核矩阵上的R行运算

Question

我有一个内核矩阵，如下所示：

kern <- matrix(c(1,0,0,1,0,0,0,1,1,0,1,1,0,0,1,0,0,1), dimnames=list(c("r1", "r1", "r3"), c("c1a", "c1b", "c2a", "c2b", "c3a", "c3b")), ncol=6, nrow=3)

> kern
   c1a c1b c2a c2b c3a c3b
r1   1   1   0   0   0   0
r2   0   0   1   1   0   0
r3   0   0   1   1   1   1

现在，我想应用行操作，使kern[,c("c1b", "c2b", "c3b")]为单位矩阵。 我知道，通过从第三行减去第二行可以轻松实现：

kern[3,] = kern[3,] - kern[2,] ，

但是R中有一个函数对我有用吗？ 我不需要在另一个线程中发布的用于减少行梯形表格的功能。

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编辑

我有一个笨拙的解决方案

sub <- kern[,c("c1b", "c2b", "c3b")]

for (i in which(colnames(kern) %in% colnames(sub))){
  ##identify which columns have more than one entry
  nonzero.row.idx <- which(kern[,i] != 0)
  while(length(nonzero.row.idx) > 1){    
    row.combinations <- combn(nonzero.row.idx, 2)
    for (j in ncol(row.combinations)){
      r1.idx <- row.combinations[1,j]
      r2.idx <- row.combinations[2,j]
      r1 <- kern[r1.idx,]
      r2 <- kern[r2.idx,]
      if (min(r1 - r2) >=0)
        kern[r1.idx, ] <- r1-r2
      else if (min(r2 - r1) >=0)
        kern[r2.idx, ] <- r2-r1
      else
        stop("Producing negative entries in row")      
      nonzero.row.idx <- which(kern[,i] != 0)
    }
  }
}      

kern[,c("c1b", "c2b", "c3b")]

我也忘记提及我不希望kern任何条目都为负数。 该代码适用于我的一些示例，但是，它很容易给其他许多矩阵带来麻烦。

Answer 1

您的分配箭头指向错误的方向。

kern <-  matrix(c(1,0,0,1,0,0,0,1,1,0,1,1,0,0,1,0,0,1), 
           dimnames=list(c("r1", "r1", "r3"), 
                         c("c1a", "c1b", "c2a", "c2b", "c3a", "c3b")),
           ncol=6, nrow=3)
sub <- kern[,c("c1b", "c2b", "c3b")]

您可以尝试复制您的大脑（或至少是我的大脑）在被要求找到正确的行以从具有离轴非零条目的行中减去时所做的事情：

id <- which( sub != 0 & row(sub) != col(sub), arr.ind=TRUE)
id
#   row col
#r3   3   2

> sub[ id[ ,"row" ], ] <- sub[id[ ,"row" ] , ] - sub[id[, "col" ], ]
> sub
   c1b c2b c3b
r1   1   0   0
r1   0   1   0
r3   0   0   1