请教大神：将一张表展开的代码

君子和而不同

如下矩阵
     A    B    C   
[1,] "A1" "B3" "C1"
[2,] "A2" "B1" "C2"
[3,] "A3" "B3" "C3"
[4,] "A1" "B2" "C3"
[5,] "A3" "B3" "C2"
[6,] "A1" "B1" "C1"
希望通过高效的代码展开为如下形态（将每个观测展开，在具体变量值里记有（1）没有（0））：因为要处理的数据有二十万行左右，每个变量的值也有几个到数十个不等。
         A    B1    B2    B3    C1    C2    C3      
[1,] "A1" "0"   "0"   "1"   "1"   "0"   "0"
[2,] "A2" "1"   "0"   "0"   "0"   "1"   "0"
[3,] "A3" "0"   "0"   "1"   "0"   "0"   "1"
[4,] "A1" "0"   "1"   "0"   "0"   "0"   "1"
[5,] "A3" "0"   "0"   "1"   "0"   "1"   "0"
[6,] "A1" "1"   "0"   "0"   "1"   "0"   "0"

多谢大神了

huhaohoo

想了半天，我只会循环... 【我是菜鸟】

#定义函数
featureExtend <- function(featureName,originalTable)
{
  originalRange <- originalTable[,featureName]
  nfeatureLevel <- nlevels(factor(originalRange))
  tmp <- matrix(0,nrow=nrow(originalTable),ncol=nfeatureLevel)
  colnames(tmp) <- levels(factor(originalRange))
  for(colIndex in 1:nfeatureLevel)
  {
    tmp[which(originalRange == colnames(tmp)[colIndex]),colIndex] <- 1;
  }
  
  return(tmp)
}

# 执行函数
> cbind(original[,"A"],featureExtend("B",original),featureExtend("C",original))

# 输出 - 少了第一列的列名
          B1  B2  B3  C1  C2  C3
[1,] "A1" "0" "0" "1" "1" "0" "0"
[2,] "A2" "1" "0" "0" "0" "1" "0"
[3,] "A3" "0" "0" "1" "0" "0" "1"
[4,] "A1" "0" "1" "0" "0" "0" "1"
[5,] "A3" "0" "0" "1" "0" "1" "0"
[6,] "A1" "1" "0" "0" "1" "0" "0"

huhaohoo

感觉可以用table直接做 继续学习吧

huhaohoo

> A <- c(1:6)
> B <- c('B3','B1','B3','B2','B3','B1')
> mytest <- cbind(A,B)
> table(mytest[,"A"],mytest[,"B"])

#输出【完美！】
    B1 B2 B3
  1  0  0  1
  2  1  0  0
  3  0  0  1
  4  0  1  0
  5  0  0  1
  6  1  0  0