I'm trying to write 5 for loop cycles with a step

19 May 2023
2 Answers
Answer Accepted
Updated 19 May 2023
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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

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Dyuman Joshi
Dyuman Joshi on 19 May 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
riki
riki on 19 May 2023
Yes exactly and for all the possible combinations of p1,p2,p3 whose sum is 1

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 Accepted Answer

Andres
Andres on 19 May 2023
Edited: Andres on 19 May 2023

0 votes

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

More Answers (1)

Dyuman Joshi
Dyuman Joshi on 19 May 2023
Ran in:

0 votes

Ran in:
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)
0 1.0000 0 0 0.4000 0 1.0000 0 0 0.7000 0 1.0000 0 0 1.0000 0 1.0000 0 0 1.3000 0 1.0000 0 0.4000 0.4000 0 1.0000 0 0.4000 0.7000 0 1.0000 0 0.4000 1.0000 0 1.0000 0 0.4000 1.3000 0 1.0000 0 0.8000 0.4000 0 1.0000 0 0.8000 0.7000 0 1.0000 0 0.8000 1.0000 0 1.0000 0 0.8000 1.3000 0 1.0000 0 1.2000 0.4000 0 1.0000 0 1.2000 0.7000 0 1.0000 0 1.2000 1.0000 0 1.0000 0 1.2000 1.3000 0 1.0000 0 1.6000 0.4000 0 1.0000 0 1.6000 0.7000 0 1.0000 0 1.6000 1.0000 0 1.0000 0 1.6000 1.3000 0 1.0000 0 2.0000 0.4000 0 1.0000 0 2.0000 0.7000 0 1.0000 0 2.0000 1.0000 0 1.0000 0 2.0000 1.3000 0.2500 0.7500 0 0 0.4000 0.2500 0.7500 0 0 0.7000 0.2500 0.7500 0 0 1.0000 0.2500 0.7500 0 0 1.3000 0.2500 0.7500 0 0.4000 0.4000 0.2500 0.7500 0 0.4000 0.7000 0.2500 0.7500 0 0.4000 1.0000 0.2500 0.7500 0 0.4000 1.3000 0.2500 0.7500 0 0.8000 0.4000 0.2500 0.7500 0 0.8000 0.7000 0.2500 0.7500 0 0.8000 1.0000 0.2500 0.7500 0 0.8000 1.3000 0.2500 0.7500 0 1.2000 0.4000 0.2500 0.7500 0 1.2000 0.7000 0.2500 0.7500 0 1.2000 1.0000 0.2500 0.7500 0 1.2000 1.3000 0.2500 0.7500 0 1.6000 0.4000 0.2500 0.7500 0 1.6000 0.7000 0.2500 0.7500 0 1.6000 1.0000 0.2500 0.7500 0 1.6000 1.3000 0.2500 0.7500 0 2.0000 0.4000 0.2500 0.7500 0 2.0000 0.7000 0.2500 0.7500 0 2.0000 1.0000 0.2500 0.7500 0 2.0000 1.3000 0.5000 0.5000 0 0 0.4000 0.5000 0.5000 0 0 0.7000 0.5000 0.5000 0 0 1.0000 0.5000 0.5000 0 0 1.3000 0.5000 0.5000 0 0.4000 0.4000 0.5000 0.5000 0 0.4000 0.7000 0.5000 0.5000 0 0.4000 1.0000 0.5000 0.5000 0 0.4000 1.3000 0.5000 0.5000 0 0.8000 0.4000 0.5000 0.5000 0 0.8000 0.7000 0.5000 0.5000 0 0.8000 1.0000 0.5000 0.5000 0 0.8000 1.3000 0.5000 0.5000 0 1.2000 0.4000 0.5000 0.5000 0 1.2000 0.7000 0.5000 0.5000 0 