Compare elements from two matrix.

Hi!

I want to compare elements for two matrix and then create another matrix with maximal element (comparing abs(x1) i abs(x2), not x1 i x2).

I wrote this:

But maybe it's possible to do it quicker and more efficient?

for i = 1:numel(x1) 
   if (abs(x1(i)) > abs(x2(i)))
       x(i) = x2(i); 
   else 
       x(i) = x1(i);
   end 
end 

Answer 1

[Tommy]

How about this?

x = x1;
idx = abs(x1) < abs(x2);
x(idx) = x2(idx);

[Nikita Zyk]

Yeah, it's good but I have 3% efficiency. Still don't understand why. Maybe you can help me?

I have to find roots of quadratic equation ax^2 + bx + c = 0 (it has to be well done for all values of a, b, c), where a,b,c - vectors, and x contains elements which absolute value is the smallest one (compare x1 and x2).

I wrote this:

function x = minroot(a,b,c) 

end
if (b < 0) 
    x1 = 2.*c./(-b + sqrt(b.^2 - 4.*a.*c));
    x2 = (-b + sqrt(b.^2 - 4.*a.*c))./2./a;
else 
    x1 = (-b - sqrt(b.^2 - 4.*a.*c))./2./a;
    x2 = 2.*c./(-b - sqrt(b.^2 - 4.*a.*c));
    
end 

x = x1;
idx = abs(x1) > abs(x2);
x(idx) = x2(idx);

[Nikita Zyk]

Forgot to add that I have testing program that gives me the percent of efficiency and accurancy. This code has 100% accurancy, but only 3% efficiency.

[Tommy]

3%? Hmm...

Well if b is an array, then

if (b < 0)

is asking whether all elements in b are less than 0. But I believe the two equations for x1 are equivalent, as are the two equations for x2 (as long as a and c are non-zero), so I don't really see the point of the if statement or the alternate form.

Does something like this improve the efficiency?

function x = minroot(a,b,c)
    sdel = sqrt(b.^2 - 4*a.*c);
    x = (-b - sdel)./(2*a);
    x2 = (-b + sdel)./(2*a);
    idx = abs(x) > abs(x2);
    x(idx) = x2(idx); 
end

If there is a possibility that one of the elements in a could be 0, then you would need to handle that case appropriately. One option would be to use the alternate form (provided c is not also 0); another would be to recognize that the sole root in that case is then -c/b. Either way, I don't see how the sign of b would matter...

[Nikita Zyk]

Yes, but accuracy wil be lesser. Why? Because of the lack of if statement for b: if b<0, so -b+sdel can effect the result, because for b~sdel we subtract two numbers that are really closer one to another.

[Tommy]

Ah okay thank you for the explanation!

It's not very pretty, but how well does this do?

function x = minroot(a,b,c)
    sdel = sqrt(b.^2 - 4*a.*c);
    
    idx = b < 0;
    x(idx) = 2*c(idx)./(-b(idx) + sdel(idx));
    x(~idx) = (-b(~idx) - sdel(~idx))./(2*a(~idx));
    x2(~idx) = 2*c(~idx)./(-b(~idx) - sdel(~idx));
    x2(idx) = (-b(idx) + sdel(idx))./(2*a(idx));
    
    idx = abs(x) > abs(x2);
    x(idx) = x2(idx);
end

[Nikita Zyk]

It's good! Thank you so much! ;)

Answer 2

[Olawale Ikuyajolu]

new_matrix = max(abs(x1),abs(x2);

[Nikita Zyk]

And in new_matrix there will be abs(x), not x. I need x.

[Olawale Ikuyajolu]

Check this accepted answer

[Nikita Zyk]

Unfortunetly, it has less efficiency. ;(