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copied!<p>Here's one way to do it for a given element of the cell matrix:</p> <pre><code>full=zeros(7200,3)+NaN; for i = 1:20 % for each day starti = (i-1)*360; % find corresponding 360 indices into full array full( starti + (1:360), 1 ) = i; % assign the day idx = find(a(:,1)==i); % find any matching data in a for that day full( starti + (1:length(idx)), 2:3 ) = a(idx,2:3); % copy matching data over end </code></pre> <p>You could probably use <code>arrayfun</code> to make this slicker, and maybe (??) faster.</p> <p>You could make this into a function and use <code>cellfun</code> to apply it to your cell.</p> <p>PS - if you ask your question at the <a href="http://www.mathworks.com/matlabcentral/newsreader/" rel="nofollow">Matlab help forums</a> you'll <strong>most definitely</strong> get a slicker & more efficient answer than this. Probably involving <code>bsxfun</code> or <code>arrayfun</code> or <code>accumarray</code> or something like that.</p> <p><strong>Update</strong> - to do this for each element in the cell array the only change is that instead of searching for <code>i</code> as the day number you calculate it based on how far allong the cell array you are. You'd do something like (untested):</p> <pre><code>for k = 1:length(cellarray) for i = 1:length(cellarray{k}) starti = (i-1)*360; % ... as before day = (k-1)*20 + i; % first cell is days 1-20, second is 21-40,... full( starti + (1:360),1 ) = day; % <-- replace i with day idx = find(a(:,1)==day); % <-- replace i with day full( starti + (1:length(idx)), 2:3 ) = a(idx,2:3); % same as before end end </code></pre>

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