R：使用单独的坐标矩阵访问矩阵的元素

Question

我有一个矩阵（expTrans），它当前被零填充。我有第二个矩阵（expCoor），第二列填充列名，第三列填充行号。我想使用第二个矩阵中的坐标将第一个矩阵中的某些元素更改为1。当我使用这个代码，R填充在具有1中提到的每一列中提到的每行（列出来相同）：

expTrans<-matrix(0, nrow(cashflows), ncol(dmat)) 
colnames(expTrans)<-substr(colnames(dmat),4,6) 
expTrans[expCoor[,3],colnames(expenses)[expCoor[,1]]]<-1 

是否有一种方法来改变仅通过各坐标所指定的行和列，而不必循环遍历整个矩阵？

下面是一个简单的例子来说明什么我谈论：

> m<-matrix(0,ncol=10,nrow=10) 
> colnames(m)<-c("a","b","c","d","e","f","g","h","i","j") 
> m 
     a b c d e f g h i j 
[1,] 0 0 0 0 0 0 0 0 0 0 
[2,] 0 0 0 0 0 0 0 0 0 0 
[3,] 0 0 0 0 0 0 0 0 0 0 
[4,] 0 0 0 0 0 0 0 0 0 0 
[5,] 0 0 0 0 0 0 0 0 0 0 
[6,] 0 0 0 0 0 0 0 0 0 0 
[7,] 0 0 0 0 0 0 0 0 0 0 
[8,] 0 0 0 0 0 0 0 0 0 0 
[9,] 0 0 0 0 0 0 0 0 0 0 
[10,] 0 0 0 0 0 0 0 0 0 0 
> ind<-cbind(sample(1:10,10),sample(1:10,10), c("a","a","b","j","c","d","e", "a", "b", "b")) 
> ind 
     [,1] [,2] [,3] 
[1,] "4" "8" "a" 
[2,] "2" "9" "a" 
[3,] "9" "3" "b" 
[4,] "3" "1" "j" 
[5,] "7" "4" "c" 
[6,] "10" "7" "d" 
[7,] "8" "6" "e" 
[8,] "1" "2" "a" 
[9,] "6" "5" "b" 
[10,] "5" "10" "b" 
> m[as.numeric(ind[,1]),ind[,3]]<-1 
> m 
     a b c d e f g h i j 
[1,] 1 1 1 1 1 0 0 0 0 1 
[2,] 1 1 1 1 1 0 0 0 0 1 
[3,] 1 1 1 1 1 0 0 0 0 1 
[4,] 1 1 1 1 1 0 0 0 0 1 
[5,] 1 1 1 1 1 0 0 0 0 1 
[6,] 1 1 1 1 1 0 0 0 0 1 
[7,] 1 1 1 1 1 0 0 0 0 1 
[8,] 1 1 1 1 1 0 0 0 0 1 
[9,] 1 1 1 1 1 0 0 0 0 1 
[10,] 1 1 1 1 1 0 0 0 0 1 

我预想列有3分1秒（行4,2,8），列b有3分1秒（9行，6列），列c有11（行7），列d有1 1（行10），列e有1 1（行8），列j有1 1（行3）。

Answer 1

有一种方法使用两列数字矩阵用于“[”或“[< - ”。如果你想要做某种查找的列和rownames的，当且仅当您提供了适当的可重复的例子，可能是可能的（

> expTrans<-matrix(0, 3,4);expTrans 
    [,1] [,2] [,3] [,4] 
[1,] 0 0 0 0 
[2,] 0 0 0 0 
[3,] 0 0 0 0 
> dmat <- matrix(c(1,3,4,2,2,2),ncol=3, byrow=TRUE);dmat 
    [,1] [,2] [,3] 
[1,] 1 3 4 
[2,] 2 2 2 
> colnames(expTrans)<- colnames(dmat); rownames(expTrans) <-rownames(dmat) 
> expTrans[cbind(dmat[,2],dmat[,3])] <- dmat[,1] 
> expTrans 
    [,1] [,2] [,3] [,4] 
[1,] 0 0 0 0 
[2,] 0 2 0 0 
[3,] 0 0 0 1 
> ?'[' 

解决方案经修订的问题：

> m[cbind(as.numeric(ind[,1]), match(ind[,3], colnames(m))) ] <- 1 
> m 
     a b c d e f g h i j 
[1,] 0 1 0 0 0 0 0 0 0 0 
[2,] 0 1 0 0 0 0 0 0 0 0 
[3,] 0 0 0 1 0 0 0 0 0 0 
[4,] 1 0 0 0 0 0 0 0 0 0 
[5,] 0 0 0 0 0 0 0 0 0 1 
[6,] 0 1 0 0 0 0 0 0 0 0 
[7,] 0 0 0 0 1 0 0 0 0 0 
[8,] 1 0 0 0 0 0 0 0 0 0 
[9,] 0 0 1 0 0 0 0 0 0 0 
[10,] 1 0 0 0 0 0 0 0 0 0