Effacer les filtres
Effacer les filtres

Concatenate content of cells containing vectors

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Matthieu
Matthieu le 17 Mar 2023
Commenté : Matthieu le 18 Mar 2023
Hello, is there a simpler way to produce this result without 'for loop' ?
Thanks a lot.
A{1,1} = [0 1] ;
A{2,1} = 2 ;
A{3,1} = [3 4 5 ];
disp(A)
{[ 0 1]} {[ 2]} {[3 4 5]}
B{1,1} = 10 ;
B{2,1} = [ ];
B{3,1} = [ 10 11 12];
disp(B)
{[ 10]} {0×0 double} {[10 11 12]}
out = horzcatcellmat(A,B);
disp(out)
{[ 0 1 10]} {[ 2]} {[3 4 5 10 11 12]}
C = {[ 100 200 300]};
disp(C)
{[100 200 300]}
out = horzcatcellmat(A,C);
disp(out)
{[ 0 1 100 200 300]} {[ 2 100 200 300]} {[3 4 5 100 200 300]}
function A = horzcatcellmat(A,B)
if nargin == 0
A{1,1} = 1 ;
A{2,1} = 2 ;
A{3,1} = [3 4 5 ];
B{1,1} = 10 ;
B{2,1} = [ ];
B{3,1} = [ 10 11 12];
out = horzcatcellmat(A,B);
disp(out)
C = {[ 100 200 300]};
out = horzcatcellmat(A,C);
disp(out)
else
if size(A,1) == 1 && size(B,1) > 1
A = repmat(A,[size(B,1),1]);
end
if size(B,1) == 1 && size(A,1) > 1
B = repmat(B,[size(A,1),1]);
end
assert(size(A,1), size(B,1),'Size of both arguments are not compatible')
sizY= size(A,1);
% --------- ORIGINAL QUESTION
for ind = 1 : sizY
A(ind) = {horzcat(A{ind},B{ind})};
end
% ---------
% EDIT WITH PROPOSED ANSWERS :
A = cellfun(@(a,b)[a,b], A,B,'UniformOutput',false); % Thx to Matt J
A = cellfun( @horzcat , A,B,'UniformOutput',false); % Thx to Stephen23
%
end
if nargout == 0
A=[];
end
end
  2 commentaires
Dyuman Joshi
Dyuman Joshi le 17 Mar 2023
Modifié(e) : Dyuman Joshi le 17 Mar 2023
You might want to address cases when the number of inputs are not equal to 2 (Unless you are sure they won't occur) and the case where A has more rows than B.
Matthieu
Matthieu le 17 Mar 2023
the 'assert' catches the case if A has not the same number of rows than B.
Indeed, the case if nargin == 1 is not handled...
However, the core question was: How to remove the for loop

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Réponse acceptée

Matt J
Matt J le 17 Mar 2023
Simpler, yes. Faster, no.
A{1,1} = 1 ;
A{2,1} = 2 ;
A{3,1} = [3 4 5 ];
B{1,1} = 10 ;
B{2,1} = [ ];
B{3,1} = [ 10 11 12];
C=cellfun(@(a,b)[a,b], A,B,'uni',0)
C = 3×1 cell array
{[ 1 10]} {[ 2]} {[3 4 5 10 11 12]}
  3 commentaires
Stephen23
Stephen23 le 18 Mar 2023
cellfun(@horzcat, A,B,'uni',0)
% ^^^^^^^^ simpler
Matthieu
Matthieu le 18 Mar 2023
Thx Stephen23, simpler to read indeed (to my point of view)

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