Asked by Thomas Veith
on 3 Jun 2019

So I have the code bleow which sovles an ODE 1000 times with random variables. The result is a 1000 by 1 cell array, the contents of which are all XX by 2 numeric arrays (column 1 solves x(1), column 2 solves x(2)). What I would like to produce is an XX by 1000 numeric array, where my 1000 columns would be comprised of all column 2 data from the numeric arrays nested in my cell array. Is this possible? I've looked through the Cell Arrays and Multilevel Indexing to Access Parts of Cells pages, and haven't been able to find what I'm looking for. Any help would be appreciated.

n = 1000;

result = cell(n,1);

for k=1:n

tspan=[1 7];A0=rand;P0=rand;g=rand;p=rand;B=rand;

[t,x] = ode45(@(t,x) [-g*x(1) + p*x(1); -x(1)*x(2)+ B*x(2)], tspan, [A0 P0]);

result{k} = x;

end

Answer by Adam Danz
on 3 Jun 2019

Edited by Adam Danz
on 3 Jun 2019

Accepted Answer

This extracts column 2 data from all elements of the cell array and arranges it in a row within a cell array as you described.

C = cellfun(@(x)x(:,2),result,'UniformOutput', false)';

Here's how to put that into a padded array

% how much padding is needed per cell?

maxLen = max(cellfun(@(x)size(x,1),result)); %max length

padNum = maxLen - cellfun(@(x)size(x,1),result);

% get the col 2 vals and pad them

C = cellfun(@(x)x(:,2),result,'UniformOutput', false);

Cpad = cellfun(@(x,p)padarray(x,p,NaN,'post'),C,num2cell(padNum),'UniformOutput',false);

% Put them into a matrix

M = cell2mat(Cpad');

Adam Danz
on 3 Jun 2019

Thomas Veith
on 3 Jun 2019

Adam Danz
on 3 Jun 2019

If they were all the same size, it would be as easy as

C = cellfun(@(x)x(:,2),result,'UniformOutput', false)';

cell2mat(C)

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Answer by madhan ravi
on 3 Jun 2019

Edited by madhan ravi
on 3 Jun 2019

result{k} = x(:,2);

To produce a numeric matrix you need to append with nan or zeros at the end.

M=max(cellfun('prodofsize', result));

Desired=cell2mat(cellfun(@(x) [x;zeros(M-numel(x),1)], result,'un', 0))

Thomas Veith
on 3 Jun 2019

madhan ravi
on 3 Jun 2019

See edited answer.

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