Question

我有两个大矩阵。矩阵A为[4144514×3]，矩阵B为[51962×17]。

A和B的前三列有标识符。在矩阵B上，3列构成唯一标识符，但这可能会在A上重复。

我想合并这两个矩阵，得到的矩阵A应该是[4144514 x 20]，即我在给出前三列标准的情况下将B的17列与矩阵A合并每个矩阵。

这是我正在做的循环：

for i=1:size(B,1)
    aux = sum(A(:,1)==B(i,1) & A(:,2)==B(i,2) & A(:,3) == B(i,3));
    A(A(:,1)==B(i,1) & A(:,2)==B(i,2) & A(:,3) == B(i,3),4:20) = repmat(B(i,:),aux,1);
end

变量aux告诉我A中有多少行符合B的条件。

循环的第二行创建一个矩阵，其B列重复aux次，并将它们放在A中的正确位置。然而，这是非常低效的。

让我用下面的小矩阵给你一个玩具示例。我将A [108 x 3]和B为[27 x 17]。

我要做的是以下内容：

A = [100, 1, 2000 ;100, 1, 2000 ;100, 1, 2000 ;100, 1, 2000 ;100, 2, 2000 ;100, 2, 2000 ;100, 2, 2000 ;100, 2, 2000 ;100, 3, 2000 ;100, 3, 2000 ;100, 3, 2000 ;100, 3, 2000 ;100, 1, 2001 ;100, 1, 2001 ;100, 1, 2001 ;100, 1, 2001 ;100, 2, 2001 ;100, 2, 2001 ;100, 2, 2001 ;100, 2, 2001 ;100, 3, 2001 ;100, 3, 2001 ;100, 3, 2001 ;100, 3, 2001 ;100, 1, 2002 ;100, 1, 2002 ;100, 1, 2002 ;100, 1, 2002 ;100, 2, 2002 ;100, 2, 2002 ;100, 2, 2002 ;100, 2, 2002 ;100, 3, 2002 ;100, 3, 2002 ;100, 3, 2002 ;100, 3, 2002 ;101, 1, 2000 ;101, 1, 2000 ;101, 1, 2000 ;101, 1, 2000 ;101, 2, 2000 ;101, 2, 2000 ;101, 2, 2000 ;101, 2, 2000 ;101, 3, 2000 ;101, 3, 2000 ;101, 3, 2000 ;101, 3, 2000 ;101, 1, 2001 ;101, 1, 2001 ;101, 1, 2001 ;101, 1, 2001 ;101, 2, 2001 ;101, 2, 2001 ;101, 2, 2001 ;101, 2, 2001 ;101, 3, 2001 ;101, 3, 2001 ;101, 3, 2001 ;101, 3, 2001 ;101, 1, 2002 ;101, 1, 2002 ;101, 1, 2002 ;101, 1, 2002 ;101, 2, 2002 ;101, 2, 2002 ;101, 2, 2002 ;101, 2, 2002 ;101, 3, 2002 ;101, 3, 2002 ;101, 3, 2002 ;101, 3, 2002 ;103, 1, 2000 ;103, 1, 2000 ;103, 1, 2000 ;103, 1, 2000 ;103, 2, 2000 ;103, 2, 2000 ;103, 2, 2000 ;103, 2, 2000 ;103, 3, 2000 ;103, 3, 2000 ;103, 3, 2000 ;103, 3, 2000 ;103, 1, 2001 ;103, 1, 2001 ;103, 1, 2001 ;103, 1, 2001 ;103, 2, 2001 ;103, 2, 2001 ;103, 2, 2001 ;103, 2, 2001 ;103, 3, 2001 ;103, 3, 2001 ;103, 3, 2001 ;103, 3, 2001 ;103, 1, 2002 ;103, 1, 2002 ;103, 1, 2002 ;103, 1, 2002 ;103, 2, 2002 ;103, 2, 2002 ;103, 2, 2002 ;103, 2, 2002 ;103, 3, 2002 ;103, 3, 2002 ;103, 3, 2002 ;103, 3, 2002];
