MATLAB: Indexing problem based on value

indexing

Hi,

I encountered a strange problem. For A= [5 4 3 2 1 0 1 2 3 4 5];

I can't do B = A(1): A(6). The return is 1×0 empty double row vector. if I do A(7): A(8) or start from index 7, I can get the output.

I am not sure what is the source of problem. I expected A(1) : A(6) = [5 4 3 2 1 0].

I tried A(1:6) and it works. what is the difference between A(1:6) and A(1): A(6) while the latter doesn't work?

Thanks.

Best Answer

B = A(1:6)

You should read about array indexing in MATLAB. MATLAB indices are always nonzero positive integers.

When you use A(1:6), it means in the array A pick indices from 1 to 6.

When you try A(1):A(6), it will try to generate an array with a difference of 1, from the value A(1) i.e 5 to A(6) i.e 0, this will work when start value is low and end value is high, in your case they are 5 and 0, so :, cannot generate an array. Where as try A(6):A(1), it will work.

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This solution takes an input vector 'v' and differentiates the indices of non-zero values in order to measure the distance between non-zero values. It then determines which distances are within a specified window 'd' and then loops through each of those groupings to calculate the maximum value within the groups and stores it in the output vector 'vNew'. If you have any trouble running the code or understanding parts of it, please leave a comment.


% Input vector (row or column)
v = [0 0 0 2 0 3 0 0 0 0 4 0 4 0 0 0 0 0 0 0 5]; 
% specify window size
d = 3; %window size. Example: [2 0 3] has window size of 3
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vNew = zeros(size(v));              %allocate the output variable
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table(v', vNew', 'VariableNames', {'Input', 'Output'})

Best Answer

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