confront values in 2 matrix

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!

[Stephen23]

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

Answer 1

[KALYAN ACHARJYA]

"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

[Dario Riviezzo]

wow it works!!! tnx a lot!

[KALYAN ACHARJYA]

My pleasure

[Dario Riviezzo]

if i have a matrix and a vector, can i sum the single elements of the vector on each row of a matrix without a cycle?

A=[ 0 1 3 5 6]

[2 3 5 8 9]

B[1 1 1 1 1]

C=[1 2 4 6 7]

[3 4 6 9 10]

[KALYAN ACHARJYA]

If the both matrices having same size (A and B)

A=[0 1 3 5 6;2 3 5 8 9]
[r c]=size(A)
B=ones(r,c)
C=A+B

If not (same size), one option to zero pad, then add.

[madhan ravi]

Kalyan is missing the uniqueness of MATLAB , it’s simply:

A+B % bsxfun() prior to 2016b

[KALYAN ACHARJYA]

Yes @Madhan

[Dario Riviezzo]

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)

[Dario Riviezzo]

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!

Answer 2

[Matt J]

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

[Stephen23]

Gives the same as KALYAN ACHARJYA 's answer:

min(A,B)

Answer 3

[Dario Riviezzo]

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!

[Matt J]

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;