Sapply 有两个条件

Question

我有两张桌子。具有如下所示的格式。其中之一是表 A：

students|Test Score|Year
A       |  100     |1993
B       |   81     |1992
C       |   92     |1992
D       |   88     |1993

我的另一个表 B 如下所示：

Class | Students | Year
1     | {A,D}    |1993
2     | {B,C}    |1992

我想在 R 中执行某种操作，通过该操作，我可以从表 A 中搜索表 B 中列下的数组中列出的学生，并将分数制成以下格式：

Class | Students | Mean Score
1     | {A,D}    |   94
2     | {B,C}    |   86.5

是否有任何公式可用于搜索，然后通过 R 中的某些操作合并这些结果？

我知道以上可以做到：

B$MeanScore <- sapply(strsplit(gsub("[{}]","", B$Students), split=","),
   function(x) mean(A$Test.Score[A$Students %in% x]))

但是我有没有办法添加第二个条件，也就是匹配年份。上课年份和考试年份。

Answer 1

在这里完全同意jogo：

A <- data.frame(students = c("A","B","C","D"), `Test Score` = c(100,81,92,88), Year = c(1993,1992,1992,1993))
A
#  students Test.Score Year
#1        A        100 1993
#2        B         81 1992
#3        C         92 1992
#4        D         88 1993

B <- data.frame(Class = c(1,2), Students = c("{A,D}","{B,C}"), Year = c(1993,1992))
B
#  Class Students Year
#1     1    {A,D} 1993
#2     2    {B,C} 1992

colnames(A) # taking note of the case sensitive "students" and "Year"
#[1] "students"   "Test.Score" "Year"   

s <- strsplit(gsub("[{}]","",B$Students), ",")
B.long <- data.frame(students = unlist(s), 
                     Class = rep(B$Class, sapply(s, length)), 
                     Year = rep(B$Year, sapply(s, length)))
B.long
#Students Class Year
#1        A     1 1993
#2        D     1 1993
#3        B     2 1992
#4        C     2 1992

Newdf <- merge.data.frame(A, B.long, c("Year","students"))
#Year students Test.Score Class
#1 1992        B         81     2
#2 1992        C         92     2
#3 1993        A        100     1
#4 1993        D         88     1

aggregate(Test.Score ~ Year + Class, Newdf, mean)
#Year Class Test.Score
#1 1993     1       94.0
#2 1992     2       86.5