Saving all output of for loop
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for i=1:1500
X1=J(i)-sqrt((D(i)^2)/((sqrt(A(i))+1)))
Y1=I(i)-A(i)*(J(i)-X1)
X2=J(i)+sqrt((D(i)^2)/((sqrt(A(i))+1)))
Y2=I(i)+A(i)*(X2-J(i))
end
coord=[X1(:),Y1(:),X2(:),Y2(:)]
%coord
writematrix(coord,'Coord_15.csv')
I am trying to save the coordinates (X1, X2, Y1,Y2) from a for loop as shown in the code above. All inputs (I, J, A, D) are columns vectors , but i only the last iteration is saved. I will appreciate any help.
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채택된 답변
Sulaymon Eshkabilov
2019년 5월 12일
Hi,
Simply add (i) after your output variables, viz. X1, Y1, X2, Y2:
for i=1:1500
X1(i)=J(i)-sqrt((D(i)^2)/((sqrt(A(i))+1)))
Y1(i)=I(i)-A(i)*(J(i)-X1)
X2(i)=J(i)+sqrt((D(i)^2)/((sqrt(A(i))+1)))
Y2(i)=I(i)+A(i)*(X2-J(i))
end
coord=[X1(:),Y1(:),X2(:),Y2(:)]
%coord
writematrix(coord,'Coord_15.csv')
Good luck
댓글 수: 4
madhan ravi
2019년 5월 16일
편집: madhan ravi
2019년 5월 16일
Then why did you accept the answer if it doesn’t solve the problem??
Sulaymon Eshkabilov
2019년 5월 16일
Before commenting any point verify what you have stated. That is the correct answer. Vectorization is another solution.
추가 답변 (1개)
madhan ravi
2019년 5월 12일
편집: madhan ravi
2019년 5월 16일
You don't need a loop , this is straight forward just vectorize your code:
Note: The other answer doesn‘t show the importance of preallocation.
X1=J-sqrt((D.^2)./((sqrt(A)+1)));
Y1=I-A.*(J-X1);
X2=J+sqrt((D.^2)./((sqrt(A)+1)));
Y2=I+A.*(X2-J);
coord=[X1(:),Y1(:),X2(:),Y2(:)];
writematrix(coord,'Coord_15.csv')
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