how to sum matrics inside a cell array?

Question

Kobi 2014-8-6

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评论： LD 2017-11-26

采纳的回答： dpb

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i have 4 matrics inside a cell and i need to sum them all currenyly i use this:

R=R{1}+R{2}+R{3}+R{4}

the original R is a cell with 4 matrics

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Answer 1

dpb 2014-8-6

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编辑：dpb 2017-4-22

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Oh, misunderstood the query...that's simply

S=plus(R{:})
S =
  1.3771    0.5139
  1.1590    0.7149
>> all(all(S==R{1}+R{2}))
ans =
   1
>>

Just used two cells here, but works for any number. As noted elsewhere sizes must be commensurate.

ERRATUM Does NOT work as written for any number of cells; plus isn't as smart as I was thinking it was/is... :(

Instead use

S=sum(cat(3,R{:}),3);

As another poster noted, all cells must contain conformantly-sized arrays.

NB: The result will be class of the content of the cell (default double, of course).

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dpb 2017-4-22

Hmmm....my bad, had tested only for two cells; I was thinking w/o checking that plus was enabled to handle a variable number of arguments but it isn't...see ERRATUM above Answer--dpb

LD 2017-11-26

Is there other methods to solve this problem? This methods is less efficient then the for-loop

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Answer 2

Evan 2014-8-6

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编辑：Evan 2014-8-6

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Are all the matrices in your cell array the same size? If so, this should work:

% Get random cell array with n cells. Here n = 3
A = {rand(4) rand(4) rand(4)}
n = size(A,2);
% Get size of matrices in cell.
matSize = size(A{1},1);
B = reshape(cell2mat(A),matSize,[],n);
% Sum 3D matrix along 3rd dimension
C = sum(B,3);

If they're different sizes, this either wont work or we need more information about how you want the different sizes handles (e.g. trimming, padding, etc.).

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Answer 3

dpb 2014-8-6

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One way, seems should be better but it's still first cup o' coffee this morning...

>> R={rand(2) rand(3)}
R = 
  [2x2 double]    [3x3 double]
>> sum(cellfun(@(x) sum(x(:)),R))
ans =
  8.7056
>>

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Kobi 2014-8-6

this is not what i need i need to + sum the matrics like this A=rand(7)+rand(7)+rand(7)+.... and to recive one matrix each and every element in one matrix is added to another corresponding element the only problem is that all of the matrics are inside a cell

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