48 Dimensional Euclidean Distance.

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Tim Mottram
Tim Mottram 2012-11-8
Hi Guys, Hopefully this will be a quicky. If I have two points in two dimensional space I find the distance as:
sqrt((x1-x2)^2 + (y1-y2)^2)
This is all good and well, but if I want the distance when each point has 48 (or otherwise) coordinates I don't really want to type out that equation. So is there some MatLab function which will let me pass it two arrays and find the resultant distance?
Thanks in advance,
Tim.
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Tim Mottram
Tim Mottram 2012-11-8
I have found the answer, Sorry, I was being silly.
Distance = sqrt(sum((cords2-cords1).^2));

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采纳的回答

Jan
Jan 2012-11-8
编辑：Jan 2012-11-8
Most of all the components should not be stored in x1, y1, z1, ... but is vectors x1=[1 x 48], x2=[1 x48]. Then some alternatives:
x1 = rand(1, 48);
x2 = rand(1, 48);
x12 = x2 - x1;
Dist1 = sqrt(sum(x12 .^ 2))
Dist2 = sqrt(x12 * x12.') % DOT product calculates the sum internally
Dist3 = norm(x12)

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