Error with Min-Max Scaling

The code below fails due to line 5, Just trying to make simple min max scaling code in range of -1 and 1

t=[ 1 5 6; 8 9 7; 2 4 5];

for i= 1:length(t)

Scale1=(t(:,i)-min(t(:,i)))/(max(t(:,i))-min(t(:,i)));

Scalef(i)=2*Scale1 -1

end

Scaled_data=Scalef

Answer 1

[Tayyab Khalil]

First thing, kindly post your code inside a code block to make it easier to read.

I got your code working by simply assigning the scaled value to a column of scalef, the error was that you were trying to assign it as if it were a single value but it is not.

t=[ 1 5 6; 8 9 7; 2 4 5];
for i= 1:length(t)
   
   Scale1=(t(:,i)-min(t(:,i)))/(max(t(:,i))-min(t(:,i)));
   Scalef(:,i)=2*Scale1 -1
end
Scaled_data=Scalef

Which gives the following output:

Scaled_data =

   -1.0000   -0.6000         0
    1.0000    1.0000    1.0000
   -0.7143   -1.0000   -1.0000

I don't know if this is what you're looking for, seems to be scaling each column between -1 and 1.

[Kyle Koutonen]

Your code did do exactly what i wanted, and yes I want to scale each row of data. The only porblem is when i implement it into matlab app.designer it no longer works. The code doesn't change, just t changes to app.t since that where the saved data is yet the Scale1 line brings up this error "Index in position 2 exceeds array bounds (must not exceed 2)" what would I need to change

for i= 1:length(app.t )

Scale1=(app.t(:,i)-min(app.t(:,i)))/(max(app.t(:,i))-min(app.t(:,i )));

Scalef(:,i)=2*Scale1 -1

%t2(:,i) = normalize(t(:,i), 'range', [-1 1 ]) ;

end

Scaled_data=Scalef

Answer 2

[Steven Lord]

t = [ 1 5 6; 8 9 7; 2 4 5]

t = 3×3

1 5 6 8 9 7 2 4 5

t2 = normalize(t, 'range', [-1 1])

t2 = 3×3

-1.0000 -0.6000 0 1.0000 1.0000 1.0000 -0.7143 -1.0000 -1.0000

You can also specify a dimension input.

t3 = normalize(t, 2, 'range', [-1 1])

t3 = 3×3

-1.0000 0.6000 1.0000 0 1.0000 -1.0000 -1.0000 0.3333 1.0000

[Kyle Koutonen]

Your code did do exactly what i wanted, and yes I want to scale each row of data. The only porblem is when i implement it into matlab app.designer it no longer works. The code doesn't change, just t changes to app.t since that where the saved data is yet the t2 line brings up this error "Index in position 2 exceeds array bounds (must not exceed 2)" what would I need to change

for i= 1:length(app.t )

t2(:,i) = normalize(app.t(:,i), 'range', [-1 1 ]) ;

end

Scaled_data=t2

app.UITable5.ColumnName= app.s;

app.UITable5.Data=Scaled_data;

[Steven Lord]

Don't use length on a non-vector especially when you want to iterate over the columns of an array.

M = zeros(5, 4);
L = length(M)

L = 5

In this case L is the number of rows in M rather than the number of columns.

numRows = size(M, 1)

numRows = 5

numCols = size(M, 2)

numCols = 4

% If using a sufficiently recent MATLAB
numRows2 = height(M)

numRows2 = 5

numCols2 = width(M) 

numCols2 = 4

In this other case, where M2 has more columns than rows, L2 is the number of columns in M2. But size, height, and width will always return the size in a specific dimension regardless of whether the array is a vector, a tall matrix, a wide matrix, or an N-dimensional array.

M2 = zeros(3, 17);
L2 = length(M2)

L2 = 17