如何计算没有环的矩阵的欧几里得长度? [英] How to calculate Euclidean length of a matrix without loops?

问题描述

似乎答案很简单，但我很沮丧.我有一个Nx3矩阵的矩阵，其中第1个第2列和第3列是第n个项目的X Y和Z坐标.我想计算从原点到项目的距离.以非矢量形式，这很容易.

It seems like the answer to this should be simple, but I am stumped. I have a matrix of Nx3 matrix where there 1st 2nd and 3rd columns are the X Y and Z coordinates of the nth item. I want to calculate the distance from the origin to the item. In a non vectorized form this is easy.

distance = norm([x y z]);

distance = norm([x y z]);

或

距离= sqrt(x ^ 2 + y ^ 2 + z ^ 2);

distance = sqrt(x^2+y^2+z^2);

但是，以矢量化形式，它并不是那么简单.当您将矩阵传递给范数时，它不再返回欧几里得长度.

However, in vectorized form its not so simple. When you pass a matrix to norm it no longer returns the Euclidean length.

distance = norm(matrix); ％不起作用

distance = norm(matrix); %doesn't work

和

距离= sqrt(x(:，1).* x(:，1)+ y(:，2).* y(:，2)+ z(:，3).* z(:，3 ))；只是看起来很乱

distance = sqrt(x(:,1).*x(:,1)+y(:,2).*y(:,2)+z(:,3).*z(:,3)); %just seems messy

有更好的方法吗?

推荐答案

尝试一下:

>> xyz = [1 2 3; 4 5 6; 7 8 9; 2 8 4]

xyz =

     1     2     3
     4     5     6
     7     8     9
     2     8     4

>> distance = sqrt(sum(xyz.^2, 2))

distance =

          3.74165738677394
          8.77496438739212
          13.9283882771841
          9.16515138991168

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