Dear Andrei Bobrov, thanks for your brilliance. Have learnt some new functions. However, with the data I am working with, replacing the first three zeros can be at different sections of the matrix. See the example below A = [0 0 0 0 1 1 0 0 0 0 0 1 1; 0 0 0 1 1 1 0 0 0 0 1 1 1; 0 0 0 1 1 1 0 0 0 0 1 1 1; 0 0 0 0 0 1 0 0 0 0 0 0 1] and this is what I expect to generate Ans = [0 1 1 1 1 1 0 0 1 1 1 1 1; 1 1 1 1 1 1 0 1 1 1 1 1 1; 1 1 1 1 1 1 0 1 1 1 1 1 1; 0 0 1 1 1 1 0 0 0 1 1 1 1]. Thanks
For loop or Array?
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Have a matrix [0 0 0 0 0 0; 0 0 0 1 1 1; 0 0 0 0 1 1; 0 0 0 0 1 1; 0 0 0 0 0 1; 0 0 0 1 1 1]. Want Matlab, in any row it encounters 1, to replace the first three zeros before the 1 with 1 so that I will have something like this [0 0 0 0 0 0; 1 1 1 1 1 1; 0 1 1 1 1 1; 0 1 1 1 1 1; 0 0 1 1 1 1; 1 1 1 1 1 1]. How do I go about this? I have a very big matrix I am dealing with.
4 Comments
Stephen23
on 31 Jul 2015
Using loops seems reasonable to me.
Your requirements exclude the use of circshift and cumsum in a trivial combination (e.g. Andrei Bobrov's initial answer). While there might be more compact solutions, using loops is likely the least obfuscated and yet also reasonably fast.
Answers (2)
Andrei Bobrov
on 29 Jul 2015
Edited: Andrei Bobrov
on 29 Jul 2015
out = cumsum(circshift(A,[0,3]),2)>0;
add
m = 3
b = [zeros(size(A,1),1),diff(A,[],2)]==1;
out = cumsum(b(:,[m+1:end,1:m])-b,2)+A;
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Stephen23
on 29 Jul 2015
Edited: Stephen23
on 29 Jul 2015
>> A = [0 0 0 0 1 1 0 0 0 0 0 1 1; 0 0 0 1 1 1 0 0 0 0 1 1 1; 0 0 0 1 1 1 0 0 0 0 1 1 1; 0 0 0 0 0 1 0 0 0 0 0 0 1]
A =
0 0 0 0 1 1 0 0 0 0 0 1 1
0 0 0 1 1 1 0 0 0 0 1 1 1
0 0 0 1 1 1 0 0 0 0 1 1 1
0 0 0 0 0 1 0 0 0 0 0 0 1
>> B = A;
>> for k = 1:3, B(:,4-k:end-k) = B(:,4-k:end-k) | A(:,4:end); end
>> B
B =
0 1 1 1 1 1 0 0 1 1 1 1 1
1 1 1 1 1 1 0 1 1 1 1 1 1
1 1 1 1 1 1 0 1 1 1 1 1 1
0 0 1 1 1 1 0 0 0 1 1 1 1
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