Add Cells/Matrices Row by Row
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I have a large number of matrices/vectors. And I need to add then row by row. So I have 100 different 50x10 matrices. Now I want to add Row 1 (first cell) + Row 1 (second cell)+...., and repeat it for all 60 rows and get a final 50x10 matrix. I have them all in cell format.
2 comentarios
Image Analyst
el 13 de Ag. de 2021
Editada: Image Analyst
el 13 de Ag. de 2021
And your question is......?
My question is why do you have 100 different matrices? Hopefully they're not all separate variables but you're just getting a new matrix in a loop that's iterating 100 times. Or are all 100 matrices in individual cells of a 1-D cell array?
My second question is why are you insisting on cell arrays when, if the matrices are all the same size 50x10, when a normal 3-D double array would be much faster and more efficient?
Next question is : can you attach your data in a .mat file, or attach your script that generates the data?
Next question is did you just try a simple for loop - should be very very simple.
By the way, 100 is not large as far as array sizes or memory usage.
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dpb
el 13 de Ag. de 2021
"matrix of about 600,000 rows and 14 columns. This data is a 6000 different bins with 100 columns each. So what I am trying to do is add (Row1 + Row101+ Row201....+..), (Row2 + Row 102+ Row 202 + ....) and get a final matrix of 100x14. "
Using your smaller example, but same logic works, the "dead ahead" solution is
>> x=rand(12,2) % some sample data
x =
0.8147 0.9572
0.9058 0.4854
0.1270 0.8003
0.9134 0.1419
0.6324 0.4218
0.0975 0.9157
0.2785 0.7922
0.5469 0.9595
0.9575 0.6557
0.9649 0.0357
0.1576 0.8491
0.9706 0.9340
>>
% the engine
N=4; % number of elements in group
M=size(x,1)/N; % number of bins -- must be evenly divisible
S=zeros(M,size(x,2); % preallocate
for i=1:M
S(i,:)=sum(x(i:N:end,:));
end
Show what we got...
>> S
S =
2.4046 2.0347
1.9682 1.4368
0.5631 2.4416
>>
>> x(1,1)+x(5,1)+x(9,1) % illustrate first term gets right answer...
ans =
2.4046
>>
2 comentarios
dpb
el 13 de Ag. de 2021
Editada: dpb
el 13 de Ag. de 2021
So, switch N from 4 to 3 -- but that's not what your example asked...
"Now what I wanted to do is to bin 4 rows together. So we have 3 bins like below, "
The logic is generic regardless of the size as long as size(data,1)/N is integer.
>> N=3; % changed mind, groups of 3, not four
>> M=size(x,1)/N; % compute # bins given N
>> S=zeros(M,2); % preallocate again
>> for i=1:M,S(i,:)=sum(x(i:N:end,:));end % and compute
>> S % what we get
S =
2.9715 1.9270
2.2426 2.7158
2.1526 3.3057
2.1568 0.9698
>> x(1,1)+x(4,1)+x(7,1)+x(10,1) % again, just check got it right...
ans =
2.9715
>>
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