# How to standardize an array so that the maximum value is 1 and minimum is -1 keeping the zero value as zero?

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Abhishek Chakraborty
on 8 Dec 2022

Edited: Bruno Luong
on 9 Dec 2022

I wanted to standardize an array so that the maximum value of the array takes the value "1" and the minimum value takes the value "-1" keeping the original value "0" in the array as "0" in the standardized array?

For example,

A=[-1 -2 -3 -4 0 1];

StdA=(A-min(A(:)))/(max(A(:))-min(A(:)))

StdA=2*StdA-1

But by doing this sort of standardization, I get:

StdA =

0.2000 -0.2000 -0.6000 -1.0000 0.6000 1.0000

However, I would want to keep 0 as 0 i.e., StdA(:,5)=0 with the maximum and minimum values of the array taking +1 and -1 values.

How to do it? Kindly help me with the same.

##### 1 Comment

Bora Eryilmaz
on 8 Dec 2022

Edited: Bora Eryilmaz
on 8 Dec 2022

### Accepted Answer

Mathieu NOE
on 8 Dec 2022

hello

this can only be accomplished if you accept that the positive and negative parts of your signal are normalized by a different scale factor

A=[-1 -2 -3 -4 0 1];

B = A;

% normalisation of the positive values

id1 = A>=0;

A1 = A(id1)

A1 = A1./max(A1);

B(id1) = A1;

% normalisation of the negative values

id2 = A<0;

A2 = A(id2)

A2 = A2./max(-A2);

B(id2) = A2;

plot(A,'-d'); hold on

plot(B,'--');

### More Answers (5)

Image Analyst
on 9 Dec 2022

Try this:

A = randi(10, 1, 18) - 5

A(A<0) = -A(A<0) / min(A(A<0)) % FIrst rescale negative numbers to -1 to 0

A(A>0) = A(A>0) / max(A(A>0)) % Next rescale positive numbers to 0 to 1

Stephen23
on 9 Dec 2022

Edited: Stephen23
on 9 Dec 2022

In just one simple step, assuming that the min<0 and max>0:

A = [-1,-2,-3,-4,0,1,2];

B = interp1([min(A),0,max(A)],[-1,0,1],A)

Or in case the values do not cross zero, the addition of ABS():

A = [-1,-2,-3,-4];

B = interp1([-abs(min(A)),0,abs(max(A))],[-1,0,1],A)

The cases min==0 or max==0 must be handled separately. Note that all of the alorithms have similar limitations.

##### 1 Comment

Stephen23
on 9 Dec 2022

Edited: Stephen23
on 9 Dec 2022

Note how this method can be easily modified to efficiently scale any number of ranges to any continuous scales.

For example, the classic "0 to 1":

A = [-1,-2,-3,-4,0,1,2];

B = interp1([min(A),max(A)],[0,1],A)

or slightly more esoteric "e to pi":

A = [-1,-2,-3,-4,0,1,2];

B = interp1([min(A),max(A)],[exp(1),pi],A)

etc. etc.

DGM
on 9 Dec 2022

Edited: DGM
on 9 Dec 2022

I'm not into statistics, so I have no idea if this has merit. I'm occasionally after preserving linearity and the center (zero) moreso than constraining zero and both extrema. If so,

% input vector

A = -10:5

% normalize with respect to zero and furthest extrema

B = A/max(abs([min(A(:)) max(A(:))]));

plot(A,B);

This is a way to normalize a nominally zero-centered signal (e.g. audio) without distorting it or adding DC bias.

Note that this still works if the data does not cross zero.

##### 2 Comments

DGM
on 9 Dec 2022

Edited: DGM
on 9 Dec 2022

Correct. As I mentioned, I am questioning whether the requested constraints are appropriate. Instead of constraining zero and both extrema, I'm constraining zero and one extrema so as to maintain linearity.

As for my oversight, depending on the expected behavior, the all-zero case can be handled easily enough. I'm sure there are other potential problems; I wasn't out to write something robust so much as to suggest an idea.

% normalize with respect to zero and furthest extrema

nf = max(abs([min(A(:)) max(A(:))]));

if nf == 0 % input is all zeros

B = A;

else

B = A/nf;

end

Bruno Luong
on 9 Dec 2022

A "smooth" mapping

% Data

a=randi([-10,10],1,5)

pp=pchip([min(a) 0 max(a)], -1:1);

normfun=@(a) ppval(pp,a)

an=normfun(a)

% Check the mapping curve

densesample = linspace(min(a),max(a));

plot(densesample,normfun(densesample))

##### 0 Comments

Matt J
on 8 Dec 2022

Edited: Matt J
on 8 Dec 2022

A=[-1 -2 -3 -4 0 1]

I=logical(A);

StdA=A;

StdA(I) = rescale(A(I),-1,1)

##### 8 Comments

Bruno Luong
on 9 Dec 2022

Edited: Bruno Luong
on 9 Dec 2022

I remove my first code with polyfit since I think it sometime give non-monotonic map, which to my mind is not "standardize" whatever that means. It but "works" if 0 falls outside the data.

I'll give IMO opinion better solution since it is unlikely to give a non-monotonic mapping.

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