B = [100, 1, 2000, 8, 7, 9, 10, 1, 2, 9, 2, 1, 3, 3, 3, 9, 7; 100, 2, 2000, 8, 2, 7, 2, 7, 5, 5, 9, 2, 7, 1, 2, 6, 4; 100, 3, 2000, 8, 8, 7, 3, 2, 8, 1, 10, 9, 8, 6, 1, 5, 7; 100, 1, 2001, 10, 10, 1, 7, 2, 5, 5, 8, 6, 5, 3, 6, 6, 4; 100, 2, 2001, 6, 7, 3, 1, 5, 3, 9, 9, 3, 8, 1, 6, 4, 4; 100, 3, 2001, 1, 5, 7, 1, 5, 4, 2, 10, 5, 4, 6, 5, 1, 10; 100, 1, 2002, 7, 4, 6, 4, 7, 8, 3, 7, 7, 8, 2, 1, 6, 2; 100, 2, 2002, 8, 7, 8, 10, 2, 10, 8, 4, 7, 5, 10, 4, 4, 2; 100, 3, 2002, 4, 9, 4, 10, 2, 4, 2, 1, 4, 10, 9, 2, 6, 9; 101, 1, 2000, 5, 9, 5, 3, 10, 1, 4, 2, 10, 2, 6, 8, 5, 4; 101, 2, 2000, 6, 1, 8, 10, 10, 7, 4, 6, 5, 2, 8, 3, 2, 1; 101, 3, 2000, 7, 7, 5, 6, 2, 8, 8, 8, 1, 6, 1, 1, 9, 7; 101, 1, 2001, 8, 4, 5, 5, 8, 7, 2, 2, 9, 8, 1, 4, 2, 1; 101, 2, 2001, 3, 7, 10, 4, 9, 9, 1, 1, 10, 7, 6, 5, 10, 9; 101, 3, 2001, 10, 8, 2, 4, 6, 1, 6, 4, 8, 10, 7, 9, 4, 7; 101, 1, 2002, 6, 9, 3, 2, 10, 8, 5, 2, 6, 9, 1, 3, 6, 6; 101, 2, 2002, 8, 3, 8, 4, 4, 3, 4, 2, 4, 7, 2, 10, 3, 3; 101, 3, 2002, 10, 2, 10, 10, 6, 5, 3, 5, 10, 1, 3, 4, 8, 5; 103, 1, 2000, 8, 3, 9, 9, 4, 3, 3, 9, 7, 7, 6, 5, 2, 6; 103, 2, 2000, 6, 7, 5, 5, 7, 10, 5, 3, 5, 4, 7, 8, 9, 7; 103, 3, 2000, 3, 4, 9, 10, 3, 10, 7, 2, 10, 3, 3, 3, 6, 6; 103, 1, 2001, 7, 7, 1, 7, 10, 7, 10, 7, 9, 8, 4, 7, 6, 2; 103, 2, 2001, 5, 5, 4, 3, 7, 7, 6, 5, 2, 5, 5, 6, 6, 5; 103, 3, 2001, 6, 3, 9, 9, 2, 10, 10, 10, 10, 7, 10, 9, 9, 8; 103, 1, 2002, 5, 10, 2, 8, 6, 5, 7, 6, 4, 3, 6, 8, 7, 4; 103, 2, 2002, 10, 7, 6, 3, 10, 4, 5, 5, 1, 3, 1, 9, 1, 5; 103, 3, 2002, 2, 1, 5, 5, 2, 8, 6, 2, 6, 6, 10, 1, 4, 9];

for i=1:size(B,1)
        aux = sum(A(:,1)==B(i,1) & A(:,2)==B(i,2) & A(:,3) == B(i,3));
        A(A(:,1)==B(i,1) & A(:,2)==B(i,2) & A(:,3) == B(i,3),4:20) = repmat(B(i,:),aux,1);
end

