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HOW TO SORT DATA IN CELL ARRAY

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Akshay Sahu
Akshay Sahu el 10 de Mzo. de 2020
Comentada: Akshay Sahu el 14 de Mzo. de 2020
A {1x4}
Such that
A = {A{1,1},A{1,2},A{1,3},A{1,4}}
A{1,1}={}
A{1,2}={0.5 0.2
1 2
3 4
NaN NaN
5 6
7 8
9 10
NaN NaN}
A{1,3}={0.5 0.8
5 6
7 8
9 10
NaN NaN
11 12
13 14
15 16
17 18
NaN NaN}
A {1,4} = {0.7 0.9
19 20
NaN NaN
25 26
27 28
29 30
31 32
NaN NaN}
I want to separate data when there is NaN NaN in the matrix, want to store values before first NaN NaN in one cell array and after NaN NaN in another cell array.
B1 = {A{1, 2}(1:3,:))}
B2 = {A{1, 3}(1:4,:))}
B3 = {A{1, 4}(1:2,:))}
C1 = {A{1, 2}(5:7,:)}
C2 = {A{1, 3}(6:9,:)}
C3 = {A{1, 4}(4:7,:)}
B = {{B1},{B2},{B}}
C = {{C1},{C2},{C3}}
  1 comentario
Luna
Luna el 10 de Mzo. de 2020
Editada: Luna el 10 de Mzo. de 2020
why are you using cell arrays inside cell arrays? What makes you understand which part belongs to B and which belongs to C?

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Respuesta aceptada

Ameer Hamza
Ameer Hamza el 10 de Mzo. de 2020
Are you looking for something like this code? Note that I have made some modifications to the datatypes and removed A{1,1} = {}, since it makes the code simpler. This code will work as long as there are two nan rows in the matrix.
A{1,1}=[0.5 0.2
1 2
3 4
NaN NaN
5 6
7 8
9 10
NaN NaN];
A{1,2}=[0.5 0.8
5 6
7 8
9 10
NaN NaN
11 12
13 14
15 16
17 18
NaN NaN];
A{1,3} = [0.7 0.9
19 20
NaN NaN
25 26
27 28
29 30
31 32
NaN NaN];
nan_rows = cellfun(@(x) find(isnan(x), 2), A, 'UniformOutput', 0);
B = cellfun(@(x, y) x(1:y(1)-1, :), A, nan_rows, 'UniformOutput', 0);
C = cellfun(@(x, y) x(y(1)+1:y(2)-1, :), A, nan_rows, 'UniformOutput', 0);
  13 comentarios
Ameer Hamza
Ameer Hamza el 14 de Mzo. de 2020
Try this. B is a cell array of cell arrays. See the result with indexing like this: B{1}{1}, B{1}{2}, B{2}{1}, ....
A{1,1}=[0.5 0.2
1 2
3 4
NaN NaN
5 6
7 8
9 10
NaN NaN
0.5 0.2
1 2
3 4
NaN NaN];
A{1,2}=[0.5 0.8
5 6
7 8
9 10
NaN NaN
11 12
13 14
15 16
17 18
NaN NaN
0.5 0.8
5 6
7 8
9 10
NaN NaN];
A{1,3} = [0.7 0.9
19 20
NaN NaN
25 26
27 28
29 30
31 32
NaN NaN
0.7 0.9
19 20
NaN NaN];
total_nan_rows = sum(isnan(A{1}(:,1)));
nan_rows = cellfun(@(x) find(isnan(x), total_nan_rows), A, 'UniformOutput', 0);
for i=1:total_nan_rows
if i==1
B{i} = cellfun(@(x, y) x(1:y(i)-1, :), A, nan_rows, 'UniformOutput', 0);
elseif i==total_nan_rows
B{i} = cellfun(@(x, y) x(y(i-1)+1:end-1, :), A, nan_rows, 'UniformOutput', 0);
else
B{i} = cellfun(@(x, y) x(y(i-1)+1:y(i)-1, :), A, nan_rows, 'UniformOutput', 0);
end
end
Akshay Sahu
Akshay Sahu el 14 de Mzo. de 2020
Yeah thanks man Ameer Hamza it's working.

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