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confront values in 2 matrix

Asked by Dario Riviezzo on 13 Sep 2019
Latest activity Edited by Matt J
on 13 Sep 2019
hi, i have 2 different matrix with same dimensions and i need to know the min value against 2 values in the same position:
A=[ 0 1 3 5 6]
[2 3 5 8 9]
B=[2 2 4 7 8]
[3 5 7 9 10]
the algorithm confront the zero in first row first column of A with the 2 in first row first column of B and show the min value (0), so for the second row first column of each matrix and so on...i used a for cycle, but i want to know if i can try without any cycle.
In my cicle i use this:
for i 1:size(A)
find(A(1,:)<B(1,:)
end
how can i confront only element by element in the same position without any cycle?
i also have to use the elements of a matrix to do a scalar product using each element in a position and inserting it into a vector:
for i=1:length(A)
for j=1:N (size of rows)
ps1(i,j)=[A(i,j) 0]*[1 0]';
end
end
there is any way to do this without cycles?
i need the value in every position and that value is putted into a vector, then i make the scalar product with [1 0].
Tnx to all for the reply!

  1 Comment

See Matt J's answer for the simplest and most efficient solution.

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3 Answers

Answer by KALYAN ACHARJYA on 13 Sep 2019
Edited by KALYAN ACHARJYA on 13 Sep 2019
 Accepted Answer

"i have 2 different matrix with same dimensions and i need to know the min value against 2 values in the same position:"
A=randi(10,[2,5]) % Random 2x5 A Matrix
B=randi(10,[2,5]) % Random 2x5 B Matrix
result=(A<=B).*A+(A>B).*B

  8 Comments

Yes @Madhan
the second matrix (B) isn't made only of 1...i wrote that for an example, but it could be anything (with the same size row of A)
mmh thinking on this problem i can simply copy the same row to make a matrix with same dimension of the first, then simply add :)
u are teaching to me how to think on matrix!

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Matt J
Answer by Matt J
on 13 Sep 2019

min(A(1,:), B(1,:))

  1 Comment

Gives the same as KALYAN ACHARJYA 's answer:
min(A,B)

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Answer by Dario Riviezzo on 13 Sep 2019

still need this:
i also have to use the elements of a matrix to do a scalar product using each element in a position and inserting it into a vector:
for i=1:length(A)
for j=1:N (size of rows)
ps1(i,j)=[A(i,j) 0]*[1 0]';
end
end
there is any way to do this without cycles?
i need the value in every position and that value is putted into a vector, then i make the scalar product with [1 0].
Tnx to all for the reply!

  1 Comment

Matt J
on 13 Sep 2019
But ps1(i,j)=[A(i,j) 0]*[1 0]'; is the same as ps1(i,j)=A(i,j). Example,
>> [3,0]*[1,0]'
ans =
3
The whole loop is therefore equivalent to the trivial single command,
ps1=A;

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