R: 使用单独的坐标矩阵访问矩阵的元素



我有一个矩阵(expTrans),它当前填充了零。我有第二个矩阵(expCoor),其中第二列用列名填充,第三列用行号填充。我想用第二个矩阵中的坐标把第一个矩阵中一些元素改成1。当我使用这个代码时,R用1填充提到的每列中提到的每一行(列显示相同):

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

有没有一种方法可以只更改每个坐标指定的行和列,而不必遍历整个矩阵?

这里有一个简单的例子来说明我所说的:

> 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

我期望列a有3个1(行4、2、8),列b有3个1秒(行9、6、5),列c有1个1(列7),列d有1个11(列10),列e有1个(列8),并且列j有1个一(列3)。

有一种方法可以为"["或"[<-"使用两列数字矩阵。如果您想对列和行名进行某种查找,这可能是可能的(如果且仅当您提供了一个适当的可复制示例。

> 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
> ?'['

修正问题的解决方案:

> 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

最新更新