# 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.
Bora Eryilmaz on 8 Dec 2022
Edited: Bora Eryilmaz on 8 Dec 2022
This cannot be done using a linear standardization function of the form y = a*x + b since you are trying to enforce 3 constraints, whereas a linear equation can only accommodate 2.

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,'--');
##### 2 CommentsShowHide 1 older comment
Mathieu NOE on 9 Dec 2022
My pleasure !

Image Analyst on 9 Dec 2022
Try this:
A = randi(10, 1, 18) - 5
A = 1×18
-2 3 2 5 -4 -1 0 0 -3 -2 5 3 2 -1 3 -3 -4 -4
A(A<0) = -A(A<0) / min(A(A<0)) % FIrst rescale negative numbers to -1 to 0
A = 1×18
-0.5000 3.0000 2.0000 5.0000 -1.0000 -0.2500 0 0 -0.7500 -0.5000 5.0000 3.0000 2.0000 -0.2500 3.0000 -0.7500 -1.0000 -1.0000
A(A>0) = A(A>0) / max(A(A>0)) % Next rescale positive numbers to 0 to 1
A = 1×18
-0.5000 0.6000 0.4000 1.0000 -1.0000 -0.2500 0 0 -0.7500 -0.5000 1.0000 0.6000 0.4000 -0.2500 0.6000 -0.7500 -1.0000 -1.0000
Abhishek Chakraborty on 9 Dec 2022
Thanks a lot.

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)
B = 1×7
-0.2500 -0.5000 -0.7500 -1.0000 0 0.5000 1.0000
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)
B = 1×4
-0.2500 -0.5000 -0.7500 -1.0000
The cases min==0 or max==0 must be handled separately. Note that all of the alorithms have similar limitations.
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)
B = 1×7
0.5000 0.3333 0.1667 0 0.6667 0.8333 1.0000
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)
B = 1×7
2.9299 2.8594 2.7888 2.7183 3.0005 3.0710 3.1416
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
A = 1×16
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 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 CommentsShowHide 1 older comment
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)
a = 1×5
2 -4 -5 2 -4
pp=pchip([min(a) 0 max(a)], -1:1);
normfun=@(a) ppval(pp,a)
normfun = function_handle with value:
@(a)ppval(pp,a)
an=normfun(a)
an = 1×5
1.0000 -0.9447 -1.0000 1.0000 -0.9447
% Check the mapping curve
densesample = linspace(min(a),max(a));
plot(densesample,normfun(densesample)) Matt J on 8 Dec 2022
Edited: Matt J on 8 Dec 2022
A=[-1 -2 -3 -4 0 1]
A = 1×6
-1 -2 -3 -4 0 1
I=logical(A);
StdA=A;
StdA(I) = rescale(A(I),-1,1)
StdA = 1×6
0.2000 -0.2000 -0.6000 -1.0000 0 1.0000
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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