Effacer les filtres
Effacer les filtres

Add each row of a matrix consecutively to all the rows of other matrix where both have equal number of columns

2 vues (au cours des 30 derniers jours)
A=[1 2;3 4;5 6;7 8];
B=[1 1;2 2];
lA=length(A);
H=zeros(1,2);
for i=1:lA
H0=A(i,:) + B;
H=cat(1,H,H0);
end
H(1,:)=[];
Please suggest something without a loop.

Réponse acceptée

Thiago Henrique Gomes Lobato
Modifié(e) : Thiago Henrique Gomes Lobato le 13 Sep 2020
You can avoid the loop by using repeated indexing:
IndexesA = 1:size(A,1);
IndexesA = sort( [IndexesA,IndexesA] ); % IndexesA=[ 1 1 2 2 3 3 4 4]
Bexpanded= repmat(B,length(IndexesA)/2,1); % Expand matrix so the sum can be vectorized
H = A(IndexesA,:)+Bexpanded
H =
2 3
3 4
4 5
5 6
6 7
7 8
8 9
9 10
Depending on the size of A this can be ca. 20x faster than the (almost) original loop:
A = randn(100,2);
B = randn(2,2);
tic
for idx=1:10000
lA=length(A);
H=zeros(lA*2,2); % I changed somethings to be a little faster
for i=1:lA
H0=A(i,:) + B;
%H=cat(1,H,H0);
H(1+2*(i-1):2*i,:)= H0;
end
%H(1,:)=[];
end
TimeLoop = toc
tic
for idx=1:10000
IndexesA = 1:size(A,1);
IndexesA = sort( [IndexesA,IndexesA] );
Bexpanded= repmat(B,length(IndexesA)/2,1);
H = A(IndexesA,:)+Bexpanded;
end
TimeVec = toc
TimeLoop =
1.7428
TimeVec =
0.0861

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