# Sum of Multiplication of column values in cell array

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Tha saliem 2018 年 1 月 4 日
コメント済み: Jan 2018 年 1 月 4 日
hey i have a cell array x
x{1,1}= {[ *3*,1,6.9,3.6,4,2.3,3,5,4.12],[ *3*,1,4,3,4,2,3.4,5,4],[ *3*,1.9,4,9.21,4,2.2,3,5,4.2],[ *2*,1.8,3.8,2,4,2.1,4,5,4.1]}
x{2,1}= {[1,1,5,3.6,4,2.3,3,2,4.12],[3,4,4,3,4,2,8,5,4],[1,5,4,7,4,2.9,3,5,4.2],[3,1.9,4.8,2,4,2.1,5,5,6]}
and similarly other elements x{3,1} x{4,1} etc. And other array y:
y= {[1,2,1.2,0.2];[0.32,0.42,0.4,0.07]}
I want to take each col value one by one and multiply it with the corresponding value in y and then add them all. Like this:
(3*1) + (3*2) + (3*1.2) + (2*0.2) [3,3,3,2 are bold values in x{1,1}]
and similarly take 2nd col of each element of x{1,1} and multiply it with corresponding values of y{1,1} and add them. When it comes to x{2,1}, it will be multiplied with y{2,1}.
Please suggest any efficient way of doing this without using multiple loops. Thank You.
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Tha saliem 2018 年 1 月 4 日
(*3*) means bold text (i think its code box so it didn't worked )

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### 採用された回答

Jan 2018 年 1 月 4 日
It is much easier to work with multi-dimensional arrays instead of nested cells. The cell arrays require loops for numerical operations, such that they are a bad choice, if you want to avoid loops. But you can convert the data:
Result = cell(size(x));
for k = 1:numel(x)
Result{k} = y{k} * cat(1, x{k}{:});
end
The actual calculation is a simple matrix multiplication.
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Tha saliem 2018 年 1 月 4 日
Ok I got it. Thank You for helping.

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### その他の回答 (2 件)

Birdman 2018 年 1 月 4 日
a=cell2mat(x{1,1}.');
b=cell2mat(y);
Then the rest is just a matrix calculation, as Jan said.
For instance, what you asked in the question is calculated as:
b(1,:)*a(:,1)
Hope this helps.
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Tha saliem 2018 年 1 月 4 日
A good approach. Thank You @Birdman.
Jan 2018 年 1 月 4 日
@Birdman: Exactly. +1. If the intermediate arrays are not huge, this is the most efficient approach.

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Guillaume 2018 年 1 月 4 日
What is the reason for the cells of x to be themselves cell arrays? Due to your multiplication requirements each vector in these subcells must be the same length. Therefore they'd be better stored as a 2D matrix instead, i.e.:
x{1, 1} = [3,1,6.9,3.6,4,2.3,3,5,4.12;
3,1,4,3,4,2,3.4,5,4;
3,1.9,4,9.21,4,2.2,3,5,4.2;
2,1.8,3.8,2,4,2.1,4,5,4.1];
x{2,1}= [1,1,5,3.6,4,2.3,3,2,4.12;
3,4,4,3,4,2,8,5,4;
1,5,4,7,4,2.9,3,5,4.2;
3,1.9,4.8,2,4,2.1,5,5,6];
You can transform your current x in the above x with
x = cellfun(@(c) vertcat(c{:}), x, 'UniformOuput', false);
With that storage, then your problem becomes a simple matrix multiplication of y cells with the x cells, which can be achieved with a single cellfun
result = cellfun(@mtimes, y, x, 'UniformOutput', false);
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Tha saliem 2018 年 1 月 4 日
Thank You for your time @Guillaume.

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