count a vector with continuous non zero elements

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]

Answer 1

[Azzi Abdelmalek]

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

[Sam]

thanks Azzi, can you explain to me what is this doing?

[Azzi Abdelmalek]

ii=strfind([0 x 0],[0 1]) % find the indices corresponding to the switch from 0 to 1

Answer 2

[the cyclist]

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.