Question

所以，我一直在寻找解决方案。但我仍然没有成功。也许你可以帮助我。

    # Minimal example
       A <- rep("A", 9)
       B <- rep("B", 4)
       schoolid <- c(A, B)
       dc <- c("a", "b", "c", "", "", "", "", "", "", "a", "b", "i", "j")
       mc <- c( "", "", "", "a", "b", "c", "", "", "", "a", "b", "i", "j")
       ec <- c( "", "", "",  "", "", "", "a", "b", "c", "a", "b", "i", "j")
       dpoints <- c(20, 15, 17,  "", "", "", "", "", "", 14, 13, 13, 12)
       mpoints <- c( "", "", "", 18, 12, 20, "", "", "", 15, 11, 14, 9)
       epoints <- c( "", "", "",  "", "", "", 13, 14, 15, 16, 21, 17, 7)
       data <- data.frame(schoolid, dc, mc, ec, dpoints, mpoints, epoints)

这是我拥有的数据集，其中包含：

    # dc ... pupilsID in Deutsch
    # mc ... pupilsID in math
    # ec ... pupilsID in English
    # dpoints, mpoints, epoints, the achieved points in the tests
    # the four pupils in school A did participate in all tests, but
    # information are spread over three rows per pupil
    # in school B everything is allright: note, that four pupils in school B have the same codes as others in school A

    View(data)
    ###############################################################################################

所以一开始我想为pupilsID设一个变量“code”，它结合了

的信息

   # the three columns dc, mc and ec
   # in the following way:
   x <- rep(c("a", "b", "c"), 3)
   data1 <- data
   data1$code <- c(x, "a", "b", "i", "j")
   View(data1)

＃

其次，对于数据框的上半部分，我想合并？像这样的行

# result is my desired dataset
   schoolid1 <- c("A", "A", "A", "B", "B", "B", "B")
   code <- c("a", "b", "c", "a", "b", "i", "j")
   dpoints1 <- c(20, 15, 17, 14, 13, 13, 12)
   mpoints1 <- c(18, 12, 20, 15, 11, 14, 9)
   epoints1 <- c(13, 14, 15, 16, 21, 17, 7)
   result <- data.frame(schoolid1, code, dpoints1, mpoints1, epoints1)
   View(result)

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

所以1.）我尝试了以下（虽然不起作用）

# i is a counting variable for the i.th row in data
   i <- 1
   for (i in 1:13){
     if (data[i, "dc"]==data[i, "mc"]==data[i, "ec"]){
       data$code <- data[i, "dc"]
     } else if (!is.na(data[i, "dc"]) & is.na(data[i, "mc"]) & is.na(data[i, "ec"])){
       data$code <- data[i, "dc"]
     } else if (is.na(data[i, "dc"]) & !is.na(data[i, "mc"]) & is.na(data[i, "ec"])){
       data$code <- data[i, "mc"]
     } else if (is.na(data[i, "dc"]) & is.na(data[i, "mc"]) & !is.na(data[i, "ec"])){
       malsehen$code <- data[i, "ec"]
     }
   i <- i+1
   }

第二个问题，我不知道

Answer 1

尝试此操作：

sdata<-split(data,data$schoolid)
Apart<-cbind(sdata$A[1:3,"schoolid",drop=FALSE],
       as.data.frame(
         Map(function(x,y) x[(seq_along(x)-1) %/% 3 ==y],sdata$A[,2:7],0:2)))
rbind(Apart,sdata$B)
#   schoolid dc mc ec dpoints mpoints epoints
#1         A  a  a  a      20      18      13
#2         A  b  b  b      15      12      14
#3         A  c  c  c      17      20      15
#10        B  a  a  a      14      15      16
#11        B  b  b  b      13      11      21
#12        B  i  i  i      13      14      17
#13        B  j  j  j      12       9       7

Answer 2

以下是使用data.table

的解决方案

library(data.table) #1.9.5+
ints<-paste0(c("d","m","e"),"points")
setDT(data)[,(ints):=lapply(.SD,function(x)as.integer(levels(x))[x]),.SDcols=ints]

# Problem 1
data[,code:=levels(dc)[pmax(as.integer(dc),as.integer(mc),as.integer(ec))]]
# Problem 2
data[,(ints):=lapply(.SD,function(x)max(x,na.rm=T)),
     by=.(schoolid,code),.SDcols=ints]
# Remove excess information
data<-unique(setkey(data,schoolid,code))[,(chars):=NULL]

> data
   schoolid dpoints mpoints epoints code
1:        A      20      18      13    a
2:        A      15      12      14    b
3:        A      17      20      15    c
4:        B      14      15      16    a
5:        B      13      11      21    b
6:        B      13      14      17    i
7:        B      12       9       7    j

注意，您应该检查identical(levels(data$dc),levels(data$ec),levels(data$mc))的定义，code的定义依赖于此。

创建一个变量中的三个并合并行

2 个答案: