I have written a function to check for the criteria you have mentioned.
You can also check out this FEX submission for the same - RunLength
%Random input
A = [1 0 0 0 0 0 2 3 1 0 0 0 0 0 0 0 2 3 3 3 0 3 1 2 2 2 2 1 1 1 3 3 3];
%Corresponding outputs
out1 = convert(A,0,4);
out2 = convert(A,[2 3],3);
% +1 for 0
% -1 for [2 3]
out = 1*out1+(-1)*out2;
vec = 1:numel(A);
%plot using stairs to get the square wave form
stairs(vec,out)
grid on
ylim([-2 2])
xticks(vec)
%% Function to check for groups of consecutive apperances above a threshold
function out = convert(in,val,thresh)
%% in - input array (expected to be a vector)
%% val - values to check for as a part of a cluster (expected to be a vector)
%% thresh - threshold for elements in a cluster (expected to be a scalar)
%% out - output (row vector) = 1 for the elements of groups that satisfy the criteria, 0 otherwise
%Check for input
if ~isvector(in)
error('Input array must be a vector');
end
%Check for values
if ~isvector(val)
error('Values must be a vector');
end
%Check for threshold
if ~isscalar(thresh)
error('Threshold must be a scalar');
end
%Convert the input to a row vector
in = reshape(in,1,[]);
%Lazy way of preallocating output
out = 0*in;
%Check for values present in the input
idx=ismember(in,val);
%Starting index of a cluster
ij1=strfind([0 idx 0],[0 1]);
%Ending index of a cluster
ij2=strfind([0 idx 0],[1 0])-1;
%Cluster with members more than a threshold
temp=find((ij2-ij1+1)>=thresh);
%Set the corresponding output as 1
for k=1:numel(temp)
out(ij1(temp(k)):ij2(temp(k)))=1;
end
end