1.2000 1.0000 0.5000 0.5000 0 1.2000 1.3000 0.5000 0.5000 0 1.6000 0.4000 0.5000 0.5000 0 1.6000 0.7000 0.5000 0.5000 0 1.6000 1.0000 0.5000 0.5000 0 1.6000 1.3000 0.5000 0.5000 0 2.0000 0.4000 0.5000 0.5000 0 2.0000 0.7000 0.5000 0.5000 0 2.0000 1.0000 0.5000 0.5000 0 2.0000 1.3000 0.7500 0.2500 0 0 0.4000 0.7500 0.2500 0 0 0.7000 0.7500 0.2500 0 0 1.0000 0.7500 0.2500 0 0 1.3000 0.7500 0.2500 0 0.4000 0.4000 0.7500 0.2500 0 0.4000 0.7000 0.7500 0.2500 0 0.4000 1.0000 0.7500 0.2500 0 0.4000 1.3000 0.7500 0.2500 0 0.8000 0.4000 0.7500 0.2500 0 0.8000 0.7000 0.7500 0.2500 0 0.8000 1.0000 0.7500 0.2500 0 0.8000 1.3000 0.7500 0.2500 0 1.2000 0.4000 0.7500 0.2500 0 1.2000 0.7000 0.7500 0.2500 0 1.2000 1.0000 0.7500 0.2500 0 1.2000 1.3000 0.7500 0.2500 0 1.6000 0.4000 0.7500 0.2500 0 1.6000 0.7000 0.7500 0.2500 0 1.6000 1.0000 0.7500 0.2500 0 1.6000 1.3000 0.7500 0.2500 0 2.0000 0.4000 0.7500 0.2500 0 2.0000 0.7000 0.7500 0.2500 0 2.0000 1.0000 0.7500 0.2500 0 2.0000 1.3000 1.0000 0 0 0 0.4000 1.0000 0 0 0 0.7000 1.0000 0 0 0 1.0000 1.0000 0 0 0 1.3000 1.0000 0 0 0.4000 0.4000 1.0000 0 0 0.4000 0.7000 1.0000 0 0 0.4000 1.0000 1.0000 0 0 0.4000 1.3000 1.0000 0 0 0.8000 0.4000 1.0000 0 0 0.8000 0.7000 1.0000 0 0 0.8000 1.0000 1.0000 0 0 0.8000 1.3000 1.0000 0 0 1.2000 0.4000 1.0000 0 0 1.2000 0.7000 1.0000 0 0 1.2000 1.0000 1.0000 0 0 1.2000 1.3000 1.0000 0 0 1.6000 0.4000 1.0000 0 0 1.6000 0.7000 1.0000 0 0 1.6000 1.0000 1.0000 0 0 1.6000 1.3000 1.0000 0 0 2.0000 0.4000 1.0000 0 0 2.0000 0.7000 1.0000 0 0 2.0000 1.0000 1.0000 0 0 2.0000 1.3000 0 0.7500 0.2500 0 0.4000 0 0.7500 0.2500 0 0.7000 0 0.7500 0.2500 0 1.0000 0 0.7500 0.2500 0 1.3000 0 0.7500 0.2500 0.4000 0.4000 0 0.7500 0.2500 0.4000 0.7000 0 0.7500 0.2500 0.4000 1.0000 0 0.7500 0.2500 0.4000 1.3000 0 0.7500 0.2500 0.8000 0.4000 0 0.7500 0.2500 0.8000 0.7000 0 0.7500 0.2500 0.8000 1.0000 0 0.7500 0.2500 0.8000 1.3000 0 0.7500 0.2500 1.2000 0.4000 0 0.7500 0.2500 1.2000 0.7000 0 0.7500 0.2500 1.2000 1.0000 0 0.7500 0.2500 1.2000 1.3000 0 0.7500 0.2500 1.6000 0.4000 0 0.7500 0.2500 1.6000 0.7000 0 0.7500 0.2500 1.6000 1.0000 0 0.7500 0.2500 1.6000 1.3000 0 0.7500 0.2500 2.0000 0.4000 0 0.7500 0.2500 2.0000 0.7000 0 0.7500 0.2500 2.0000 1.0000 0 0.7500 0.2500 2.0000 1.3000 0.2500 0.5000 0.2500 0 0.4000 0.2500 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Asked:

riki
riki
on 19 May 2023

Answered:

Dyuman Joshi
Dyuman Joshi
on 19 May 2023

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