好吧,您只需要找到边缘时间和它们之间的矩阵索引。因此,如果您的数字是 datenum 格式,则一个单位与一天相同,这意味着我们可以从 30 和 30 个单位跳到最后,如下所示:
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
此处getNearestIdx
讨论了该函数,它为您提供距输入值最近的点,而无需检查所有可能的点。
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
注意:这是伪代码,可能包含错误。请尝试将其调整为您的问题。