divide each column of an array by its norm of the column

If there is array with size (2*3*128) and I want to divid each column over its norm.

How to do that?

For example this is part of the matrix

val(:,:,1) =

-0.0401 -0.2077 0.4750

-0.3867 -1.7742 3.3940

val(:,:,2) =

-0.0259 -0.1569 0.3964

-0.2504 -1.3830 3.1329

I want each column such as -0.0401 to divid it over its norm and same for other column

-0.3867

2nd column -0.2077 divid it over its norm

-1.7742

[Torsten]

Since your matrix is 3d, it's difficult to say what you mean by "columns".

Better specify the matrix dimension over which you want to take the norm.

[M]

@Torsten , For example this is part of the matrix

val(:,:,1) =

-0.0401 -0.2077 0.4750

-0.3867 -1.7742 3.3940

val(:,:,2) =

-0.0259 -0.1569 0.3964

-0.2504 -1.3830 3.1329

I want each column such as -0.0401 to divid it over its norm and same for other column

-0.3867

2nd column -0.2077 divid it over its norm

-1.7742

[Image Analyst]

When you say "each column such as -0.0401" that confuses us. A column does not have just a single value in it. From what you've displayed and said, you'd have 2 rows, 3 columns (with a "plane" of 2-by-128 in each column), and 128 "slices".

I assume you want to take the norm of each column? So that would be the norm of each 2-by-128 matrix. What would you consider to be the "norm" of that matrix?

[Torsten]

@Image Analyst

OP listed the second element of the column in the next line.

[M]

@Image Analyst as @Torsten said the second element of the column listed in the next line

[Image Analyst]

Arrays are indexed like (row, column, slice). So by column, you don't mean dimension 2

val = 5*rand(2,3,128);
col2 = val(:, 2, :); % Get the second column -- all values of dimension 2
whos col2

  Name      Size               Bytes  Class     Attributes

  col2      2x1x128             2048  double              

col2 = squeeze(col2) % Turn from 3-D array into 2-D matrix

col2 = 2×128

1.6494 1.7078 0.5240 4.1775 4.6723 0.8532 1.7875 2.8801 0.8354 4.9773 2.5038 4.2715 2.1833 2.8106 2.6810 0.2129 3.3235 1.1497 2.5837 4.5390 3.5507 3.4277 1.2060 3.0198 3.1786 1.5361 1.3881 4.4988 3.9208 2.5136 2.8813 2.0693 3.8127 1.7947 3.8008 4.7853 4.6437 3.5783 3.8805 4.8341 2.9445 0.0719 0.9767 1.8386 0.4462 2.3776 0.3446 2.2689 0.9295 1.4432 4.7759 4.8002 0.3142 1.1288 1.1407 2.2899 1.8586 0.8476 2.4115 2.9267

whos col2

  Name      Size             Bytes  Class     Attributes

  col2      2x128             2048  double              

but you mean like a vertical vector perpendicular to the slices going down through the 128 slices. Like one vector of 128 would be val(1,1,:), and another vector of 128 would be val(1,2,:), and so on. Is that right?

Answer 1

[Torsten]

A = A./vecnorm(A,2)

[M]

@Torsten Thanks

Answer 2

[Image Analyst]

Is this possibly what you're trying to describe?

val = 5 * rand(2, 3, 128); % Sample data.
[rows, columns, slices] = size(val);
% Get norms
theNorms = zeros(rows, columns);
for col = 1 : columns
    for row = 1 : rows
        v = squeeze(val(row, col, :));
        theNorms(row, col) = norm(v);
    end
end
% Divide each value by the norm for that (row, column) location.
output = zeros(rows, columns, slices);
for slice = 1 : slices
    for col = 1 : columns
        for row = 1 : rows
            thisNorm = theNorms(row, col);
            output(row, col, slice) = val(row, col, slice) / thisNorm;
        end
    end
end

[M]

@Image Analyst , Actually this is just what i am trying to do : A = A./vecnorm(A,2)

Thanks