CODEWITHSUNDEEP
rmatrix
989
989

Reputation: 12937

Convert matrix to three defined columns in R

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

Upvotes: 2

Views: 1381

Answers (3)

Aaron - mostly inactive
Aaron - mostly inactive

Reputation: 37734

Just use row and col.

> data.frame(m=as.vector(m), row=as.vector(row(m)), col=as.vector(col(m)))
   m row col
1  5   1   1
2  1   2   1
3  3   3   1
4  2   4   1
5  1   5   1
...

Subset, sort, and print as desired.

> tmp <- out[order(out$m, out$row), ]
> print(subset(tmp, m==1), row.names=FALSE)
 m row col
 1   2   1
 1   3   2
 1   4   2
 1   5   1
 1   6   2

Upvotes: 2

akrun
akrun

Reputation: 886938

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.

Upvotes: 3

akuiper
akuiper

Reputation: 214927

A convenient way to transform it is to use sparseMatrix from Matrix library, since your desired output is very close to the representation of sparse Matrix:

library(Matrix)
summary(Matrix(m, sparse = T))
# 7 x 5 sparse Matrix of class "dgCMatrix", with 23 entries 
#    i j x
# 1  1 1 5
# 2  2 1 1
# 3  3 1 3
# 4  4 1 2
# 5  5 1 1
# 6  6 1 4
# 7  7 1 5
# 8  1 2 2
# 9  2 2 5
# 10 3 2 1
# 11 4 2 1
# 12 5 2 5
# 13 6 2 1
# 14 7 2 4
# 15 2 3 4
# 16 3 3 5
# 17 4 3 5
# 18 5 3 3
# 19 6 3 2
# 20 2 4 3
# 21 4 4 3
# 22 5 4 2
# 23 6 4 3

To see it better:

summary(Matrix(m, sparse = T)) %>% dplyr::arrange(x)
#    i j x
# 1  2 1 1
# 2  5 1 1
# 3  3 2 1
# 4  4 2 1
# 5  6 2 1
# 6  4 1 2
# 7  1 2 2
# 8  6 3 2
# 9  5 4 2
# 10 3 1 3
# 11 5 3 3
# 12 2 4 3
# 13 4 4 3
# 14 6 4 3
# 15 6 1 4
# 16 7 2 4
# 17 2 3 4
# 18 1 1 5
# 19 7 1 5
# 20 2 2 5
# 21 5 2 5
# 22 3 3 5
# 23 4 3 5

Upvotes: 4

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