r - 根据最高百分比将行名分配给列名

Question

voyages =(
VIC0016,
VIC0016,
VIC0016,
VIC0016,
VIC0016,
VIC0016,
Truck,
VIC0016,
VIC0016,
VIC0016,
JUL0983,
BB11356,
VIC0022,
VIC0022,
ISK1981,
ISK1981,
ISK1981,
ISK1981,
ISK1981,
ISK1981,
ISK1981,
ISK1981,
ISK1981,
ISK1981,
ISK1981,
ISK1981)

clusters = (5,
5,
5,
4,
4,
4,
1,
3,
4,
3,
5,
2,
4,
5,
6,
6,
6,
6,
6,
6,
6,
6,
6,
6,
6,
6)

>calculate.confusion <- function(voyages, clusters)  
{
  d <- data.frame(voyages, clusters)  
  td <- as.data.frame(table(d))  
  # convert the raw counts into percentage of each voyage number  
  pc <- matrix(ncol=max(clusters),nrow=0)  
  for (i in 1:11) # 11 different voyage numbers  
  {  
    total <- sum(td[td$voyages==td$voyages[i],3])   
    #,3 is the third column, showing the frequencies  
    pc <- rbind(pc, td[td$voyages==td$voyages[i],3]/total)  
  }   
  rownames(pc) <- td[1:11,1]  
  colnames(pc)<-1:11  
  return(pc)  
}  

有了上面的数据框（数字是百分比），我怎样才能用行的名称替换列名 [1:11]，这样：

• 在该行内，该行中百分比最高的列以该行命名
• 每个行名都使用一次

希望有人能帮助我。

Answer 1

这应该有助于：

# sample data
df <- data.frame(a = c(1,2,3), b = c(3,2,1), c = c(2,3,1))
colnames(df)
# [1] "a" "b" "c"
for(i in 1:nrow(df)) {colnames(df)[df[i, ] == max(df[i, ])] <- rownames(df)[i]}
colnames(df)
# [1] "3" "1" "2"