How to do the following nested for loop?

M
M
23 Juil 2023
2 Réponses
Réponse acceptée
Mise à jour 15 Août 2023
3 Vues (30 jours)

0 votes

How to do the following nested for loop?
if there is a vector P = [1 1 1 1 1 1]
and I want to change P in each loop as the following:
first outer loop
inner loop:
first it:
P = [0.1 1 1 1 1 1]
2nd it:
P = [0.2 1 1 1 1 1]
3rd
P = [0.3 1 1 1 1 1]
and so on until reach 0.9
2nd outer loop
P = [1 0.1 1 1 1 1]
2nd it:
P = [1 0.2 1 1 1 1]
3rd
P = [1 0.3 1 1 1 1]
and so on until reach 0.9
3nd outer loop
change the 3rd index as the above.. until reach the 6th outer loop

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 Réponse acceptée

Voss
Voss le 23 Juil 2023
Ran in:

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Ran in:
format shortg
P = [1 1 1 1 1 1];
P_orig = P;
for ii = 1:numel(P)
fprintf('outer loop iteration %d\n',ii);
P = P_orig;
for jj = 1:9
fprintf('inner loop iteration %d\n',jj);
P(ii) = jj/10;
disp(P)
end
end
outer loop iteration 1
inner loop iteration 1
0.1 1 1 1 1 1
inner loop iteration 2
0.2 1 1 1 1 1
inner loop iteration 3
0.3 1 1 1 1 1
inner loop iteration 4
0.4 1 1 1 1 1
inner loop iteration 5
0.5 1 1 1 1 1
inner loop iteration 6
0.6 1 1 1 1 1
inner loop iteration 7
0.7 1 1 1 1 1
inner loop iteration 8
0.8 1 1 1 1 1
inner loop iteration 9
0.9 1 1 1 1 1
outer loop iteration 2
inner loop iteration 1
1 0.1 1 1 1 1
inner loop iteration 2
1 0.2 1 1 1 1
inner loop iteration 3
1 0.3 1 1 1 1
inner loop iteration 4
1 0.4 1 1 1 1
inner loop iteration 5
1 0.5 1 1 1 1
inner loop iteration 6
1 0.6 1 1 1 1
inner loop iteration 7
1 0.7 1 1 1 1
inner loop iteration 8
1 0.8 1 1 1 1
inner loop iteration 9
1 0.9 1 1 1 1
outer loop iteration 3
inner loop iteration 1
1 1 0.1 1 1 1
inner loop iteration 2
1 1 0.2 1 1 1
inner loop iteration 3
1 1 0.3 1 1 1
inner loop iteration 4
1 1 0.4 1 1 1
inner loop iteration 5
1 1 0.5 1 1 1
inner loop iteration 6
1 1 0.6 1 1 1
inner loop iteration 7
1 1 0.7 1 1 1
inner loop iteration 8
1 1 0.8 1 1 1
inner loop iteration 9
1 1 0.9 1 1 1
outer loop iteration 4
inner loop iteration 1
1 1 1 0.1 1 1
inner loop iteration 2
1 1 1 0.2 1 1
inner loop iteration 3
1 1 1 0.3 1 1
inner loop iteration 4
1 1 1 0.4 1 1
inner loop iteration 5
1 1 1 0.5 1 1
inner loop iteration 6
1 1 1 0.6 1 1
inner loop iteration 7
1 1 1 0.7 1 1
inner loop iteration 8
1 1 1 0.8 1 1
inner loop iteration 9
1 1 1 0.9 1 1
outer loop iteration 5
inner loop iteration 1
1 1 1 1 0.1 1
inner loop iteration 2
1 1 1 1 0.2 1
inner loop iteration 3
1 1 1 1 0.3 1
inner loop iteration 4
1 1 1 1 0.4 1
inner loop iteration 5
1 1 1 1 0.5 1
inner loop iteration 6
1 1 1 1 0.6 1
inner loop iteration 7
1 1 1 1 0.7 1
inner loop iteration 8
1 1 1 1 0.8 1
inner loop iteration 9
1 1 1 1 0.9 1
outer loop iteration 6
inner loop iteration 1
1 1 1 1 1 0.1
inner loop iteration 2
1 1 1 1 1 0.2
inner loop iteration 3
1 1 1 1 1 0.3
inner loop iteration 4
1 1 1 1 1 0.4
inner loop iteration 5
1 1 1 1 1 0.5
inner loop iteration 6
1 1 1 1 1 0.6
inner loop iteration 7
1 1 1 1 1 0.7
inner loop iteration 8
1 1 1 1 1 0.8
inner loop iteration 9
1 1 1 1 1 0.9

