R: create diagonal from 1 to 10 in matrix -> returns only last value-CodePudding

I need to modify a matrix filled with random numbers so that the resulting matrix has a diagonal with the numbers 1 to 10 and everywhere else there are supposed to be zeros. I am almost done but the diagonal only shows the last value 10 instead of the numbers 1 to 10. I know I need to cache the result somehow but I don't know how to do that.

rand_mat = matrix(sample(1:50, 100, replace = TRUE), nrow=10, ncol=10)
#10x10 matrix filled with random numbers


for (i in 1:nrow(rand_mat)) {
  #for each row
  for (j in 1:ncol(rand_mat)) {
    #for each column
    for (o in 1:10){
      #go through numbers 1 to 10 for the diagonal
      
      if(i == j){
        #if row matches column -> diagonal
      
      rand_mat[i,j] = o
      #assign numbers 1 to 10 in the diagonal
      

      }else{
        
        rand_mat[i,j] = 0
        #if location is not in the diagonal assign 0
        
      }
      
    }
    
  }
  
}

This is the current result:

> print(rand_mat)
      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
 [1,]   10    0    0    0    0    0    0    0    0     0
 [2,]    0   10    0    0    0    0    0    0    0     0
 [3,]    0    0   10    0    0    0    0    0    0     0
 [4,]    0    0    0   10    0    0    0    0    0     0
 [5,]    0    0    0    0   10    0    0    0    0     0
 [6,]    0    0    0    0    0   10    0    0    0     0
 [7,]    0    0    0    0    0    0   10    0    0     0
 [8,]    0    0    0    0    0    0    0   10    0     0
 [9,]    0    0    0    0    0    0    0    0   10     0
[10,]    0    0    0    0    0    0    0    0    0    10
>

CodePudding user response：

Your innermost loop is unnecessary. When you are looping through j, you are looping through each cell in row i. So you need only assign the value i to the cell if i == j and 0 otherwise. You don't need the variable o at all.

So a fixed-up version of your code might be:

rand_mat = matrix(sample(1:50, 100, replace = TRUE), nrow=10, ncol=10)

for (i in 1:nrow(rand_mat)) {
  #for each row
  for (j in 1:ncol(rand_mat)) {
    #for each column
      if(i == j) {
        #if row matches column -> diagonal
      
         rand_mat[i,j] <- i
         #assign row number in the diagonal
      } else {
        
        rand_mat[i,j] = 0
        #if location is not in the diagonal assign 0
        
      }
   }
}

rand_mat
#>       [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
#>  [1,]    1    0    0    0    0    0    0    0    0     0
#>  [2,]    0    2    0    0    0    0    0    0    0     0
#>  [3,]    0    0    3    0    0    0    0    0    0     0
#>  [4,]    0    0    0    4    0    0    0    0    0     0
#>  [5,]    0    0    0    0    5    0    0    0    0     0
#>  [6,]    0    0    0    0    0    6    0    0    0     0
#>  [7,]    0    0    0    0    0    0    7    0    0     0
#>  [8,]    0    0    0    0    0    0    0    8    0     0
#>  [9,]    0    0    0    0    0    0    0    0    9     0
#> [10,]    0    0    0    0    0    0    0    0    0    10

Or, more succinctly:

for (i in 1:nrow(rand_mat)) 
  for (j in 1:ncol(rand_mat)) 
      rand_mat[i, j] <- if(i == j) i else 0

^{Created on 2022-05-14 by the reprex package (v2.0.1)}

CodePudding user response：

Using only functions you used in your question:

rand_mat = matrix(sample(1:50, 100, replace = TRUE), nrow=10, ncol=10)
rand_mat <- matrix(0, 10, 10)        # Make all the elements 0.
for(i in 1:10) rand_mat[i, i] <- i   # Replace the diagonal elements

CodePudding user response：

Although this is not an efficent way to do this (as noted in comments), you can fix your code by removing the innermost loop and replacing it with an assignment:

for (i in 1:nrow(rand_mat)) {
  for (j in 1:ncol(rand_mat)) {
    rand_mat[i,j] = if(i == j) i else 0 
      }
    }

CodePudding user response：

I got it thanks to @dcarlson Their code was definitely a lot shorter but I really wanted to use my code to see what I was doing wrong. Turns out I had to write rand_mat[o,o] = o instead of rand_mat[i,j] = o in the if-else-loop. So the code now looks like this:

rand_mat = matrix(sample(1:50, 100, replace = TRUE), nrow=10, ncol=10)
#10x10 matrix filled with random numbers


for (i in 1:nrow(rand_mat)) {
  #for each row
  for (j in 1:ncol(rand_mat)) {
    #for each column
    for (o in 1:10){
      #go through numbers 1 to 10 for the diagonal
      
      if(i == j){
        #if row matches column -> diagonal
      
      rand_mat[o,o] = o
      #assign numbers 1 to 10 in the diagonal

      }else{
        
        rand_mat[i,j] = 0
        #if location is not in the diagonal assign 0
        
      }
      
    }
    
  }
  
}

and the result looks like when I tried it with the code from @dcarlson

> print(rand_mat)
      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
 [1,]    1    0    0    0    0    0    0    0    0     0
 [2,]    0    2    0    0    0    0    0    0    0     0
 [3,]    0    0    3    0    0    0    0    0    0     0
 [4,]    0    0    0    4    0    0    0    0    0     0
 [5,]    0    0    0    0    5    0    0    0    0     0
 [6,]    0    0    0    0    0    6    0    0    0     0
 [7,]    0    0    0    0    0    0    7    0    0     0
 [8,]    0    0    0    0    0    0    0    8    0     0
 [9,]    0    0    0    0    0    0    0    0    9     0
[10,]    0    0    0    0    0    0    0    0    0    10
>

But as I learnt from @Allan Cameron you don't need the variable o so you don't even need the third for-loop:

rand_mat = matrix(sample(1:50, 100, replace = TRUE), nrow=10, ncol=10)

for (i in 1:nrow(rand_mat)) {
  #for each row
  for (j in 1:ncol(rand_mat)) {
    #for each column
    if(i == j) {
      #if row matches column -> diagonal
      
      rand_mat[i,j] <- i
      #assign row number in the diagonal
      
    } else {
      
      rand_mat[i,j] = 0
      #if location is not in the diagonal assign 0
      
    }
  }
}

The result is the same. Thank you for your help!