FOR loop with IF condition alternative
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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)?
采纳的回答
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
2014-8-5
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.
更多回答（2 个）
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
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