How to find specific elements in a matrix

Good Day,

I have a matrix: R55 = [2,4,6,9,11;7,8,9,10,6;1,2,6,4,5;,6,9,8,5,6;1,6,5,7,5];

I would like to find the specific locations of all the 6's (rows and colums) using a for loop and using logical matrices rather than using just the "find" function.

Answer 1

[Voss]

R55 = [2,4,6,9,11;7,8,9,10,6;1,2,6,4,5;,6,9,8,5,6;1,6,5,7,5]

R55 = 5×5

2 4 6 9 11 7 8 9 10 6 1 2 6 4 5 6 9 8 5 6 1 6 5 7 5

val = 6;

Using a for loop:

rows = [];
columns = [];
m = size(R55,1);
for ii = 1:numel(R55)
    if R55(ii) == val
        rows(end+1) = mod(ii-1,m)+1;
        columns(end+1) = ceil(ii/m);
    end
end
rows, columns

rows = 1×6

4 5 1 3 2 4

columns = 1×6

1 2 3 3 5 5

Using two nested for loops:

rows = [];
columns = [];
[m,n] = size(R55);
for jj = 1:n
    for ii = 1:m
        if R55(ii,jj) == val
            rows(end+1) = ii;
            columns(end+1) = jj;
        end
    end
end
rows, columns

rows = 1×6

4 5 1 3 2 4

columns = 1×6

1 2 3 3 5 5

Using a logical matrix, without find:

is_val = R55 == val

is_val = 5×5 logical array
   0   0   1   0   0
   0   0   0   0   1
   0   0   1   0   0
   1   0   0   0   1
   0   1   0   0   0

idx = 1:numel(R55);
[rows,columns] = ind2sub(size(R55),idx(is_val))

rows = 1×6

4 5 1 3 2 4

columns = 1×6

1 2 3 3 5 5

Answer 2

[Walter Roberson]

This kind of task is usually easier to write with nested for loops over size() of the array. If you are required to use a single for loop then loop over the linear indices and ind2sub to translate to row and column