在多列中查找最新出现的值并在 R 的另一列中返回值答案

【问题标题】：Find latest occurrence of value in multiple columns and return value in another column in R在多列中查找最新出现的值并在 R 的另一列中返回值
【发布时间】：2022-10-06 00:12:52
【问题描述】：

我有一个像这样的数据框：

home_team <- c(\'Team A\', \'Team B\', \'Team C\', \'Team D\', \'Team B\', \'Team F\')
away_team <- c(\'Team B\', \'Team C\', \'Team D\', \'Team A\', \'Team F\', \'Team A\')
home_team_score_pre <- c(300, 150, 600, 800, 50, 450)
away_team_score_pre <- c(550, 340, 100, 208, 412, 18)
winning_team <- c(\'Team A\', \'Team C\', \'Team C\', \'Team D\', \'Team F\', \'Team F\')
res <- c(16, 25, 11, 4, 22, 9) 
home_team_score_change <- c(16, -25, 11, 4, -22, 9) 
away_team_score_change <- c(-16, 25, -11, -4, 22, -9)
home_team_score_post <- c(316, 125, 611, 804, 28, 459)
away_team_score_post <- c(534, 365, 89, 204, 434, 9)

输出：

  home_team away_team home_team_score_pre away_team_score_pre winning_team res home_team_score_change away_team_score_change
1    Team A    Team B                 300                 550       Team A  16                     16                    -16
2    Team B    Team C                 150                 340       Team C  25                    -25                     25
3    Team C    Team D                 600                 100       Team C  11                     11                    -11
4    Team D    Team A                 800                 208       Team D   4                      4                     -4
5    Team B    Team F                  50                 412       Team F  22                    -22                     22
6    Team F    Team A                 450                  18       Team F   9                      9                     -9
  home_team_score_post away_team_score_post
1                  316                  534
2                  125                  365
3                  611                   89
4                  804                  204
5                   28                  434
6                  459                    9

每支球队在比赛开始前都有一个分数（home_team_score_pre 和away_team_score_pre）。

比赛结束后，调整后的分数是结果 (res)，这取决于他们是赢还是输了比赛。例如。第 1 行主队是 Team A 获胜，res 是 16，所以 Team A 的得分增加了 16，而 Team B 输了，他们的得分减去了 16。总体结果是赛后得分（home_team_score_post 和away_team_score_post）。

我想要做的是更新home_team_score_pre 和away_team_score_pre，方法是查找每支球队上一场比赛，然后在home_team_score_post 或away_team_score_post 上输入值）。

因此，例如，如果接下来的两行是：

  home_team away_team home_team_score_pre away_team_score_pre
1    Team C    Team B  
2    Team A    Team F

然后我想查找\'Team C\'（第 3 行）进行的最后一场比赛，并将home_score_post (611) 输入home_team_score_pre column。

同样，对于Team B（第 5 行）玩的最后一场游戏，将home_score_post (28) 输入home_team_score_pre column。

这些值可以来自任一列，因为它是球队最后一场比赛的值，所以他们可能是主队或客队。

另外，如果球队正在玩它的第一场比赛（因此没有以前的值），我想输入一个默认值 100。

因此，最终输出将是：

home_team away_team home_team_score_pre away_team_score_pre winning_team res home_team_score_change away_team_score_change
1    Team A    Team B                 100                100       Team A  16                     16                    -16
2    Team B    Team C                 84                 100       Team C  25                    -25                     25
3    Team C    Team D                 125                100       Team C  11                     11                    -11
4    Team D    Team A                 89                 116       Team D   4                      4                     -4
5    Team B    Team F                 59                 100       Team F  22                    -22                     22
6    Team F    Team A                 122                112       Team F   9                      9                     -9
7    Team C    Team B                 136                37        Team B  12                      49                      131
8    Team B    Team F                 49                 131       Team F  10                      0                      0
  home_team_score_post away_team_score_post
1                  116                  84
2                  59                   125
3                  136                  89
4                  93                   112
5                  37                   122
6                  131                  103
7                  124                  49
8                  39                   141

你想要的最终输出是什么？将其包含在您的问题中
此外，您的问题主题谈到查找以前的值，但您表达的方式似乎找到了最新的值而不是以前的值。哪个是哪个？
抱歉，我添加了最终输出作为示例。是的，最新值是我正在寻找的。感谢您的澄清。

标签： r

【解决方案1】：

更新 - 2022-10-05

如果您有如下所示的起始数据 (df)：

  home_team away_team winning_team res
1    Team A    Team B       Team A  16
2    Team B    Team C       Team C  25
3    Team C    Team D       Team C  11
4    Team D    Team A       Team D   4
5    Team B    Team F       Team F  22
6    Team F    Team A       Team F   9

