[R] Comparing matrices in R - matrixB %in% matrixA

[Charles Novaes de Santana]

My apologies, because I sent the message before finishing it. i am very
sorry about this. Please find below my message (I use to write the messages
from the end to the beginning... sorry :)).

Dear all,

I am trying to compare two matrices, in order to find in which rows of a
matrix A I can find the same values as in matrix B. I am trying to do it
for matrices with around 2500 elements, but please find below a toy example:

A = matrix(1:10,nrow=5)
B = A[-c(1,2,3),];

So
> A
     [,1] [,2]
[1,]    1    6
[2,]    2    7
[3,]    3    8
[4,]    4    9
[5,]    5   10

and
> B
     [,1] [,2]
[1,]    4    9
[2,]    5   10

I would like to compare A and B in order to find in which rows of A I can
find the  rows of B. Something similar to %in% with one dimensional arrays.
In the example above, the answer should be 4 and 5.

I did a function to do it (see it below), it gives me the correct answer
for this toy example, but the excess of for-loops makes it extremely slow
for larger matrices. I was wondering if there is a better way to do this
kind of comparison. Any idea? Sorry if it is a stupid question.

matbinmata<-function(B,A){
    res<-c();
    rowsB = length(B[,1]);
    rowsA = length(A[,1]);
    colsB = length(B[1,]);
    colsA = length(A[1,]);
    for (i in 1:rowsB){
        for (j in 1:colsB){
            for (k in 1:rowsA){
                for (l in 1:colsA){
                    if(A[k,l]==B[i,j]){res<-c(res,k);}
                }
            }
        }
    }
    return(unique(sort(res)));
}


Best,

Charles

On Fri, Oct 31, 2014 at 2:12 PM, Charles Novaes de Santana <
charles.santana em gmail.com> wrote:

> A = matrix(1:10,nrow=5)
> B = A[-c(1,2,3),];
>
> So
> > A
>      [,1] [,2]
> [1,]    1    6
> [2,]    2    7
> [3,]    3    8
> [4,]    4    9
> [5,]    5   10
>
> and
> > B
>      [,1] [,2]
> [1,]    4    9
> [2,]    5   10
>
> I would like to compare A and B in order to find in which rows of A I can
> find the  rows of B. Something similar to %in% with one dimensional arrays.
> In the example above, the answer should be 4 and 5.
>
> I did a function to do it (see it below), it gives me the correct answer
> for this toy example, but the excess of for-loops makes it extremely slow
> for larger matrices. I was wondering if there is a better way to do this
> kind of comparison. Any idea? Sorry if it is a stupid question.
>
> matbinmata<-function(B,A){
>     res<-c();
>     rowsB = length(B[,1]);
>     rowsA = length(A[,1]);
>     colsB = length(B[1,]);
>     colsA = length(A[1,]);
>     for (i in 1:rowsB){
>         for (j in 1:colsB){
>             for (k in 1:rowsA){
>                 for (l in 1:colsA){
>                     if(A[k,l]==B[i,j]){res<-c(res,k);}
>                 }
>             }
>         }
>     }
>     return(unique(sort(res)));
> }
>
>
> Best,
>
> Charles
>
>
> --
> Um axé! :)
>
> --
> Charles Novaes de Santana, PhD
> http://www.imedea.uib-csic.es/~charles
>



-- 
Um axé! :)

--
Charles Novaes de Santana, PhD
http://www.imedea.uib-csic.es/~charles

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