Plus de réponses (1)

Davide Masiello
Davide Masiello le 23 Juil 2023
Déplacé(e) : Star Strider le 23 Juil 2023
Ran in:

0 votes

Ran in:
Must you do this with a nested loop?
Example
P0 = [(0.1:0.1:0.9)',ones(9,5)];
P1 = P0;
for i = 1:5
P1 = [P1;circshift(P0,i,2)];
end
disp(P1)
0.1000 1.0000 1.0000 1.0000 1.0000 1.0000 0.2000 1.0000 1.0000 1.0000 1.0000 1.0000 0.3000 1.0000 1.0000 1.0000 1.0000 1.0000 0.4000 1.0000 1.0000 1.0000 1.0000 1.0000 0.5000 1.0000 1.0000 1.0000 1.0000 1.0000 0.6000 1.0000 1.0000 1.0000 1.0000 1.0000 0.7000 1.0000 1.0000 1.0000 1.0000 1.0000 0.8000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.1000 1.0000 1.0000 1.0000 1.0000 1.0000 0.2000 1.0000 1.0000 1.0000 1.0000 1.0000 0.3000 1.0000 1.0000 1.0000 1.0000 1.0000 0.4000 1.0000 1.0000 1.0000 1.0000 1.0000 0.5000 1.0000 1.0000 1.0000 1.0000 1.0000 0.6000 1.0000 1.0000 1.0000 1.0000 1.0000 0.7000 1.0000 1.0000 1.0000 1.0000 1.0000 0.8000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.1000 1.0000 1.0000 1.0000 1.0000 1.0000 0.2000 1.0000 1.0000 1.0000 1.0000 1.0000 0.3000 1.0000 1.0000 1.0000 1.0000 1.0000 0.4000 1.0000 1.0000 1.0000 1.0000 1.0000 0.5000 1.0000 1.0000 1.0000 1.0000 1.0000 0.6000 1.0000 1.0000 1.0000 1.0000 1.0000 0.7000 1.0000 1.0000 1.0000 1.0000 1.0000 0.8000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.1000 1.0000 1.0000 1.0000 1.0000 1.0000 0.2000 1.0000 1.0000 1.0000 1.0000 1.0000 0.3000 1.0000 1.0000 1.0000 1.0000 1.0000 0.4000 1.0000 1.0000 1.0000 1.0000 1.0000 0.5000 1.0000 1.0000 1.0000 1.0000 1.0000 0.6000 1.0000 1.0000 1.0000 1.0000 1.0000 0.7000 1.0000 1.0000 1.0000 1.0000 1.0000 0.8000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.1000 1.0000 1.0000 1.0000 1.0000 1.0000 0.2000 1.0000 1.0000 1.0000 1.0000 1.0000 0.3000 1.0000 1.0000 1.0000 1.0000 1.0000 0.4000 1.0000 1.0000 1.0000 1.0000 1.0000 0.5000 1.0000 1.0000 1.0000 1.0000 1.0000 0.6000 1.0000 1.0000 1.0000 1.0000 1.0000 0.7000 1.0000 1.0000 1.0000 1.0000 1.0000 0.8000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.1000 1.0000 1.0000 1.0000 1.0000 1.0000 0.2000 1.0000 1.0000 1.0000 1.0000 1.0000 0.3000 1.0000 1.0000 1.0000 1.0000 1.0000 0.4000 1.0000 1.0000 1.0000 1.0000 1.0000 0.5000 1.0000 1.0000 1.0000 1.0000 1.0000 0.6000 1.0000 1.0000 1.0000 1.0000 1.0000 0.7000 1.0000 1.0000 1.0000 1.0000 1.0000 0.8000 1.0000 1.0000 1.0000 1.0000 1.0000 0.9000
After you produced P1, at each iteration of your code you can just call the next row of the array P1

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Question posée :

M
M
le 23 Juil 2023

Commenté :

Voss
Voss
le 15 Août 2023

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