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

Question

Vijay Shekhawat ，2020-9-13

0
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Direct link to this question

https://ww2.mathworks.cn/matlabcentral/answers/592876-add-each-row-of-a-matrix-consecutively-to-all-the-rows-of-other-matrix-where-both-have-equal-number

编辑： Thiago Henrique Gomes Lobato ，2020-9-13

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.

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Answer 1

Thiago Henrique Gomes Lobato 2020-9-13

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/592876-add-each-row-of-a-matrix-consecutively-to-all-the-rows-of-other-matrix-where-both-have-equal-number#answer_493813

编辑：Thiago Henrique Gomes Lobato 2020-9-13

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