Sum two cells and write to a third cell

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User05
User05 le 11 Mai 2015
Commenté : User05 le 12 Mai 2015
I have three cells A{10}, B{10} and C{10} of size 5x5. On each cell I want to do: A(:,1) = B(:,1) + C(:,1) I might also want to do: A(2:4,5) = B(2:4,5) + C(2:4,5)
Currently if I just write something like above, I will get error as Index exceeds matrix dimensions.
Any ideas on how this can be done (as fast as possible)

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Alfonso Nieto-Castanon
Alfonso Nieto-Castanon le 11 Mai 2015
Modifié(e) : Alfonso Nieto-Castanon le 11 Mai 2015
Probaby a loop is going to be fastest, e.g.:
for n=1:numel(A)
A{n}(:,1)=B{n}(:,1)+C{n}(:,1);
end
but if you really dislike for loops you could always do something like:
A = cellfun(@(a,b,c)[b(:,1)+c(:,1) a(:,2:end)], A, B, C, 'uni',0);
  3 commentaires
Alfonso Nieto-Castanon
Alfonso Nieto-Castanon le 11 Mai 2015
Modifié(e) : Alfonso Nieto-Castanon le 11 Mai 2015
not sure what you mean by:
in1{1} = tin1(:,1) + tin2(:,1);
that will change the variable in1 instead of the variable out, it will make the first element of that cell array a 10 by 1 vector instead of a 10 by 10 matrix, and it will only modify that first element of the in1 cell array and none of the other elements. Is that what you really want?
If, instead, what you want is to change all five elements of out, but only the first column of each, and keep the rest of those 10x10 matrices the same as they were, then use the original code again (but first initialize out to something that shows you unambiguously what happened, e.g.
out = repmat({randn(10)}, 1,5);
User05
User05 le 12 Mai 2015
Sorry that in1 was typo. Initializing out to something in beginning works. Below code achieves what I want to do:
in1 = cell(1,5);
in1(:) = {int32(zeros(10,10))};
in2 = cell(1,5);
in2(:) = {int32(ones(10,10))};
out = cell(1,5);
out(:) = {int32(zeros(10,10))};
for n=1:5
out{n}(:,1)=in1{n}(:,1)+in2{n}(:,1);
end

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Plus de réponses (1)

Walter Roberson
Walter Roberson le 11 Mai 2015
Modifié(e) : Walter Roberson le 12 Mai 2015
cellfun()
However, the method to do it "as fast as possible" is likely to involve creating some mex code written in C and optimizing the heck out of it. It would probably take you a few months to learn how to do properly -- I know the resident expert James Tursa is still learning about using mex after what must be at least 8 years of working with it.
  3 commentaires
Walter Roberson
Walter Roberson le 11 Mai 2015
Use an anonymous function.
A = cellfun(@(a, b, c) [a(:,1:end-1), b(:,1) + c(:,1)], A, B, C, 'Uniform', false);
User05
User05 le 12 Mai 2015
I tried this anonymous function suggestion, but I dont quite get whats it exactly doing. My code:
in1 = cell(1,5);
in1(:) = {int32(zeros(10,10))};
in2 = cell(1,5);
in2(:) = {int32(ones(10,10))};
out = cell(1,5);
out(:) = {int32(zeros(10,10))};
out = cellfun(@(a, b, c) [a(:,1:end-1), b(:,1) + c(:,1)], out, in1, in2, 'Uniform', false);
After execution I see that 10th col of each cell in out is getting assigned value 1. What I wanted and expected was the 1st col should be all 1. So isnt that what is suppose to happen when I write b(:,1)+c(:,1)? Also what is a(:,1:end-1) doing?

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