How to normalize the values of matrix to make it greater than zero and less than one?

hello,

ho to normalize the values of a matrix to make it greater than zero and less than one?

regards,

[madhan ravi]

See my answer below

[Sarah A]

yes I am outside now, I need to try this on my laptop. thanks for answering I will accept the answer if it works.

Answer 1

[Matt J]

 B=(A-min(A(:)))/(max(A(:))-min(A(:))) ;

[Sarah A]

no this gives some values equal to one. I want all the values be: 0<x<1

[Matt J]

 delta=0.01;
 B=(A-min(A(:)))/(max(A(:))-min(A(:))) ;

 B=B*(1-2*delta)+delta;

[Sarah A]

It works! thanks a lot :)

[Walter Roberson]

 delta = max(eps(A(:));
 B=(A-min(A(:))+delta)/(max(A(:))-min(A(:))+delta) ;

Answer 2

[madhan ravi]

Try :

Np = normalize(A,'norm',1) % A your matrix

[Sarah A]

This error occur:

Undefined function 'normalize' for input arguments of type 'double'.

when I searched about "normalized" function I got this:

function ydata = normalize(this,ydata,Xlims,varargin)

%NORMALIZE  Scales Y data to normalized time data range.

%

%  Input arguments:

%    * YDATA is the Y data to be normalized

%    * XLIMS are the X limits for the axes of interest

%    * The last argument(s) is either an absolute index or a pair

%      of row/column indices specifying the axes location in the 

%      axes grid.

%  Author(s): P. Gahinet



[ns,ny,nu] = size(this.Amplitude);

if ny>0

     [ymin,ymax,FlatY] = ydataspan(this.Time,this.Amplitude(:,varargin{:}),Xlims);

     ydata = (ydata - (ymin+ymax)/2)/((ymax-ymin)/2+FlatY);

end

Copyright 1986-2004 The MathWorks, Inc.

Answer 3

[Matt J]

 delta=0.01;
 B = rescale(A,delta,1-delta);

[Sarah A]

The same problem I got this error:

Undefined function 'rescale' for input arguments of type 'double'.

Answer 4

[Matt J]

A(:)=0.5;

[Sarah A]

this going to replace all the values in the matrix A by 0.5

[Matt J]

Yep. Therefore all elements will be >0 and <1 as required.

[Sarah A]

no I want to normalize here where the output of normalization is greater than zero and less than one.

[Matt J]

But hopefully you can see now how vague that is. There are a million answers.