找到一个组成矩阵的指数位置 [英] finding index-positions of a composed-matrix
问题描述
我需要用我的code帮助。在code被用于查找的 minumin 的平方距离的问题。我通过一个实例，提供我的code，我相信这将是解释什么，我需要最简单的方法。
I need help with my code. The code is used to find the minumin of a square-distance problem. I am providing my code through an example, I believe this will be the easiest way to explain what I need.
clear all
clc
x=10.8; % is a fixed value
y=34; % is a fixed value
z=12; % is a fixed value
A = [11 14 1; 5 8 18; 10 8 19; 13 20 16]; % a (4x3) matrix
B = [2 3 10; 6 15 16; 7 3 15; 14 14 19]; % a (4x3) matrix
我创建了一个新的矩阵 C 这在本下列方式组成：
C1 = bsxfun(@minus, A(:,1)',B(:,1));
C1=C1(:); % this is the first column of the new matrix C
C2 = bsxfun(@minus, A(:,2)',B(:,2));
C2=C2(:); % this is the second column of the new matrix C
C3 = bsxfun(@minus, A(:,3)',B(:,3));
C3=C3(:); % this is the third column of the new matrix C
C = [C1 C2 C3]; % the new matrix C of size (16x3)
C 在这种方式中形成！而这也正是我的意思是，当我在我的标题写了组成的矩阵的
C has to be formed in this way! And this is what I meant when I wrote in my title a composed-matrix
然后：
[d,p] = min((C(:,1)-x).^2 + (C(:,2)-y).^2 + (C(:,3)-z).^2);
d = sqrt(d);
outputs:
d = 18.0289;
p = 13;
让我满足该的分的问题的距离（d）和位置（P）。
Gives me the distance (d) and position (p) which satisfies this min problem.
我的问题：
我需要找到哪些 A和B 的组合给了我这个 P 价值，换句话说，我需要的指数从'A，B'这给了我这样的优化的 C1，C2，C3 ：
MY PROBLEM:
I need to find which combinations of A and B has given my this p value, in other words I need the index from ´A,B´ which gives me this optimal C1,C2,C3:
C1 = bsxfun(@minus, A(?,1)',B(?,1));
C2 = bsxfun(@minus, A(?,2)',B(?,2));
C3 = bsxfun(@minus, A(?,3)',B(?,3));
的？是索引位置，我需要，在这种情况下，矩阵A的指数位置和B的索引位置。
The ? is the index position I need, in this case the index position of the matrix A and the index position of B.
手工计算我有如下图所示：
Calculated by hand I have the following illustration:
我知道：
C = [9 11 -9
5 -1 -15
4 11 -14
-3 0 -18
3 5 8
-1 -7 2
-2 5 3
-9 -6 -1
8 5 9
4 -7 3
3 5 4
-4 -6 0
11 17 6
7 5 0
6 17 1
-1 6 -3]
我知道，我的最佳指数的在13位给出。该指数持仓又回到了：
And I know that my optimal index is given in the position 13th. This index positions goes back to:
[13-2 20-3 16-10]
这是 A（4 :) - B（1：）
我需要一个code，可以帮助我找到A和B这个指标
I need a code which can help me to find this indexes from A and B
在此先感谢！
PS。我使用的常微分方程参数估计问题code。
PS. I am using the code in parameter estimation problems of ODEs.
推荐答案
第一种情况：矢量矩阵情况下
subvals = bsxfun(@minus,A,[x y z])
[distance,index] = min(sqrt(sum(subvals.^2,2)))
情况二：两个矩阵的情况下
subvals = bsxfun(@minus,A,permute(B,[3 2 1]));
[distances,indices] = min(sqrt(sum(subvals.^2,2)),[],3)
测试的第二种情况：
%%// Get some random data into A and B
A = randi(20,8,3)
B = randi(20,4,3)
%%// Just to test out out code for correctness,
%%// let us make any one one row of B, say 3rd row equal to
%%// any one row of A, say the 6th row -
B(3,:) = A(6,:)
%%// Use the earlier code
subvals = bsxfun(@minus,A,permute(B,[3 2 1]));
[distances,indices] = min(sqrt(sum(subvals.^2,2)),[],3)
%%// Get the minimum row index for A and B
[~,min_rowA] = min(distances)
min_rowB = indices(min_rowA)
验证
min_rowA =
6
min_rowB =
3
修改1 [响应张贴在问题简单的例子]：
标题说，你有兴趣在寻找两个矩阵的差异，然后找到一个向量[X Y Z]这之间的最短距离。所以，我希望这是你所需要的 -
The title says you are interested in finding the difference of two matrices and then find the shortest distance between it to a vector [x y z]. So I am hoping this is what you need -
x=10.8;
y=34;
z=12;
A = [11 14 1; 5 8 18; 10 8 19; 13 20 16];
B = [2 3 10; 6 15 16; 7 3 15; 14 14 19];
C = A -B; %%// Distance of two vectors as posted in title
subvals = bsxfun(@minus,C,[x y z])
[distance,index] = min(sqrt(sum(subvals.^2,2)))
输出
distance =
31.0780
index =
3
编辑2：当你这样做之后 -
Edit 2: After you have done this -
[d,p] = min((C(:,1)-x).^2 + (C(:,2)-y).^2 + (C(:,3)-z).^2);
如果你想找到A和B对应的指数，你可以做到这一点 -
If you are looking to find the corresponding indices of A and B , you may do this -
[minindex_alongB,minindex_alongA] = ind2sub(size(A),p)
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