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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，例如kern[,c("c1b", "c2b", "c3b")]单位矩阵。我知道这很容易通过从第三行中减去第二行来完成：

kern[3,] = kern[3,] - kern[2,],

但是R中是否有一个函数可以为我做到这一点？在另一个线程中发布的缩减行梯形表格的功能不是我需要的。

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")]

另外我忘了提到我不希望任何条目kern是负面的。此代码适用于我的几个示例，但是，它很容易对许多其他矩阵造成麻烦。

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1 回答 1

1

你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（或至少我的大脑）在被要求从具有离轴非零条目的行中找到要减去的正确行时所做的事情：

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
于 2013-03-21T20:35:45.460 回答