The first step below counts all consecutive numbers. The second step eliminates the 0-counts.
A = [0 0 1 1 1 1 0 0 0 -1 -1 -1 0 1 1 0 ];
changeIdx = [1,diff(A)]~=0;
counts = histcounts(cumsum(changeIdx),1:sum(changeIdx)+1);
consecutiveCounts = [A(changeIdx)',counts']
% Result:
consecutiveCounts =
0 2 %First there are 2 zeros
1 4 %then 4 ones
0 3 %then 3 zeros
-1 3 %then 3 negative ones
0 1 % etc...
1 2
0 1
To get the vector you described, eliminate the 0-counts and multiply by the sign of the value in A so counts of negative numbers are negative.
consecutiveCounts(consecutiveCounts(:,1)~=0,2)' .* sign(consecutiveCounts(consecutiveCounts(:,1)~=0,1))'
ans =
4 -3 2
