Compare elements from two matrix.

2020 5 13
2 个回答
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更新时间：2020 5 15
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0 个投票

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

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 采纳的回答

Tommy
Tommy 2020-5-13

0 个投票

How about this?
x = x1;
idx = abs(x1) < abs(x2);
x(idx) = x2(idx);

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Nikita Zyk
Nikita Zyk 2020-5-14
编辑：Nikita Zyk 2020-5-14
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
Nikita Zyk 2020-5-14
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
Tommy 2020-5-14
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
Nikita Zyk 2020-5-14
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
Tommy 2020-5-14
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
Nikita Zyk 2020-5-15
It's good! Thank you so much! ;)

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更多回答（1 个）

Olawale Ikuyajolu
Olawale Ikuyajolu 2020-5-13

0 个投票

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

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