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