storing data in a matrix from a nested loop

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rajesh regar
rajesh regar 2019-9-11
评论： rajesh regar 2019-9-11
I want to save the vector made from d12 each time i.e. d12 is 4*3 matrix. d12 in nested loop
d1 = [0 0 2
0 0 -2 ];
d2 = [0 0 15
0 0 13 ];
for i = 1:2
for j = 1:2
d12() = [d1(i,:)-d2(j,:)];
end
end

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采纳的回答

rajesh regar
rajesh regar 2019-9-11
d1 = [0 0 2
0 0 -2 ];
d2 = [0 0 15
0 0 13 ];
k = 1;
for i = 1:2
for j = 1:2
d12(k,:) = [d1(i,:)-d2(j,:)];
k = k+1;
end
end
you can initiate k as 1 then increase it by one

更多回答（2 个）

Stephen23
Stephen23 2019-9-11
编辑：Stephen23 2019-9-11
No loops required:
>> d1 = [0,0,2;0,0,-2]
d1 =
0 0 2
0 0 -2
>> d2 = [0,0,15;0,0,13]
d2 =
0 0 15
0 0 13
>> out = repelem(d1,size(d2,1),1) - repmat(d2,size(d1,1),1)
out =
0 0 -13
0 0 -11
0 0 -17
0 0 -15
This gives exactly the sme output as your nested loops:
>> isequal(out,d12)
ans =
1
  2 个评论
rajesh regar
rajesh regar 2019-9-11
Thank you, I have to use loop for solving my problem because I need to define more variable and finally I will put in formula. Also the size of nested loop in my case is aroung 1000*1000 i.e. need to vary i = 1:1000 and j = 1:1000.

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madhan ravi
madhan ravi 2019-9-11
编辑：madhan ravi 2019-9-11
[m,n]=size(d1);
[m1,n1]=size(d2);
z = permute(reshape(d1.',1,n,m) - d2,[2,1,3])
d12 = reshape(z,n,[]).'
% Or if you prefer a loop
k = 1;
d12 = zeros(m+m1,n); % preallocate
for ii = 1:m % don't hardcode the sizes , use size() instead
for jj = 1:m1
d12(k,:) = d1(ii,:) - d2(jj,:);
k = k+1;
end
end

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