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Eliminating zero's between rising and falling edge

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pr
pr on 20 Mar 2014
Answered: Roger Stafford on 21 Mar 2014
A series with 1's and 0's. a = [0 0 0 1 1 1 0 0 1 1 1 0 0 0 0 0 0 1 1 0 1 1 0 0 0 0 0 ]; and so on... Now I would like to eliminate the zeros in between 1's i.e debouncing and the desired output is: [0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0] . I achieved this using while loop. But if I run the code for about 100 thousand elements the program get's hanged and moreover takes lot of time. For eg: the index for rising edge is 4 and the falling edge is 11, now the zero's between them should be made equal to 1. In the sequence there are atleast 5 or more zeros between the two rising edges.
Would any one suggest the easiest way to solve this?

Accepted Answer

Roger Stafford
Roger Stafford on 21 Mar 2014
It isn't clear just how many consecutive zeros are required for them not to be changed to ones. Your statement "there are atleast 5 or more zeros between the two rising edges" hints at maybe 5, but you should state it clearly to avoid misunderstandings. Here is a rather cumbersome vectorized method, and I don't guarantee it is better than your 'while' loop.
f = find(diff([0,a,0])~=0);
f1 = f(2:2:end-2);
f2 = f(3:2:end-1);
t = (f2-f1)<5;
b = zeros(size(a));
b(f1(t)) = 1;
b(f2(t)) = -1;
b = a+cumsum(b);

More Answers (1)

Image Analyst
Image Analyst on 20 Mar 2014
If you have the Image Processing Toolbox, it's trivial - just a single line of code:
b = ~bwareaopen(~a, 3);
where 3 is the longest length of 0's that you want to allow. 100 thousand elements is no problem - this is just a very small fraction of the number of elements it usually works on (tens of millions of elements or pixels), so it will be pretty fast. If you don't have that toolbox , there are some trickier, more complicated ways to do it. Let us know and someone will probably work it out for you.

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