我正在尝试编写一个循环，用于过滤由特定团队与AwayTeam一起玩的HomeTeam。所以最后我可以有一个专栏，告诉我一个特定球队在过去对阵AwayTeam时赢得/输掉比赛的数量。所以原始数据集如下所示：
Date HomeTeam AwayTeam FTHG FTAG FTR Draw Lost Win
(chr) (chr) (chr) (int) (int) (chr) (dbl) (dbl) (dbl)
1 17/08/2002 Blackburn Sunderland 0 0 D 1 0 0
2 17/08/2002 Charlton Chelsea 2 3 A 0 1 0
3 17/08/2002 Everton Tottenham 2 2 D 1 0 0
4 17/08/2002 Fulham Bolton 4 1 H 0 0 1
5 17/08/2002 Leeds Man City 3 0 H 0 0 1
6 17/08/2002 Man United West Brom 1 0 H 0 0 1
我先写了这段代码：
arsenal <- football %>%
select(1:10) %>%
filter(football$HomeTeam == "Arsenal" & football$AwayTeam == "Birmingham" & Date >= 17/08/2002)
birmingham <- football %>%
select(Date, HomeTeam, AwayTeam, FTHG, FTAG, FTR) %>%
filter(football$HomeTeam == "Birmingham" & football$AwayTeam == "Arsenal" & Date >= 17/08/2002)
它产生了以下内容：
Date HomeTeam AwayTeam FTHG FTAG FTR Draw Lost Win
(chr) (chr) (chr) (int) (int) (chr) (dbl) (dbl) (dbl)
1 18/08/2002 Arsenal Birmingham 2 0 H 0 0 1
2 01/05/04 Arsenal Birmingham 0 0 D 1 0 0
3 04/12/04 Arsenal Birmingham 3 0 H 0 0 1
4 02/10/05 Arsenal Birmingham 1 0 H 0 0 1
5 12/01/08 Arsenal Birmingham 1 1 D 1 0 0
6 17/10/09 Arsenal Birmingham 3 1 H 0 0 1
Date HomeTeam AwayTeam FTHG FTAG FTR
(chr) (chr) (chr) (int) (int) (chr)
1 12/01/2003 Birmingham Arsenal 0 4 A
2 22/11/03 Birmingham Arsenal 0 3 A
3 15/05/05 Birmingham Arsenal 2 1 H
4 04/02/06 Birmingham Arsenal 0 2 A
5 23/02/08 Birmingham Arsenal 2 2 D
6 27/03/10 Birmingham Arsenal 1 1 D
所以我想为所有团队自动化这个过程，就像上面的例子一样。我已经产生了以下循环，但它不起作用：
list_of_teams <- c("Arsenal", "Aston Villa", "Manchester City", "Liverpool")
for(i in 1:length(list_of_teams)){ football_1 <- football %>%
select(1:10) %>%
filter(football$HomeTeam == list_of_teams[i] & football$AwayTeam == list_of_teams[i+1])
for(j in 1:length(list_of_teams[-i])){
football$HomeTeam == list_of_teams[j] & football$AwayTeam == list_of_teams[-i]
print(c(list_of_teams[[i]], list_of_teams[-i][[j]]))
}
}
计划是创建空数据框，然后根据产生的循环将其存储在那里。但是，最终它只会存储最初创建的list_of_teams变量中的团队。 问题是：为什么循环忽略/或没有选择列1:10并相应地过滤HomeTeam和AwayTeam以及我们如何解决这个问题？ 提前感谢您，我是R的新手并且编程任何帮助将不胜感激。