您可以使用以下方法对其进行转换

加载库，并添加一列表示游戏编号

library(data.table)
setDT(df)
df[, game:=.I]

融长，生成这是否是球队的第一场比赛以及是否获胜的指标列，并设置初始值（即第一场比赛为100），并为第一场比赛发布得分游戏

df_long = melt(df, id=c("game", "res", "winning_team")) %>% 
  .[, fgame:=min(game)==game, value] %>% 
  .[fgame==1, score_pre:= 100] %>% 
  .[,win:=winning_team == value] %>% 
  .[, score_post:= fifelse(win,res,-res)+score_pre] %>% 
  .[order(value,game)]

使用 for 循环填充团队所有其他出场的前一个得分和后一个得分

for(i in 1:nrow(df_long)) {
  if(df_long[i,is.na(score_pre)]) {
    df_long[i,score_pre:=df_long[i-1,score_post]]
    df_long[i,score_post:=df_long[i,score_pre] + fifelse(df_long[i,win],res,-res)]
  }
}

将数据转换回宽格式，并重命名列

dcast(df_long, game+res+winning_team~variable, value.var = c("score_pre", "score_post","value")) %>% 
  .[, .(home_team = value_home_team, away_team=value_away_team, 
        home_team_score_pre = score_pre_home_team,
        away_team_score_pre = score_pre_away_team,
        winning_team, res,
        home_team_score_post = score_post_home_team,
        away_team_score_post = score_post_away_team)]

输出：

   home_team away_team home_team_score_pre away_team_score_pre winning_team   res home_team_score_post away_team_score_post
      <char>    <char>               <num>               <num>       <char> <num>                <num>                <num>
1:    Team A    Team B                 100                 100       Team A    16                  116                   84
2:    Team B    Team C                  84                 100       Team C    25                   59                  125
3:    Team C    Team D                 125                 100       Team C    11                  136                   89
4:    Team D    Team A                  89                 116       Team D     4                   93                  112
5:    Team B    Team F                  59                 100       Team F    22                   37                  122
6:    Team F    Team A                 122                 112       Team F     9                  131                  103

输入：

df = structure(list(home_team = c("Team A", "Team B", "Team C", "Team D", 
"Team B", "Team F"), away_team = c("Team B", "Team C", "Team D", 
"Team A", "Team F", "Team A"), winning_team = c("Team A", "Team C", 
"Team C", "Team D", "Team F", "Team F"), res = c(16, 25, 11, 
4, 22, 9)), class = "data.frame", row.names = c(NA, -6L))

以前的解决方案

您可以使用如下辅助函数：

# Helper Function return row
prepare_next_row <- function(df, home, away) {
  r = data.table(home_team = home, away_team=away)
  lrh = last(df[home_team==home | away_team == home])
  lra = last(df[home_team==away | away_team == away])
  r[, home_team_score_pre:=fifelse(lrh$home_team==home,lrh$home_team_score_post, lrh$away_team_score_post)]
  r[, away_team_score_pre:=fifelse(lra$home_team==away,lra$home_team_score_post, lra$away_team_score_post)]
  r[]
}

现在，当你想要一个新行时，你只需调用它，连同rbind()

rbind(
  df, 
  prepare_next_row(df, "Team C", "Team B"),
  prepare_next_row(df, "Team A", "Team F"),
  fill=TRUE
)

输出：

   home_team away_team home_team_score_pre away_team_score_pre winning_team   res home_team_score_change away_team_score_change home_team_score_post away_team_score_post
      <char>    <char>               <num>               <num>       <char> <num>                  <num>                  <num>                <num>                <num>
1:    Team A    Team B                 300                 550       Team A    16                     16                    -16                  316                  534
2:    Team B    Team C                 150                 340       Team C    25                    -25                     25                  125                  365
3:    Team C    Team D                 600                 100       Team C    11                     11                    -11                  611                   89
4:    Team D    Team A                 800                 208       Team D     4                      4                     -4                  804                  204
5:    Team B    Team F                  50                 412       Team F    22                    -22                     22                   28                  434
6:    Team F    Team A                 450                  18       Team F     9                      9                     -9                  459                    9
7:    Team C    Team B                 611                  28         <NA>    NA                     NA                     NA                   NA                   NA
8:    Team A    Team F                   9                 459         <NA>    NA                     NA                     NA                   NA                   NA

【讨论】：

这太棒了。如果我已经拥有数据，但我不想添加新行，该怎么办？而是根据我所拥有的进行更新？抱歉，如果我在最初的帖子中没有明确说明这一点。
好吧，我不明白 - 我现在已经大大改变了我的解决方案。