Asked by Dario Riviezzo
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!

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

KALYAN ACHARJYA
on 13 Sep 2019

Yes @Madhan

Dario Riviezzo
on 13 Sep 2019

Dario Riviezzo
on 13 Sep 2019

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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Answer by Matt J
on 13 Sep 2019

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

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!

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## Stephen Cobeldick (view profile)

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https://uk.mathworks.com/matlabcentral/answers/480242-confront-values-in-2-matrix#comment_745561

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