# Could anyone help me how to get the result as desired in the following code.

1 vue (au cours des 30 derniers jours)
jaah navi le 3 Nov 2021
Commenté : jaah navi le 3 Nov 2021
A=1:20;
while ~isempty(A)
B=reshape(A,[],2);
B(:,end)=B(end:-1:1,end);
for k=1:size(B,1)
C=B(k,:)
A=[];
end
end
The code executes and gives the following result
C = 1 20
C = 2 19
C = 3 18
C = 4 17
C = 5 16
C = 6 15
C = 7 14
C = 8 13
C = 9 12
C = 10 11
But I want to have the result in the following manner
C = 1 20 4 17 7 14 10 11
C = 2 19 5 16 8 13
C = 3 18 6 15 9 12
##### 3 commentairesAfficher 1 commentaire plus ancienMasquer 1 commentaire plus ancien
jaah navi le 3 Nov 2021
With respect to the above code I am having 20 elements. The reason I am saving it to C every time is that, I need to compare the performance metrices based on the elements
C = 1 20 4 17 7 14 10 11
C = 2 19 5 16 8 13
C = 3 18 6 15 9 12
that have been stored in C.
If you need those specific values to run something else, then there is not much point in coding the order. Just store as cells that you can use
C={[1 20 4 17 7 14 10 11],[2 19 5 16 8 13 ],[3 18 6 15 9 12]}
And then use them in your code like
for k=1:3
C{k}
% ...
%...
end

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### Réponse acceptée

Mathieu NOE le 3 Nov 2021
hello
after so guess work, I finally change the code to this
clc
A=1:20;
B=reshape(A,[],2);
B(:,end)=B(end:-1:1,end);
for ci = 1:3
C = [];
ind1 = ci:3:size(B,1);
for k=1:length(ind1)
ind2 = ci+(k-1)*3;
C = [C B(ind2,:)];
end
disp(C)
end
and this is the result , as expected :
1 20 4 17 7 14 10 11
2 19 5 16 8 13
3 18 6 15 9 12
maybe those 3 lines should be stored in 3 cells ?
##### 1 commentaireAfficher -1 commentaires plus anciensMasquer -1 commentaires plus anciens
jaah navi le 3 Nov 2021

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