How could normalize a matrix between 0 and 1.

I have a matrix 14x15536 how it shows in the picture, and i would like to normalize each row between 0 and 1.

How could I do it??

Thanks in advance.

Answer 1

[Stephan]

result = normalize(x,2,'range')

[Edu Gomez]

Thanks for you quick answer.

Always I try to use "normalize" I recive this error: Undefined function 'normalize' for input arguments of type 'double'.

Function "normalize" is for normalizating between -1 and 1, is it true??

How could I solve it??

Thanks again in advance.

[Stephan]

Which release do you use?

[Edu Gomez]

Matlab R2015a

[Stephan]

Then use this function:

function result = myNormalizer(x)
result = zeros(size(x));
for k = 1:size(x,1)
    result(k,:) = (x(k,:) - min(x(k,:))) ./ max(x(k,:) - min(x(k,:)));
end
end

[Edu Gomez]

Thank you so much Stephan, now the algorithm goes right. Only I have seen one thing more to implement, and the row 4 is always 2.5, so it is impossible to the algorithm identify min and max, and it shows "NaN". I don´t know if I have to give for this value 1 or 0.

I have thought this, what do you think about??

Thanks for you help and your time in advance. Im very pleased tu you, and sorry for my poor english, I need to improve it.

function result = Normalize(x)
%  result = zeros(size(x));
[varFil,~]=size(x);

for i=1:varFil
  if min(x(i,:))==max(x(i,:))
     x(i,:)=1; % I don´t know if it is correct.
  else
    result(i,:) = (x(i,:) - min(x(i,:))) ./ max(x(i,:) - min(x(i,:)));
  end
end
end

But I see this:

[Stephan]

I dont think that you can say it is correct to set this equal to one. It would also not be correct to set it to zero. You do a division of zero by zero - this leads to a NaN value. However, the inbuilt normalize function treats this with setting the values equal to zero.

[Adam]

You can do it in vectorised form without the need of a for loop as:

rowMin = min( x, [], 2 );
result = ( x - rowMin ) ./ ( max( x, [], 2 ) - rowMin );

You will still get the same divide by 0 problem though if you have a constant row because you could 'normalise' it to any value between 0 and 1 equally so you have to just majke a choice depending on your usage.

[Edu Gomez]

Thank you for your answer, i would like to try also this opction in order to learn more about Matlab.

So, I see this error with that:

function result = Normalizar(x)

rowMin = min( x, [], 2 );
result = ( x - rowMin ) ./ ( max( x, [], 2 ) - rowMin );
end

I have seen this error several times, and I dont´t know how interpretate this (-) with dimensions, I don´t know how Matlab works for doing: ( x - rowMin ).

And one more question, whats means " ./ " ????

Thanks in advance

[Adam]

./ is point-wise division rather than matrix division

[Stephen23]

"And one more question, whats means " ./ " ????"

https://www.mathworks.com/help/matlab/matlab_prog/array-vs-matrix-operations.html

[Jan]

Edu Gomez uses R2015a, so no auto-expanding, which was introduced in R2016b. Then bsxfun is required:

rowMin = min(x, [], 2);
result = bsxfun(@minus, x, rowMin) ./ bsxfun(@minus, max(x, [], 2), rowMin);

Answer 2

[Edu Gomez]

I want say thanks to both for you time and your help, Its very rewarding to have your help for this. Im doing a master thesis and I need a little help with Matlab sometimes.

Thank you very much :))