How to construct a binary matrix reporting 1 in case of equal rows of two arrays of different dimensions?

MRC
2014 8 6
2 个回答
回答已采纳
更新时间：2014 8 6
3 次查看（30 天）

0 个投票

Hi, I have a matrix nxk
A=[ 1 2; 3 4; 5 6; 7 8]
and a matrix B mxk (m can be > = or < n)
B=[ 2 3; 4 5; 1 2; 5 6; 10 23; 7 8]
Each row of B is different.
I want to construct a matrix C nxm in which the hi-th element is 1 if A(h,:)=B(i,:), i.e.
C=[0 0 1 0 0 0; 0 0 0 0 0 0; 0 0 0 1 0 0; 0 0 0 0 0 1];
without looping.

请先登录，再回答此问题。

 采纳的回答

Ashish Gudla
Ashish Gudla 2014-8-6
编辑：Ashish Gudla 2014-8-6

0 个投票

You could create the 'C' matrix with all zeros and then, find the positions where its supposed to be '1' and replace it.
To find the positions where there would be '1', you can use "ismember" function with rows( see doc ) to get the lowest index of each row in A that appears in B.
[a1,~] = size(A);
[b1,~] = size(B);
[~,t2]=ismember(A,B,'rows');
t1 = 1:a1;
t1= t1(t2~=0); %ignore zero indices
t2 = t2(t2~=0);
C = zeros(a1,b1);
i = sub2ind(size(C),t1,t2');
C(i) =1;

更多回答（1 个）

Chris Turnes
Chris Turnes 2014-8-6
编辑：Chris Turnes 2014-8-6

1 个投票

One way to do this without looping would be to exploit the property that two vectors w and v are equal if and only if
Using this, you could construct each term and compare:
>> IP = A*B.';
>> An = sum(abs(A).^2, 2); % abs only necessary if data is complex
>> Bn = sum(abs(B).^2, 2);
>> C = (IP == An*ones(1, length(Bn))) & (IP == ones(length(An), 1)*Bn.');
This is a way of vectorizing the operation, but it will use much more memory than looping would.

类别

在 帮助中心 和 File Exchange 中查找有关 Resizing and Reshaping Matrices 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by