trying to combine cells
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I have 3 cells [1x5000] , [1x3000] and [1x2000] and I want to combine these in one cell [1x10000]. With small dimensions you can try the copy paste but now this is very slow procedure.
Do you have any ideas?
Thanks in advance!
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Respuesta aceptada
  Image Analyst
      
      
 el 24 de Mayo de 2013
        Study this example:
% Original cell array, with 3 elements (cells).
ca = {ones(1,5000), rand(1,3000), zeros(1,2000)}
% Construct new cell with one element comprised of the
% contents of the other cell array's cells:
new_ca = {[ca{1}, ca{2}, ca{3}]}
2 comentarios
  Andrei Bobrov
      
      
 el 24 de Mayo de 2013
				new_ca = {cat(2,ca{:})};
new_ca = {[ca{:}]};
  Image Analyst
      
      
 el 24 de Mayo de 2013
				
      Editada: Image Analyst
      
      
 el 24 de Mayo de 2013
  
			nicolas said "Something like this,(I know that is wrong) C= [X(1, 1:5000),Y(1, 5001:8000), Z(1, 8001:10000)] Any ideas? "
Then just do this:
% Original cell array, with 3 elements (cells).
ca = {ones(1,99000), rand(1,99000), zeros(1,99000)}
% Construct new cell with one element comprised of certain specific elements 
% of the contents of the other cell array's cells:
new_ca = {[ca{1}(1:5000), ca{2}(5001:8000), ca{3}(8001:10000)]}
If you have 3 separate cells instead of a cell array, then just concatenate them and use the same solution as above.
% Three separate original cells.
ca1 = {ones(1,99000)};
ca2 = {rand(1,99000)};
ca3 = {zeros(1,99000)};
% Combine, then use same solution as above.
ca = [ca1, ca2, ca3]
% Construct new cell with one element comprised of the
% contents of the other cell array's cells:
new_ca = {[ca{1}(1:5000), ca{2}(5001:8000), ca{3}(8001:10000)]}
Más respuestas (5)
  yagnesh
      
 el 24 de Mayo de 2013
        suppose x is [1x5000] , y is [1x3000] and z is [1x2000]
xyz=[x;y;z]
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  Andrei Bobrov
      
      
 el 24 de Mayo de 2013
        A = {X,Y,Z};
ii = hankel([0 5000],[5000 8000 10000]);
C = arrayfun(@(x)A{x}(ii(1,x)+1:ii(2,x)),1:numel(A),'un',0);
out = {[C{:}]};
0 comentarios
  nicolas
 el 24 de Mayo de 2013
        2 comentarios
  Image Analyst
      
      
 el 24 de Mayo de 2013
				Yes, cell arrays are almost always complicated. Can't you work with just normal old reliable numerical arrays? It would be so much simpler.
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