Asked by Andrew
on 7 Jun 2011

Hi,

I would like to be able to create a cell array, with each cell containing the same 1 by 3 vector that I specify. For example, suppose that I have the following:

A=[1 2 3];

and suppose that I want to obtain the following:

mycell={[1 2 3],[1 2 3],[1 2 3];[1 2 3],[1 2 3],[1 2 3]};

Is there an elegant way to do this? It is as if I would like a sort of "repcell" function, similar to the repmat function. I am not sure if one exists, though.

I could do this:

tic mycell=cell(2,3); for i=1:1:size(mycell,1) for j=1:1:size(mycell,2) mycell{i,j}=A; end end toc

and I get "Elapsed time is 0.000956 seconds." Or I could do:

tic myarray=repmat(A,2,3); mycell=mat2cell(myarray,[1 1],[3 3 3]); toc

and I get "Elapsed time is 0.029993 seconds."

Either one of these seems like it will work fine for me, even when I want my cell array to be very large. Still, I am wondering, is there a more elegant way to do this (even in terms of code readability)?

Thank you very much for your time.

Andrew DeYoung

Carnegie Mellon University

Answer by Matt Fig
on 7 Jun 2011

Accepted answer

mycell = repmat({A},2,3)

Your MAT2CELL method properly pre-allocates a cell array, in the sense James Tursa is talking about. It can be written as a one-liner:

mycell = mat2cell(repmat(A,M,N),ones(1,M),ones(1,N)*length(A));

So does this, and it is much faster than the MAT2CELL method:

mycell = cellfun(@double,repmat({A},M,N),'Un',0);

Answer by Fangjun Jiang
on 7 Jun 2011

A=[1 2 3]; mycell=cell(2,3); mycell(:)={A};

Show 1 older comment

Fangjun Jiang
on 7 Jun 2011

James Tursa
on 7 Jun 2011

@Andrew: I am still not clear what you are doing downstream with mycell. Are you *replacing* cells with a completely different variables, or are you replacing only certain *elements* of the cell contents with different value? i.e., are you doing something like:

mycell{k} = something_else

Or are you doing something like:

mycell{k}(m) = some_value

If it is the former, then pre-allocating mycell with variable contents ala repmat and the like makes no sense whatsoever, and you would be better off pre-allocating an empty cell array and then filling the contents downstream as you go.

Answer by Fangjun Jiang
on 7 Jun 2011

I am not trying to win a competition but here is the example. It looks like it will be faster to pre-allocate using hidden reference than using the repmat().

>> clear all; A=1:3; tic; mycell=cell(10000,1); mycell(:)={A}; for k=length(mycell) mycell{k}=rand(1,3); end toc;

clear all; A=1:3; tic; mycell=repmat({A},10000,1); for k=length(mycell) mycell{k}=rand(1,3); end toc; Elapsed time is 0.003057 seconds. Elapsed time is 0.012636 seconds.

And it is still the case if the loop is mycell{k}(2)=rand;

Show 7 older comments

Sean de Wolski
on 8 Jun 2011

http://www.mathworks.com/matlabcentral/newsreader/view_thread/292183

cell array preallocation thread on CSSM.

James Tursa
on 8 Jun 2011

The data storage layout for a cell array is *exactly* the same as the data storage layout for all other variables ... the data is contiguous in memory. But the "data" for a cell array in this context is variable pointers. So, yes, the actual content of the cells themselves are not in general contiguous in memory (the variables being pointed to are not contiguous), but the *pointers* to these variables, which is what is physically stored in the cell array data area, *are* contiguous in memory. It is *that* memory that gets constantly copied and re-copied in a loop that increases the size of the cell array if you don't pre-allocate the cell array. The L in your above code has nothing to do with this since it only affects the cell variables and not the physical cell array data area. e.g., to see the effect isolated:

n = 100000;

A = 1:3;

clear mycell1

tic

mycell1 = cell(1,n); % pre-allocated

for k=1:n

mycell1{k} = A; % not growing in a loop

end

toc

clear mycell2

tic

for k=1:n

mycell2{k} = A; % growing in a loop

end

toc

isequal(mycell1,mycell2)

Elapsed time is 0.049412 seconds.

Elapsed time is 27.087694 seconds.

ans =

1

Fangjun Jiang
on 8 Jun 2011

Opportunities for recent engineering grads.

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