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Function that finds elements from matrix, and saves the indices of the elements to another matrix

3 messages · Tuomas Koponen, Bert Gunter, Rui Barradas

Tuomas Koponen
# 
Hi all dear R-list users,

This might sound a silly problem, but but for one reason or another it has
proved unsolvable to me.

I need to solve the following task. I have tried to use two nested for
loops as a solution, but have not been able to made it work. It would be
great, if someone could formulate a code for me:

I need to create a function X that takes as its input a 10x10 matrix, which
elements consists of ones and zeros. The function X must find the elements
in the matrix A that are "ones", and save the indices of those elements in
2-column matrix, in a following way: the first column has a row number, and
the second a column number.

The answer should be formulated in a way, that it uses two nested for
loops. The matrix to be returned has a maximum of 100 rows, so one
possibility exists for a 100x2 matrix filled with zeros in the index and
finally separate those rows with zeros.

I have tried the following code:

A<-matrix(1:0, nrow = 10, ncol = 10)

X <- data.frame()
for (i in seq(1, nrow(A))) {
for (j in seq(1,ncol(A))) {
if (A [i,j] > 0) {

new_row <- data.frame(Row_number= rownames(A)[i], Column_number=
colnames(A)[j])

X <- rbind(X, new_row)

  }

  }
}

I only manage to get "Error: object 'X' not found", as a result.

thanks in advance for your help.

Br. Tuomas
Bert Gunter
# 
Is this a homework problem? We try not to do others' homework here.

Incidentally, this can easily be done much more efficiently without any
for() loops.

Bert Gunter

"The trouble with having an open mind is that people keep coming along and
sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )


On Mon, Sep 14, 2020 at 11:50 AM Tuomas Koponen <tuomas.o.koponen at gmail.com>
wrote:

  
  


  


      

    

    

      
      

        


  

        

      Rui Barradas
    

    

      
      #
    

  


  

      
Hello,

This is simple:


which(A == 1, arr.ind = TRUE)


Hope this helps,

Rui Barradas

?s 12:03 de 14/09/20, Tuomas Koponen escreveu: