Concatenate content of cells containing vectors
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Matthieu
am 17 Mär. 2023
Kommentiert: Matthieu
am 18 Mär. 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)
B{1,1} = 10 ;
B{2,1} = [ ];
B{3,1} = [ 10 11 12];
disp(B)
out = horzcatcellmat(A,B);
disp(out)
C = {[ 100 200 300]};
disp(C)
out = horzcatcellmat(A,C);
disp(out)
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 Kommentare
Dyuman Joshi
am 17 Mär. 2023
Bearbeitet: Dyuman Joshi
am 17 Mär. 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.
Akzeptierte Antwort
Matt J
am 17 Mär. 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)
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