使用这个较小的示例，代码运行得非常快。但是只要矩阵变得像我的那么大，就需要很长时间。有没有更快的方法呢？

Answer 1

一个非常简单的矢量化任务，ismember完成所有工作。

%your example
A = [100, 1, 2000 ;100, 1, 2000 ;100, 1, 2000 ;100, 1, 2000 ;100, 2, 2000 ;100, 2, 2000 ;100, 2, 2000 ;100, 2, 2000 ;100, 3, 2000 ;100, 3, 2000 ;100, 3, 2000 ;100, 3, 2000 ;100, 1, 2001 ;100, 1, 2001 ;100, 1, 2001 ;100, 1, 2001 ;100, 2, 2001 ;100, 2, 2001 ;100, 2, 2001 ;100, 2, 2001 ;100, 3, 2001 ;100, 3, 2001 ;100, 3, 2001 ;100, 3, 2001 ;100, 1, 2002 ;100, 1, 2002 ;100, 1, 2002 ;100, 1, 2002 ;100, 2, 2002 ;100, 2, 2002 ;100, 2, 2002 ;100, 2, 2002 ;100, 3, 2002 ;100, 3, 2002 ;100, 3, 2002 ;100, 3, 2002 ;101, 1, 2000 ;101, 1, 2000 ;101, 1, 2000 ;101, 1, 2000 ;101, 2, 2000 ;101, 2, 2000 ;101, 2, 2000 ;101, 2, 2000 ;101, 3, 2000 ;101, 3, 2000 ;101, 3, 2000 ;101, 3, 2000 ;101, 1, 2001 ;101, 1, 2001 ;101, 1, 2001 ;101, 1, 2001 ;101, 2, 2001 ;101, 2, 2001 ;101, 2, 2001 ;101, 2, 2001 ;101, 3, 2001 ;101, 3, 2001 ;101, 3, 2001 ;101, 3, 2001 ;101, 1, 2002 ;101, 1, 2002 ;101, 1, 2002 ;101, 1, 2002 ;101, 2, 2002 ;101, 2, 2002 ;101, 2, 2002 ;101, 2, 2002 ;101, 3, 2002 ;101, 3, 2002 ;101, 3, 2002 ;101, 3, 2002 ;103, 1, 2000 ;103, 1, 2000 ;103, 1, 2000 ;103, 1, 2000 ;103, 2, 2000 ;103, 2, 2000 ;103, 2, 2000 ;103, 2, 2000 ;103, 3, 2000 ;103, 3, 2000 ;103, 3, 2000 ;103, 3, 2000 ;103, 1, 2001 ;103, 1, 2001 ;103, 1, 2001 ;103, 1, 2001 ;103, 2, 2001 ;103, 2, 2001 ;103, 2, 2001 ;103, 2, 2001 ;103, 3, 2001 ;103, 3, 2001 ;103, 3, 2001 ;103, 3, 2001 ;103, 1, 2002 ;103, 1, 2002 ;103, 1, 2002 ;103, 1, 2002 ;103, 2, 2002 ;103, 2, 2002 ;103, 2, 2002 ;103, 2, 2002 ;103, 3, 2002 ;103, 3, 2002 ;103, 3, 2002 ;103, 3, 2002];
B = [100, 1, 2000, 8, 7, 9, 10, 1, 2, 9, 2, 1, 3, 3, 3, 9, 7; 100, 2, 2000, 8, 2, 7, 2, 7, 5, 5, 9, 2, 7, 1, 2, 6, 4; 100, 3, 2000, 8, 8, 7, 3, 2, 8, 1, 10, 9, 8, 6, 1, 5, 7; 100, 1, 2001, 10, 10, 1, 7, 2, 5, 5, 8, 6, 5, 3, 6, 6, 4; 100, 2, 2001, 6, 7, 3, 1, 5, 3, 9, 9, 3, 8, 1, 6, 4, 4; 100, 3, 2001, 1, 5, 7, 1, 5, 4, 2, 10, 5, 4, 6, 5, 1, 10; 100, 1, 2002, 7, 4, 6, 4, 7, 8, 3, 7, 7, 8, 2, 1, 6, 2; 100, 2, 2002, 8, 7, 8, 10, 2, 10, 8, 4, 7, 5, 10, 4, 4, 2; 100, 3, 2002, 4, 9, 4, 10, 2, 4, 2, 1, 4, 10, 9, 2, 6, 9; 101, 1, 2000, 5, 9, 5, 3, 10, 1, 4, 2, 10, 2, 6, 8, 5, 4; 101, 2, 2000, 6, 1, 8, 10, 10, 7, 4, 6, 5, 2, 8, 3, 2, 1; 101, 3, 2000, 7, 7, 5, 6, 2, 8, 8, 8, 1, 6, 1, 1, 9, 7; 101, 1, 2001, 8, 4, 5, 5, 8, 7, 2, 2, 9, 8, 1, 4, 2, 1; 101, 2, 2001, 3, 7, 10, 4, 9, 9, 1, 1, 10, 7, 6, 5, 10, 9; 101, 3, 2001, 10, 8, 2, 4, 6, 1, 6, 4, 8, 10, 7, 9, 4, 7; 101, 1, 2002, 6, 9, 3, 2, 10, 8, 5, 2, 6, 9, 1, 3, 6, 6; 101, 2, 2002, 8, 3, 8, 4, 4, 3, 4, 2, 4, 7, 2, 10, 3, 3; 101, 3, 2002, 10, 2, 10, 10, 6, 5, 3, 5, 10, 1, 3, 4, 8, 5; 103, 1, 2000, 8, 3, 9, 9, 4, 3, 3, 9, 7, 7, 6, 5, 2, 6; 103, 2, 2000, 6, 7, 5, 5, 7, 10, 5, 3, 5, 4, 7, 8, 9, 7; 103, 3, 2000, 3, 4, 9, 10, 3, 10, 7, 2, 10, 3, 3, 3, 6, 6; 103, 1, 2001, 7, 7, 1, 7, 10, 7, 10, 7, 9, 8, 4, 7, 6, 2; 103, 2, 2001, 5, 5, 4, 3, 7, 7, 6, 5, 2, 5, 5, 6, 6, 5; 103, 3, 2001, 6, 3, 9, 9, 2, 10, 10, 10, 10, 7, 10, 9, 9, 8; 103, 1, 2002, 5, 10, 2, 8, 6, 5, 7, 6, 4, 3, 6, 8, 7, 4; 103, 2, 2002, 10, 7, 6, 3, 10, 4, 5, 5, 1, 3, 1, 9, 1, 5; 103, 3, 2002, 2, 1, 5, 5, 2, 8, 6, 2, 6, 6, 10, 1, 4, 9];
%reference code, changed output to C to preserve input
C=A;
for i=1:size(B,1)
        aux = sum(A(:,1)==B(i,1) & A(:,2)==B(i,2) & A(:,3) == B(i,3));
        C(A(:,1)==B(i,1) & A(:,2)==B(i,2) & A(:,3) == B(i,3),4:20) = repmat(B(i,:),aux,1);
end
%vectorized version
[~,lia]=ismember(A(:,1:3),B(:,1:3),'rows');
C2=nan(numel(lia),size(B,2)+3);
C2(lia>0,:)=B(lia(lia>0),[1,2,3,1:end]);
toc;
tic;
%simplified vectorized version, assuming there is no need to duplicate the first three rows:
[~,lia]=ismember(A(:,1:3),B(:,1:3),'rows');
C3=nan(numel(lia),size(B,2));
C3(lia>0,:)=B(lia(lia>0),:);
toc;

将性能与接近真实数据的一些示例数据进行比较（相同的数据很大）：

n=100000 %4144514
m=10000 %51962
B=rand(10000,17);
B=unique(B,'rows');
A=B(randi([1 size(B,1)],100000,1),1:3);

将执行时间从14秒减少到0.05秒

Matlab - 非常低效的循环

1 个答案: