# 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 am 8 Dez. 2022
Bearbeitet: Bruno Luong am 9 Dez. 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.
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Bora Eryilmaz am 8 Dez. 2022
Bearbeitet: Bora Eryilmaz am 8 Dez. 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.

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### Akzeptierte Antwort

Mathieu NOE am 8 Dez. 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,'--');
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Abhishek Chakraborty am 9 Dez. 2022
Wow. This is such an interesting way to handle this problem. I also checked it for the 2-D case and it works perfectly:
A = [1.37 -3.299 0.1;5.33 0 7.1;-8.222 -0.09 -22];
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,'--');
A1 =
1.3700
5.3300
0
0.1000
7.1000
A2 =
-8.2220
-3.2990
-0.0900
-22.0000
>> B
B =
0.1930 -0.1500 0.0141
0.7507 0 1.0000
-0.3737 -0.0041 -1.0000
Thanks a lot.
Mathieu NOE am 9 Dez. 2022
My pleasure !

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### Weitere Antworten (5)

Image Analyst am 9 Dez. 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
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Abhishek Chakraborty am 9 Dez. 2022
Thanks a lot.

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Stephen23 am 9 Dez. 2022
Bearbeitet: Stephen23 am 9 Dez. 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.
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Stephen23 am 9 Dez. 2022
Bearbeitet: Stephen23 am 9 Dez. 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.

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DGM am 9 Dez. 2022
Bearbeitet: DGM am 9 Dez. 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.
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Stephen23 am 9 Dez. 2022
Bearbeitet: Stephen23 am 9 Dez. 2022
A = [-1,-2,-3,-4,0,1];
B = A/max(abs([min(A(:)),max(A(:))]))
B = 1×6
-0.2500 -0.5000 -0.7500 -1.0000 0 0.2500
DGM am 9 Dez. 2022
Bearbeitet: DGM am 9 Dez. 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

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Bruno Luong am 9 Dez. 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))
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Matt J am 8 Dez. 2022
Bearbeitet: Matt J am 8 Dez. 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
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DGM am 9 Dez. 2022
You're right. There isn't any "making 1/0 work", but those cases can be conditionally handled. Since normalize() and similar tools handle those cases by returning zeros, that would suffice for the edit that I made.
I don't know what's appropriate either. I'm here surrounded by a lot of people who know more than I do about what's meaningful in terms of statistics or signal processing. I just want to know if the answer that's needed isn't an answer to the original question (an XY problem). I suppose I'm also kind of throwing out an idea for future readers, regardless of OP's decision.
FWIW, I removed my original comment, since you've updated your answer and I guess Bruno deleted his some time before I even posted the comment.
Bruno Luong am 9 Dez. 2022
Bearbeitet: Bruno Luong am 9 Dez. 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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