将矩阵转换为R中三个定义的列(Convert matrix to three defined columns in R)

鉴于m ：

m <- structure(c(5, 1, 3, 2, 1, 4, 5, 2, 5, 1, 1, 5, 1, 4, 0, 4, 5, 5, 3, 2, 0, 0, 3, 0, 3, 2, 3, 0, 0, 0, 0, 0, 0, 0, 0), .Dim = c(7L, 5L)) # [,1] [,2] [,3] [,4] [,5] # [1,] 5 2 0 0 0 # [2,] 1 5 4 3 0 # [3,] 3 1 5 0 0 # [4,] 2 1 5 3 0 # [5,] 1 5 3 2 0 # [6,] 4 1 2 3 0 # [7,] 5 4 0 0 0

考虑元素1 ，它出现在5行(2, 3, 4, ,5, 6) ，并且相应的列方向索引是(1, 2, 2, 1, 2) 1,2,2,1,2 (1, 2, 2, 1, 2) 。我想有以下几点：

1 2 1 1 3 2 1 4 2 1 5 1 1 6 2

作为另一个例子，考虑元素2 ，它出现在4行(1, 4, 5, 6) ，并且相应的列方向索引是(2, 1, 4, 3) 2,1,4,3 (2, 1, 4, 3)并且我们具有：

1 2 1 1 3 2 1 4 2 1 5 1 1 6 2 2 1 2 2 4 1 2 5 4 2 6 3

我想要的是所有1-5的n*3矩阵。 优选在基础R中

Given m:

m <- structure(c(5, 1, 3, 2, 1, 4, 5, 2, 5, 1, 1, 5, 1, 4, 0, 4, 5, 5, 3, 2, 0, 0, 3, 0, 3, 2, 3, 0, 0, 0, 0, 0, 0, 0, 0), .Dim = c(7L, 5L)) # [,1] [,2] [,3] [,4] [,5] # [1,] 5 2 0 0 0 # [2,] 1 5 4 3 0 # [3,] 3 1 5 0 0 # [4,] 2 1 5 3 0 # [5,] 1 5 3 2 0 # [6,] 4 1 2 3 0 # [7,] 5 4 0 0 0

Consider the element 1, it appears in 5 rows (2, 3, 4, ,5, 6) and the respective column-wise indices are (1, 2, 2, 1, 2). I would like to have the following:

1 2 1 1 3 2 1 4 2 1 5 1 1 6 2

As another example, consider the element 2, it appears in 4 rows (1, 4, 5, 6) and the respective column-wise indices are (2, 1, 4, 3) and we have:

1 2 1 1 3 2 1 4 2 1 5 1 1 6 2 2 1 2 2 4 1 2 5 4 2 6 3

What I want is a n*3 matrix for all 1-5. Preferably in base R

最满意答案

我们可以使用arr.ind=TRUE

cbind(val= 1, which(m==1, arr.ind=TRUE)) # val row col #[1,] 1 2 1 #[2,] 1 5 1 #[3,] 1 3 2 #[4,] 1 4 2 #[5,] 1 6 2

对于多个案例，如@RHertel所述

for(i in 1:5) print(cbind(i,which(m==i, arr.ind=TRUE)))

或与lapply

do.call(rbind, lapply(1:2, function(i) { m1 <-cbind(val=i,which(m==i, arr.ind=TRUE)) m1[order(m1[,2]),]})) # val row col #[1,] 1 2 1 #[2,] 1 3 2 #[3,] 1 4 2 #[4,] 1 5 1 #[5,] 1 6 2 #[6,] 2 1 2 #[7,] 2 4 1 #[8,] 2 5 4 #[9,] 2 6 3

正如OP提到有关base R解决方案，上述内容将有所帮助。但是，如果有人想要一个紧凑的解决方案，

library(reshape2) melt(m)

然后将感兴趣的值分组。

We can use which with arr.ind=TRUE

cbind(val= 1, which(m==1, arr.ind=TRUE)) # val row col #[1,] 1 2 1 #[2,] 1 5 1 #[3,] 1 3 2 #[4,] 1 4 2 #[5,] 1 6 2

For multiple cases, as @RHertel mentioned

for(i in 1:5) print(cbind(i,which(m==i, arr.ind=TRUE)))

Or with lapply

do.call(rbind, lapply(1:2, function(i) { m1 <-cbind(val=i,which(m==i, arr.ind=TRUE)) m1[order(m1[,2]),]})) # val row col #[1,] 1 2 1 #[2,] 1 3 2 #[3,] 1 4 2 #[4,] 1 5 1 #[5,] 1 6 2 #[6,] 2 1 2 #[7,] 2 4 1 #[8,] 2 5 4 #[9,] 2 6 3

As the OP mentioned about base R solutions, the above would help. But, in case, if somebody wants a compact solution,

library(reshape2) melt(m)

and then subset the values of interest.

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