Question

我有四类产品：

A类： {X800818， X822707， X822708， X870082， X800810， X800323， X800835， X893890， X822541， X800831}
B类： {X830742， X841223， X841449， X870138， X810352， X870146， X800850， X841236， X811712， X893314}
C类： {X893609， X890188， X893313， X841271， X891250， X811820， X728538， X727220， X960804， X728904}
D类： {X727345， X800875， X727302， X870426， X729002， X727300， X759042， X728495， X897198， X790190}

我有一个二进制数据框，每行包含一个或一组属于不同类别的产品：这是一个例子：

    X800818 X822707 X822708 X870082 X800810 X800323 X800835 X893890 X822541 X800831 X830742 X841223 X841449 

1          0       0       0       1       0       0       0       0       0       1       1       0       0

2          1       0       0       0       0       0       0       0       0       0       1       0       0

3          1       0       0       0       0       0       0       0       0       0       0       0       0

4          0       0       0       0       0       0       0       0       0       0       0       0       1

我想得到这样的结果：

2 A，1 B
1 A，1 B
1 A
1 B

如何让R从二进制数据框中扣除此结果？有任何建议吗？

Answer 1

我尝试制作一些可重现的代码，但我的解决方案并不尽可能干净：

#Data Creation

A <- c("X800818","X822707","X822708","X870082","X800810","X800323","X800835","X893890","X822541","X800831")
B <- c("X830742","X841223","X841449","X870138","X810352","X870146","X800850","X841236","X811712","X893314")
C <- c("X893609","X890188","X893313","X841271","X891250","X811820","X728538","X727220","X960804","X728904")
D <- c("X727345","X800875","X727302","X870426","X729002","X727300","X759042","X728495","X897198","X790190")
df <-data.frame(c(1,0,0,1),c(0,0,0,0),c(1,1,1,1),c(1,0,1,1),c(1,0,0,1),c(0,1,0,0),c(1,0,1,1))
names(df) <- c("X800818","X800323","X841223","X811820","X960804","X727300","X728495")

#Transforming binaries to letters

for( col in names(df)) {
 for(L in LETTERS[1:4]) {
  if(col %in% get(L)) df[df[,col] == 1,col] <- L
 }
}

#Transpose 

tdf <- data.frame(t(df))

#Get Results df
res <- NULL
for (col in names(tdf)) {
levels(tdf[,col]) <- c("0",LETTERS[1:4])
res <- rbind(res,table(tdf[,col]))
}


     0 A B C D
[1,] 2 1 1 2 1
[2,] 5 1 1 0 0
[3,] 4 1 1 1 0
[4,] 2 1 1 2 1

希望这会有所帮助

如何从二进制数据框中扣除结果？

1 个答案: