Question

BigList = rand(20, 3)
LittleList = rand(5, 3)

我想在大列表中的每一行找到最近的＆＃39;小列表中的行，由欧几里德范数定义（即k = 3维中相应值之间的平方距离之和）。

我可以看到如何使用两个循环来做到这一点，但似乎应该有一种更好的方法来使用内置矩阵运算来做到这一点。

Answer 1

正确的方法当然是使用nearest-neighbor searching algorithms 但是，如果您的维度不是太高而且您的数据集不是很大，那么只需使用bsxfun：

d = bsxfun( @minus, permute( bigList, [1 3 2] ), permute( littleList, [3 1 2] ) ); %//diff in third dimension
d = sum( d.^2, 3 ); %// sq euclidean distance
[minDist minIdx] = min( d, [], 2 );

除了提出的here矩阵乘法方法之外，还有另一个没有循环的矩阵乘法

nb = sum( bigList.^2, 2 ); %// norm of bigList's items
nl = sum( littleList.^2, 2 ); %// norm of littleList's items
d = bsxfun( @sum, nb, nl.' ) - 2 * bigList * littleList'; %// all the distances

这种方法背后的观察是欧几里德距离（L2范数）

|| a - b ||^2 = ||a||^2 + ||b||^2 - 2<a,b>

<a,b>是两个向量的点积。

Answer 2

您可以使用bsxfun：

执行此操作

d = squeeze(sum((bsxfun(@minus, BigList, permute(LittleList, [3 2 1]))).^2, 2));
[~, ind] = min(d,[],2);

Answer 3

方法＃1

内置MATLAB函数 pdist2 ，找到"Pairwise distance between two sets of observations"。有了它，您可以计算欧氏距离矩阵，然后找到距离矩阵中适当维度的最小值索引，它代表bigList中每行littleList的“最接近”。

这是带衬里的单行 -

[~,minIdx] = min(pdist2(bigList,littleList),[],2); %// minIdx is what you are after

方法＃2

如果你关心性能，这里有一个利用fast matrix multiplication in MATLAB的方法此处提供的大部分代码均来自this smart solution。

dim = 3;
numA = size(bigList,1);
numB = size(littleList,1);

helpA = zeros(numA,3*dim);
helpB = zeros(numB,3*dim);
for idx = 1:dim
    helpA(:,3*idx-2:3*idx) = [ones(numA,1), -2*bigList(:,idx), bigList(:,idx).^2 ];
    helpB(:,3*idx-2:3*idx) = [littleList(:,idx).^2 ,    littleList(:,idx), ones(numB,1)];
end
[~,minIdx] = min(helpA * helpB',[],2); %//'# minIdx is what you are after

基准

基准代码 -

N1 = 1750; N2 = 4*N1; %/ datasize
littleList = rand(N1, 3);
bigList = rand(N2, 3);

for k = 1:50000
    tic(); elapsed = toc(); %// Warm up tic/toc
end

disp('------------- With squeeze + bsxfun + permute based approach [LuisMendo]')
tic
d = squeeze(sum((bsxfun(@minus, bigList, permute(littleList, [3 2 1]))).^2, 2));
[~, ind] = min(d,[],2);
toc,  clear d ind

disp('------------- With double permutes + bsxfun based approach [Shai]')
tic
d = bsxfun( @minus, permute( bigList, [1 3 2] ), permute( littleList, [3 1 2] ) ); %//diff in third dimension
d = sum( d.^2, 3 ); %// sq euclidean distance
[~,minIdx] = min( d, [], 2 );
toc
clear d minIdx

disp('------------- With bsxfun + matrix-multiplication based approach [Shai]')
tic
nb = sum( bigList.^2, 2 ); %// norm of bigList's items
nl = sum( littleList.^2, 2 ); %// norm of littleList's items
d = bsxfun(@plus, nb, nl.' ) - 2 * bigList * littleList'; %// all the distances
[~,minIdx] = min(d,[],2);
toc, clear nb nl d minIdx

disp('------------- With matrix multiplication based approach  [Divakar]')
tic
dim = 3;
numA = size(bigList,1);
numB = size(littleList,1);

helpA = zeros(numA,3*dim);
helpB = zeros(numB,3*dim);
for idx = 1:dim
    helpA(:,3*idx-2:3*idx) = [ones(numA,1), -2*bigList(:,idx), bigList(:,idx).^2 ];
    helpB(:,3*idx-2:3*idx) = [littleList(:,idx).^2 ,    littleList(:,idx), ones(numB,1)];
end
[~,minIdx] = min(helpA * helpB',[],2);
toc, clear dim numA numB helpA helpB idx minIdx

disp('------------- With pdist2 based approach [Divakar]')
tic
[~,minIdx] = min(pdist2(bigList,littleList),[],2);
toc, clear minIdx

基准测试结果 -

------------- With squeeze + bsxfun + permute based approach [LuisMendo]
Elapsed time is 0.718529 seconds.
------------- With double permutes + bsxfun based approach [Shai]
Elapsed time is 0.971690 seconds.
------------- With bsxfun + matrix-multiplication based approach [Shai]
Elapsed time is 0.328442 seconds.
------------- With matrix multiplication based approach  [Divakar]
Elapsed time is 0.159092 seconds.
------------- With pdist2 based approach [Divakar]
Elapsed time is 0.310850 seconds.

快速结论：Shai第二种方法的运行时间是bsxfun和矩阵乘法的组合，与基于pdist2的方法非常接近，没有明显的赢家可以在这两者之间做出决定。

如何矢量化从矢量中找出最近的点

3 个答案:

方法＃1

方法＃2

基准