CODEWITHSUNDEEP
rsortingmatrix
user2797174
user2797174

Reputation: 167

sort a matrix based on another matrix in r

I have two matrices:

A = matrix(c(2,1,3,6,4,3,8,1,6,2,9,5), ncol=4)
B = matrix(c(20,10,30,60,40,30,80,10,60,20,90,50), ncol=4)

Now, I have sorted the matrix A by rows:

A_sorted=t(apply(A,1,sort))

2 2 6 8
1 1 4 9
3 3 5 6

Now, I want to sort the B matrix in the same order as of A, so that the new matrix 'B_sorted' would be:

20 20 60 80
10 10 40 90
30 30 50 60

I have found a similar answer from the past Sort one matrix based on another matrix but the codes from the answer do not work with a matrix of different dimensions.

Upvotes: 1

Views: 1934

Answers (3)

akuiper
akuiper

Reputation: 214917

You can create the index that sort the matrix by row with order(row(A), A) and then reorder the original matrix with it; later on add dimension to sorted data with matrix:

matrix(B[order(row(A), A)], byrow = TRUE, ncol = ncol(B))

#     [,1] [,2] [,3] [,4]
#[1,]   20   20   60   80
#[2,]   10   10   40   90
#[3,]   30   30   50   60

Upvotes: 2

akrun
akrun

Reputation: 886998

We can loop through the sequence of rows with for loop and assign the rows of 'B' based on the order of 'A' rows

for(i in seq_len(nrow(A))) B[i,] <- B[i,][order(A[i,])]
B
#     [,1] [,2] [,3] [,4]
#[1,]   20   20   60   80
#[2,]   10   10   40   90
#[3,]   30   30   50   60

Upvotes: 2

d.b
d.b

Reputation: 32548

t(sapply(1:NROW(A), function(i) B[i,][order(A[i,])]))
#     [,1] [,2] [,3] [,4]
#[1,]   20   20   60   80
#[2,]   10   10   40   90
#[3,]   30   30   50   60

Upvotes: 1

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