Smart for loop for summing cell array matrices

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federico nutarelli
回答： Steven Lord 2021-7-7
Hi all,
I have an empty cell test_set_class and a cell matrix class_test made of 100x30 matrices (i.e. each element of classs_test is a 1229x119 matrix). What I would like to do is to perform the following
N_RIPETIZIONI=100;
for RIPETIZIONE = 1:N_RIPETIZIONI
test_set_class{RIPETIZIONE}=class_test{RIPETIZIONE,1}+class_test{RIPETIZIONE,2}+class_test{RIPETIZIONE,3}+class_test{RIPETIZIONE,4}+...
class_test{RIPETIZIONE,5}+class_test{RIPETIZIONE,6}+class_test{RIPETIZIONE,7}+class_test{RIPETIZIONE,8}+class_test{RIPETIZIONE,9}+class_test{RIPETIZIONE,10}+...
class_test{RIPETIZIONE,11}+class_test{RIPETIZIONE,12}+class_test{RIPETIZIONE,13}+class_test{RIPETIZIONE,14}+...
class_test{RIPETIZIONE,15}+class_test{RIPETIZIONE,16}+class_test{RIPETIZIONE,17}+class_test{RIPETIZIONE,18}+class_test{RIPETIZIONE,19}+class_test{RIPETIZIONE,20}+...
class_test{RIPETIZIONE,21}+class_test{RIPETIZIONE,22}+class_test{RIPETIZIONE,23}+class_test{RIPETIZIONE,24}+...
class_test{RIPETIZIONE,25}+class_test{RIPETIZIONE,26}+class_test{RIPETIZIONE,27}+class_test{RIPETIZIONE,28}+class_test{RIPETIZIONE,29}+class_test{RIPETIZIONE,30};
end
but looping also on the rows. I tried something like:
for RIPETIZIONE = 1:N_RIPETIZIONI
for cols =2:N_cols
test_set_class{RIPETIZIONE}=class_test{RIPETIZIONE,1}+class_test{RIPETIZIONE,cols};
end
end
but I am not sure that it does what I want.
Any help please?

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采纳的回答

Steven Lord
Steven Lord 2021-7-7
Ran in:
a cell matrix class_test made of 100x30 matrices (i.e. each element of classs_test is a 1229x119 matrix)
In this case, where all the elements inside the cells of class_test are the same type and size, I would consider a different data structure.
c = cell(3, 4);
A = zeros(2, 5, 3, 4);
for row = 1:size(c, 1)
for col = 1:size(c, 2)
c{row, col} = rand(2, 5);
A(:, :, row, col) = c{row, col};
end
end
celldisp(c)
c{1,1} = 0.3306 0.6299 0.6771 0.9105 0.2659 0.1979 0.7490 0.9951 0.4654 0.7687 c{2,1} = 0.8961 0.4946 0.2456 0.7889 0.0785 0.1325 0.1188 0.3285 0.9613 0.8639 c{3,1} = 0.5348 0.0686 0.0419 0.2903 0.8682 0.8177 0.3323 0.1261 0.0416 0.5009 c{1,2} = 0.0622 0.6295 0.1948 0.7948 0.0402 0.0928 0.0580 0.8608 0.9771 0.7861 c{2,2} = 0.7431 0.1538 0.2800 0.8544 0.1884 0.3732 0.6756 0.6714 0.9956 0.1708 c{3,2} = 0.6587 0.9058 0.2324 0.9266 0.6938 0.0685 0.2263 0.9513 0.7382 0.4114 c{1,3} = 0.3168 0.6707 0.1110 0.2844 0.2740 0.2383 0.9724 0.8853 0.0172 0.1679 c{2,3} = 0.2090 0.4523 0.5057 0.4638 0.1971 0.8271 0.7511 0.1102 0.2951 0.3152 c{3,3} = 0.5004 0.7998 0.9737 0.7259 0.6527 0.4693 0.9873 0.5821 0.1988 0.9433 c{1,4} = 0.6058 0.4562 0.5678 0.8480 0.7427 0.4811 0.5137 0.6755 0.2705 0.6834 c{2,4} = 0.4182 0.0767 0.7408 0.0116 0.6074 0.6580 0.3813 0.8827 0.1208 0.7691 c{3,4} = 0.0462 0.5263 0.3671 0.2239 0.2728 0.7184 0.0575 0.6601 0.5686 0.4865
disp(A)
(:,:,1,1) = 0.3306 0.6299 0.6771 0.9105 0.2659 0.1979 0.7490 0.9951 0.4654 0.7687 (:,:,2,1) = 0.8961 0.4946 0.2456 0.7889 0.0785 0.1325 0.1188 0.3285 0.9613 0.8639 (:,:,3,1) = 0.5348 0.0686 0.0419 0.2903 0.8682 0.8177 0.3323 0.1261 0.0416 0.5009 (:,:,1,2) = 0.0622 0.6295 0.1948 0.7948 0.0402 0.0928 0.0580 0.8608 0.9771 0.7861 (:,:,2,2) = 0.7431 0.1538 0.2800 0.8544 0.1884 0.3732 0.6756 0.6714 0.9956 0.1708 (:,:,3,2) = 0.6587 0.9058 0.2324 0.9266 0.6938 0.0685 0.2263 0.9513 0.7382 0.4114 (:,:,1,3) = 0.3168 0.6707 0.1110 0.2844 0.2740 0.2383 0.9724 0.8853 0.0172 0.1679 (:,:,2,3) = 0.2090 0.4523 0.5057 0.4638 0.1971 0.8271 0.7511 0.1102 0.2951 0.3152 (:,:,3,3) = 0.5004 0.7998 0.9737 0.7259 0.6527 0.4693 0.9873 0.5821 0.1988 0.9433 (:,:,1,4) = 0.6058 0.4562 0.5678 0.8480 0.7427 0.4811 0.5137 0.6755 0.2705 0.6834 (:,:,2,4) = 0.4182 0.0767 0.7408 0.0116 0.6074 0.6580 0.3813 0.8827 0.1208 0.7691 (:,:,3,4) = 0.0462 0.5263 0.3671 0.2239 0.2728 0.7184 0.0575 0.6601 0.5686 0.4865
What does making a 4-dimensional array A get us? Compare how to sum up the matrices:
result1 = zeros(2, 5);
for k = 1:numel(c)
result1 = result1 + c{k};
end
result1
result1 = 2×5
5.3219 5.8641 4.9380 7.1229 4.8817 5.0749 5.8233 7.7292 5.6502 6.8673
result2 = sum(A, [3 4]) % sum over dimensions 3 and 4
result2 = 2×5
5.3219 5.8641 4.9380 7.1229 4.8817 5.0749 5.8233 7.7292 5.6502 6.8673

更多回答（1 个）

Jan
Jan 2021-7-7
Yes, this does, what you expect. Simply compare the results of both methods.
This is faster and looks cleaner:
for RIPETIZIONE = 1:N_RIPETIZIONI
S = 0;
for cols =2:N_cols
S = S + class_test{RIPETIZIONE, cols};
end
test_set_class{RIPETIZIONE} = S;
end

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