find the largest array

21 views (last 30 days)
Richard on 2 Apr 2012
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);
data1{1}{3} = rand(32,1);
time1{1}{3} = rand(32,1);
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
newD{i}{ii} = interp1(t{i}{ii},p{i}{ii},x);
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?

Accepted Answer

Image Analyst
Image Analyst on 2 Apr 2012
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;
% Report the max number of times.
maxNumTimes = max(numberOfTimes)
Richard on 3 Apr 2012
solved it! thanks for your help

Sign in to comment.

More Answers (1)

John D'Errico
John D'Errico on 3 Apr 2012
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));
  1 Comment
Jan on 3 Apr 2012
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".

Sign in to comment.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by