Well, all you need is to find the edge times and the matrix indexes in between them. So, if your numbers are at datenum format, one unit is the same as one day, which means that we can jump from 30 and 30 units until we get as close as we can to the end, as follows:
startTime = originalMatrix(1,4);
endTime = originalMatrix(end,4);
edgeTimes = startTime:30:endTime;
% And then loop though the edges checking for samples that complete a cycle:
nEdges = numel(edgeTimes);
totalMeasures = size(originalMatrix,1);
subMatrixes = cell(1,nEdges);
prevEdgeIdx = 0;
for curEdgeIdx = 1:nEdges
nearIdx=getNearestIdx(originalMatrix(:,4),edgeTimes(curEdgeIdx));
if originalMatrix(nearIdx,4)>edgeTimes(curEdgeIdx)
nearIdx = nearIdx-1;
end
if nearIdx>0 && nearIdx<=totalMeasures
subMatrix{curEdgeIdx} = originalMatrix(prevEdgeIdx+1:curEdgeIdx,:);
prevEdgeIdx=curEdgeIdx;
else
error('For some reason the edge was not inbound.');
end
end
% Now we check for the remaining days after the edges which does not complete a 30 day cycle:
if curEdgeIdx<totalMeasures
subMatrix{end+1} = originalMatrix(curEdgeIdx+1:end,:);
end
The function getNearestIdx
was discussed here and it gives you the nearest point from the input values without checking all possible points.
function vIdx = getNearestIdx(values,point)
if isempty(values) || ~numel(values)
vIdx = [];
return
end
vIdx = 1+round((point-values(1))*(numel(values)-1)...
/(values(end)-values(1)));
if vIdx < 1, vIdx = []; end
if vIdx > numel(values), vIdx = []; end
end
Note: This is pseudocode and may contain errors. Please try to adjust it into your problem.