find the largest array

Consider the following example:

clear all
data = {rand(1,5),rand(1,4),rand(1,4),rand(1,6)};
data1 = cellfun(@(x) cell(size(x)),data,'UniformOutput',0);
time1 = cellfun(@(x) cell(size(x)),data,'UniformOutput',0);

for i = 1:length(data1);
    a1{i} = length(data1{i});
    for ii = 1:a1{i};
        data1{i}{ii} = rand(26,1);
        time1{i}{ii} = rand(26,1);
    end
end
data1{1}{3} = rand(32,1);
time1{1}{3} = rand(32,1);

      %interpolation
  for i = 1:length(data1);
      a1{i} = length(data1{i});
      for ii = 1:a1{i};
          t{i}{ii} = time1{i}{ii}; %original time
        p{i}{ii} = data1{i}{ii};%data
        x=time1{1}{3};%altered time - this need to be the data with the most
            %measurements 
        newD{i}{ii} = interp1(t{i}{ii},p{i}{ii},x);
    end
end

With this script I am trying to interpolate the time series of a set of data onto another series which has the maximum number of measurements. So, from the example above I would like to replace the line

x = time1{1}{3};

with a command which finds the cell with the maximum number of measurements (in this case time1{1}{3}) which can then be used to calculate 'newD'. How would I go about doing this?

Answer 1

[Image Analyst]

It might not be the most compact way, but hopefully it's understandable, or at least as understandable as you can be with cell arrays:

count = 1;
for cellNumber = 1 : length(time1)
  % for each cell in the time1 cell array...
  numberOfArraysInCell = length(time1{cellNumber})
  ca = time1{1, cellNumber}
  % Get the number of times for each cell.
  for arrayNumber = 1 : numberOfArraysInCell
    numberOfTimes(count) = size(ca{3}, 1);
    count = count + 1;
  end
end
% Report the max number of times.
maxNumTimes = max(numberOfTimes)

[Richard]

The problem that I am having is not finding the maximum number of times, but finding which cells correspond to that maximum number of times i.e. which cells have 'maxNumTimes' of measurements!

[Richard]

solved it! thanks for your help

Answer 2

[John D'Errico]

Start to think as you should in MATLAB, using tools like cellfun to do the work for you instead of loops.

[maxcellsize,maxcellind] = max(cellfun(@numel,time1));

[Jan]

Or, my usual comment about CELLFUN:

max(cellfun('prodofsize', time1))

The documentation claims, that the string commands of CELLFUN are kept for backward compatibility only. But in fact they are "much" faster than using a function handle, when the processed cell is "large".