# 2 different for loops at the same time ?

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Gökçe Öter on 19 May 2018
Commented: Gökçe Öter on 19 May 2018
Hi, I've just tried to make 2 for loops at the same time but I couldn't achieve that properly. Assume that "A" can be any matrix. I upload example.xlsx file as an .txt file.
for ii= 5:5:length(A);
for jj= 4:5:length(A);
M=A(ii,11)*(10^A(jj,1))
end
end
When I try this code I get 16 M as an answer, I want to take 4 answers. Does anybody have any clue that can help ?

Walter Roberson on 19 May 2018
iivals = 5:5:length(A);
jjvals = 4:5:length(A);
numii = length(iivals);
numjj = length(jjvals);
M = zeros(numii, numjj);
for ii_idx = 1 : numii
ii = iivals(ii_idx);
for jj_idx = 1 : numjj
jj = jjvals(jj_idx);
M(ii_idx, jj_idx) = A(ii,11)*(10^A(jj,1))
end
end
Gökçe Öter on 19 May 2018
Thank you so much !

Majid Farzaneh on 19 May 2018
Edited: Majid Farzaneh on 19 May 2018
Hi, It's better use size(A,1) [for number of rows] or size(A,2) [for number of columns] instead of length(A). Because it's may not exactly what you want. At the M=A(ii,11)*(10^A(jj,1)) if you want to multiply values element by element you should use '.*' instead of '*'.
for ii= 5:5:size(A,1);
for jj= 4:5:size(A,1);
M=A(ii,11).*(10^A(jj,1))
end
end
Gökçe Öter on 19 May 2018
Thanks for your effort, but with this code i get 16 matrix answers, I just want to get 4 value as an answer
0.1312*1.0e+24 0.3681*1.0e+24 0.6380*1.0e+24 0.4334*1.0e+24
which are multiplication of two elements of A matrix.

Rik on 19 May 2018
You must index M.
jj_= 4:5:length(A);
ii_= 5:5:length(A);
M=zeros(numel(ii),numel(jj));%pre-allocate
for ii=1:numel(ii_)
for jj=1:numel(jj_)
M(ii,jj)=A(ii_(ii),11)*(10^A(jj_(jj),1));
end
end

Ameer Hamza on 19 May 2018
Edited: Ameer Hamza on 19 May 2018
You don't need a for loop. Replace it with this:
M = A(5:5:length(A), 11).*10.^(A(4:5:length(A), 1));
This will work in R2016b and later.
Ameer Hamza on 19 May 2018
As you can see, that your final answer is the first column of of 4x4 matrix. So you can get it like this
M = A(5:5:length(A), 11)*10.^(A(4:5:length(A), 1))';
M = M(:,1);
It will give you your required 4 numbers

R2017b

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