iterating values of a vector under conditions

조회 수: 3 (최근 30일)
FATEMEH
FATEMEH 2012년 9월 12일
I have a for loop to make the elements of vector v. I want to do iteration i+1 only for the elements with zero value in iteration i. (meaning that if I get v=[0,1,0,1] with i=1, I want to do iteration i=2 only for the 1st and 3rd elements.I need to break the loop whenever all the elements of v are filled with non zero values.
v=[0;0;0;0] for i=1:maxit
v=[process1(i)...;process2(i)...;process3(i)...;process4(i)...]
end

채택된 답변

Matt Tearle
Matt Tearle 2012년 9월 12일
편집: Matt Tearle 2012년 9월 12일
If I understand you correctly, you have a procedure you want to iteratively apply to a vector, but only to the elements of that vector that are 0.
Here's some code that applies the "3n+1" procedure to the elements of a vector that are greater than 1:
% starting vector
v = 1:10;
% iterate
for k = 1:8
% exclusion criterion
idx = v>1;
% extract just the elements that satisfy the criterion
w = v(idx);
% and do something to them
isodd = (mod(w,2)==1);
w(~isodd) = 0.5*w(~isodd);
w(isodd) = 3*w(isodd) + 1;
% update just the altered values
v(idx) = w
end
Obviously, you would change the test to idx = v==0 and the "do something" section would be whatever you want.

추가 답변 (2개)

Matt Fig
Matt Fig 2012년 9월 12일
편집: Matt Fig 2012년 9월 12일
I am not quite sure what you want, but perhaps you should look at the CONTINUE keyword. If this isn't what you need, you should come up with an example input and an expected output...
  댓글 수: 3
Matt Fig
Matt Fig 2012년 9월 12일
편집: Matt Fig 2012년 9월 12일
What are you trying to achieve by this in general? This is silly to do in a loop, because you know what values of the index will make v give a zero.
v = [0;0;0;0];
ii = 1;
while all(~v)
v = [ii-1;ii-2;ii-3;ii-4];
end
FATEMEH
FATEMEH 2012년 9월 12일
The thing is that I don't want to update non zero elements.
This is a simple example. I need the loop actually. In the real code for each i, I read a separate file to get the v values.

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Azzi Abdelmalek
Azzi Abdelmalek 2012년 9월 12일
편집: Azzi Abdelmalek 2012년 9월 12일
v=[0,1,0,1]
idx=find(v==0)
p=process(idx)

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