iterating values of a vector under conditions

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FATEMEH
FATEMEH 2012 年 9 月 12 日
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

採用された回答

Matt Tearle
Matt Tearle 2012 年 9 月 12 日
編集済み: Matt Tearle 2012 年 9 月 12 日
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.

その他の回答 (2 件)

Matt Fig
Matt Fig 2012 年 9 月 12 日
編集済み: Matt Fig 2012 年 9 月 12 日
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...
  3 件のコメント
Matt Fig
Matt Fig 2012 年 9 月 12 日
編集済み: Matt Fig 2012 年 9 月 12 日
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 2012 年 9 月 12 日
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 2012 年 9 月 12 日
編集済み: Azzi Abdelmalek 2012 年 9 月 12 日
v=[0,1,0,1]
idx=find(v==0)
p=process(idx)

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