count a vector with continuous non zero elements
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Is there a way to count non-zero elements in vector x below x=[0 1 1 1 0 0 1 1 1 1 1 0 0 0 0 1 1 1 1 1 1 1 0 1]
so that the output y will be y=[3 5 7 1]
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Azzi Abdelmalek
2014-5-16
x=[0 1 1 1 0 0 1 1 1 1 1 0 0 0 0 1 1 1 1 1 1 1 0 1]
ii=strfind([0 x 0],[0 1])
jj=strfind([0 x 0],[1 0])
out=jj-ii
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Azzi Abdelmalek
2014-5-16
ii=strfind([0 x 0],[0 1]) % find the indices corresponding to the switch from 0 to 1
the cyclist
2014-5-16
编辑:the cyclist
2014-5-16
Here's a similar approach:
y = diff(find([0 x 0]==0))-1;
y(y==0) = []
The first line identifies the zero locations, and then the "distance" between them. This distance is the length of the string of ones.
The second line removes the instances where that distance is zero (i.e. no 1's), since you are not interested in those.
Note that the reason you need to do this operation on "[0 x 0]" instead of just x is to be able to identify a string of ones at the beginning and end.
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