I'm trying to write 5 for loop cycles with a step
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riki
el 19 de Mayo de 2023
Respondida: Dyuman Joshi
el 19 de Mayo de 2023
Hello, I would like to know how could I write 5 loop cycle with a step that resolve the following problem. I have 3 vectors p1,p2,p3 that increments, from 0 to 1, with a step of 0.25. Also, I have 2 parameters (b, errore) that increments with a different step b = [0:0.4:2] ed errore = [0.4:0.3:1.5]. The sum of p1,p2,p3 must be 1 in order to get b and err parameters
I would like to have some different combination of the 5 parameters (p1,p2,p3,b,errore) so that:
- p1 + p2 + p3 = 1 --> b = 0 , errore = 0.4 with p1=0.25, p2=0.5, p3=0.25 (it's just an example) --> then I would like to save these 5 parameters in the empty arrays.
- p1 + p2 + p3 = 1 --> b = 0 , errore = 0.7 with p1=0.25, p2=0.5, p3=0.25
And so on for every step increment of every combination of p1,p2,p3 possible. In this case, I would get a matrix like this:
How can I do it in a smart way? I'm pretty new to MATLAB so I'm still learning
p1= [0 0 0 0];
p2= [0 0 0 0];
p3= [0 0 0 0];
b = [0 0 0 0 0 0];
errore = [0 0 0 0];
A=zeros(25,25)
for i=1:4
p1 = p1 + 0.25;
for j=1:4
p2 = p2 + 0.25;
for k=1:4
p3 = p3 + 0.25;
if p1+p2+p3 == 1
for l=1:6
b(l)=b(l)+0.4;
x(:,l)=b(l);
elseif p1(i)+p2(i)+p3(i) > 1
warning('sum exceeding 1')
else
p3 = p3 + 0.25;
end
end
end
end
2 comentarios
Dyuman Joshi
el 19 de Mayo de 2023
Do you want to save all the values of b and errore for every (p1, p2, p3) whose sum is equal to 1?
%p1 p2 p3 b errore
[0.25 0.25 0.5 0 0.4;
0.25 0.25 0.5 0 0.7;
0.25 0.25 0.5 0 1; ...
...
0.25 0.25 0.5 0.4 0.4
0.25 0.25 0.5 0.4 0.7;
0.25 0.25 0.5 0.4 1; ...
%and so on, like this
Respuesta aceptada
Andres
el 19 de Mayo de 2023
Editada: Andres
el 19 de Mayo de 2023
There are many different ways to generate the desired matrix A. You may loop over two variables only and calculate the remaining p3, and generate the combinations of b and errore (that are always the same) beforehand.
% parameters to construct p<n>
pMin = 0;
pInc = 0.25;
pMax = 1;
pSum = 1; % allowed sum of p<n>
% the other two parameters
b = 0:0.4:2;
errore = 0.4:0.3:1.5;
% number of elements of each p<n>
pNum = (pMax-pMin)/pInc+1;
% number of elements of b and errore
bNum = numel(b);
eNum = numel(errore);
% number of valid p1, p2, p3 combinations (p3 = pSum-p1-p2)
pNumCom = (pNum)*(pNum-1)/2;
% number of b, errore combinations
beNumCombP = bNum*eNum;
% combine all b elements with all errore elements separately before
bE = [reshape(repmat(b, [eNum, 1]), [beNumCombP, 1]), ...
repmat(errore(:), [bNum, 1])];
% initialize A and its row index k
A = zeros(pNumCom*beNumCombP,5);
k = 0;
% loop only over p1 and p2 and calculate p3
for p1 = pMin:pInc:pMax
for p2 = pMin:pInc:(pSum-p1)
p3 = pSum-p1-p2;
P = repmat([p1, p2, p3], [beNumCombP, 1]);
A(k+1 : k+beNumCombP, :) = [P, bE];
k = k + beNumCombP;
end
end
% test:
all(sum(A(:, 1:3), 2) == pSum)
% logical
%
% 1
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Dyuman Joshi
el 19 de Mayo de 2023
Vectorization ftw!
%Define variables
b = [0:0.4:2];
errore = [0.4:0.3:1.5];
%To obtain the combination according to the condition
%define p1, p2, p3 as grids
[p1, p2, p3] = meshgrid(0:0.25:1);
%Indices to for which the sum is equal to 1
idx = find(p1+p2+p3==1);
%Corresponding combinations
[E, B, IDX] = ndgrid(errore, b, idx);
IDX = IDX(:);
%Output
out = [p1(IDX) p2(IDX) p3(IDX) B(:) E(:)];
disp(out)
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