.rank.l
    d_ij_l <- old.rank.l - old.rank.w
    #
    mu_ij_w <- 1 / (1 + 10 ^ ((-1 * d_ij_w)/400))
    new.rank.w <- round( old.rank.w + (k_factor * mv_mult * (w_w - mu_ij_w)))
    #
    mu_ij_l <- 1 / (1 + 10 ^ ((-1 * d_ij_l)/400))
    new.rank.l <- round( old.rank.l + (k_factor * mv_mult * (w_l - mu_ij_l)))
    #
    print (sprintf("Rank : W = %d L = %d",new.rank.w,new.rank.l))
    rankings[rankings$Team == winner,week.no+2] <- new.rank.w
    rankings[rankings$Team == loser,week.no+2] <- new.rank.l
  } 
  
###################################################################3  
  # if team didn't play, carry forward early ratings
  # not needed for wildcard, division et al
#  for (i in 1:nrow(rankings)) {
#    if (is.na(rankings[i,week.no+2])) {
#      rankings[i,week.no+2] <- rankings[i,week.no+1]
#    }
#    if (rankings[i,week.no+2] < 1) {
#      rankings[i,week.no+2] <- rankings[i,week.no+1]
#    }
#    }
##################################################################

}
# week #1 ranking
rankings %>% select(Team,Week.1) %>% arrange(-Week.1)
# week #17 ranking
rankings %>% select(Team,Week.17) %>% arrange(-Week.17)
#  

我收到了错误

Error in x[[jj]][iseq] <- vjj : replacement has length zero

我已经尝试修改 for 索引和调试，但找不到问题所在

Answer 1

我将附和 Ronak 的评论，即提供最少可重复的示例的好处，以便我们可以更好地帮助您（这意味着我们不必梳理许多代码行来查找错误源）！

在讨论解决方案之前，我只想指出一些风格编码“最佳”实践：

• 如果您选择使用 dplyr（以及随后的 tidyverse），那么坚持使用它通常在风格上更加一致。您的代码在 dplyr 和 base 之间切换，这会使代码更难阅读，但这并不是世界末日
• 当您使用read_xlsx() 时，请注意它是从readxl 包调用的，该包在您的示例中没有作为库加载，也没有通过命名空间引用，即readxl::read_xlsx()。确保您的示例完全可以通过您使用的包重现是一种很好的做法

代码重写

我已经重写了您在dplyr 中的部分代码，直到错误的来源为止。

data_clean <- data %>%
  mutate(across(c("PtsW":"TOL"), as.numeric))

teams <- data_clean %>% 
  distinct(Winner.tie) %>% 
  select(Winner.tie)
rankings <- teams

但是，您可以尝试自行实施的一些修复是：

• 以dplyr 样式重写代码的其余部分

• 使用purrr 系列map() 函数来替换for 循环。我还没有找到明确的证据证明purrr::map()普遍比 for 循环快，而且我个人认为使用 for 循环从根本上是有益的。

错误来源

我相信错误的根源来自您的这部分代码：

winner <- week.data[i,"Winner.tie"]
loser <- week.data[i,"Loser.tie"]

当我打印loser 时，我没有在内部循环的第一次迭代（i = 1）中获得值“Green Bay Packers”，而是得到了一个小标题，这就是您的代码通过索引所做的。其余代码失败，因为您随后尝试将 tibble 传递到此语句中：old.rank.l <- rankings[rankings$Team == loser,week.no+1] 并返回一个空值 (numeric(0))。

解决方案

这是代码重写，直到错误的来源。

winner <- week.data[i,"Winner.tie"] %>% pull()
loser <- week.data[i,"Loser.tie"] %>% pull()
old.rank.w <- rankings %>%
  filter(Team == winner) %>%
  .[[week.no + 1]]
old.rank.l <- rankings %>%
  filter(Team == loser) %>%
  .[[week.no + 1]]

进行此更改后，运行其余代码应该可以工作。如果您有任何问题，请告诉我

结论性想法

使用print() 语句查看您的代码在哪里出错，以及您的对象是否包含您期望的值。当您开始功能化您的代码时，browser() 是一个有用的调试工具。