How can I solve if condition in for loop matrix?
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Dina Maulina
am 30 Mai 2019
Kommentiert: Dina Maulina
am 28 Jun. 2019
I want to make an iteration that when the component of a matrix A has less value than the component of matrix B, it will continue the process to the next iteration.
A =
[ 2 5 9 12
6 4 13 1
3 2 19 5]
B=
[ 5
5
5]
in the first column, the condition is not met so it cannot proceed to the second column. and in the second column, conditions have been met so that it can proceed to the third column.
I tried in for loop but, the A values always stuck in first column even condition is right.
Thank you.
3 Kommentare
Guillaume
am 30 Mai 2019
then the loop
What loop? Remember that we have no idea what you're doing. If you don't show us your code, how can we tell us how to fix it?
Most likely, whatever loop you're using is not even necessary.
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Murugan C
am 30 Mai 2019
Bearbeitet: Guillaume
am 30 Mai 2019
Hi
use below code for your query.
Here, B will check all rows, each column in A is less than, then it will allow to next iteration.
A = [ 2 5 9 12; 6 4 13 1; 3 2 19 5]
B = [5; 5 ;5];
[row,col] = size(A);
for ii = 1 : col
% fprintf('Current Column is : %d\n', ii);
C1 = A(:,ii) < B;
if all(C1,1)
fprintf('Current Column is : %d, Next Column is : %d\n', ii, ii+1);
continue;
else
fprintf('Stopped at %d Column because unsatisfied B input \n', ii);
break;
end
end
3 Kommentare
Murugan C
am 30 Mai 2019
Thanks Guillaume.
I knew, continue will not take any action in 'for' loop. Any how I will try avoid it..
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