FOR loop with IF condition alternative

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Agent Cooper
Agent Cooper 2014-8-5
回答： Agent Cooper 2014-8-5
I have the following code
I = [1:1:10; 10:10:100]'
a = size(I)
limit1 = 1
limit2 = 3
limit3 = 7
limit4 = 10
for j = 1:a(2)
X = I(:,j)
for k = 1:a(1)
if (k>limit1 && k<limit2) || (k>limit3 && k<limit4)
Y = 0
else Y = X(k)
end
B(k) = Y
end
C{j} = B'
end
that replaces the 2nd, the 8th and the 9th elements on each column with 0. For the example above, the [1 2 3 4 5 6 7 8 9 10; 10 20 30 40 50 60 70 80 90 100] matrix is replaced with the [1 0 3 4 5 6 7 0 0 10; 10 0 30 40 50 60 70 0 0 100] matrix.
Could anyone give me a hint on how to do this without using the FOR loop and the IF condition (because they are very slow when processing big amounts of data)?

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采纳的回答

Hikaru
Hikaru 2014-8-5
Try this
I = [1:1:10; 10:10:100]';
I(2,:) = 0;
I(8,:) = 0;
I(9,:) = 0;
  2 个评论
Agent Cooper
Agent Cooper 2014-8-5
Dear Hikaru,
Of course this is a solution to my simple example. But, please, consider the case in which you do not have only 3 indexes, but 10 or more and placed in different intervals (e.g. 84 to 98 and 758 to 767). I need something that works automatically, like the FOR loop.
Hikaru
Hikaru 2014-8-5
Sorry, I failed to notice it.
One way I can think of is something like this.
I = [1:1:10; 10:10:100]';
a = size(I);
limit = [2,3,9]; % specify the rows you want to be 0 here
for i=1:a(2)
I(limit(i),:) = 0;
end
I'm assuming you want the whole row be 0, but let me know if that's not the case.

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更多回答（2 个）

Amir
Amir 2014-8-5
编辑：Amir 2014-8-5
Please try this code. You will only need to specify your limits (L matrix)
I = [1:1:10; 10:10:100]';
L = [1 3 7 10]; % limit Matrix
NewL=vec2mat(L,2);
R=[]; %Removed rows
for i=1:size(NewL,1)
R=[R NewL(i,1)+1:NewL(i,2)-1];
end
I(R,:)=0; % or I(R,:)=[] if you want to remove those rows
Agent Cooper
Agent Cooper 2014-8-5
Thank you very much, Hikaru and Amir. Both answers were of great use.

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