How to change all elements of a vector V to achieve the condition sum(V) equal with a upper and lower boundaries?

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Rahim Rahim
Rahim Rahim 2022-12-29
评论： DGM 2022-12-29
I have a vector V, for example, V=[0.1 0.002 0.5 0.2 0.1 0.003 0.4].
The boundaries of all elements of the matrix V should always be between 0.01 and 0.8.
I want to create a function to change the elements of the vector V where the sum of V becomes equal to one.
I am looking to create a function in Matlab V= Editor(V, lp, up );
  • where lp: is the lower boundary, in my example 0.01
  • and up: is the upper boundary, in my example 0.8
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Rahim Rahim
Rahim Rahim 2022-12-29
I want to change all the variable of the vector !
@Rik I know that, but I want to change the variable of V and the V after the modification shold alwaus respects the up and lower bounderies

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采纳的回答

DGM
DGM 2022-12-29
编辑：DGM 2022-12-29
Ran in:
Well with requirements like that, I guess this is fair game.
% input
V = [0.1 0.002 0.5 0.2 0.1 0.003 0.4];
% parameters
limits = [0.01 0.8];
outsum = 1;
% constraints cannot be met generally
if outsum/numel(V) < limits(1)
error('too many elements')
elseif outsum/numel(V) > limits(2)
error('too few elements')
end
% i'm going to be super lazy
sumv = sum(V);
while abs(sumv-outsum) > 1E-9 % or some arbitrary tolerance
scale = outsum/sumv;
V = min(max(V*scale,limits(1)),limits(2));
sumv = sum(V);
end
V
V = 1×7
0.0754 0.0100 0.3769 0.1508 0.0754 0.0100 0.3015
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Rahim Rahim
Rahim Rahim 2022-12-29
@DGM Thank you so much
But I want everytime the function generat a different matrix ... ( many solution )
and thank you so much
DGM
DGM 2022-12-29
Nothing about the question suggested that the output was random. If the output is random, what's the purpose of V? Does the output depend on V? If so, how?

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更多回答（2 个）

Matt J
Matt J 2022-12-29
编辑：Matt J 2022-12-29
Ran in:
V=[0.1 0.002 0.5 0.2 0.1 0.003 0.4];
lp=0.01;
up=0.8;
n=numel(V);
V(:)=(1-n*lp)/n+lp
V = 1×7
0.1429 0.1429 0.1429 0.1429 0.1429 0.1429 0.1429
sum(V)
ans = 1.0000
V>=lp
ans = 1×7 logical array
1 1 1 1 1 1 1
V<=up
ans = 1×7 logical array
1 1 1 1 1 1 1
Walter Roberson
Walter Roberson 2022-12-29
if the requirement is that you apply a linear transform to the elements of V producing a new vector W such that sum(W) == 1 and the elements respect the upper and lower bounds, then the problem cannot generally be solved.
A simple proof is that V might consist of more than 100 elements. With the minimum being 0.01 then the sum would have to be more than 100*0.01 == 1
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Walter Roberson
Walter Roberson 2022-12-29
That said
syms a
w = (v-0.01)*a + 0.01;
sola = solve(sum(w)==1,a);
w = simplify(w, a, sola)
sola can be expressed analytically so it can be computed without the symbolic toolbox
Walter Roberson
Walter Roberson 2022-12-29
each entry w(k) is a*(v(k)-0.01)+0.01 which is a*v(k) - a*0.01 + 0.01
Let N = length(v). Then sum(w) is a*sum(v) - N*a*0.01 + N*0.01 = a*(sum(v) - N*0.01) + N*0.01. The unknown is a and the sum is 1 so a = (1-N*0.01) / (sum(v) - N*0.01)
Note that if the original values in v could be negative or less than 0.01 then the denominator could be 0 which would be a problem. If the denominator is negative because entries in v are less than 0.01 then I do not promise at the moment that the transformed values are within the required limits.

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