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Javad
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combining matrix rows with repeated elements

Javad
さんによって質問されました 2014 年 7 月 31 日
最新アクティビティ Javad
さんによって コメントされました 2014 年 8 月 8 日
I have a matrix which its some rows have one same elements. I want to combine this rows and save the result in a cell array without repeating the elements. What is the easiest way to do so? For example, I have
A = [1 2;3 4;2 5;5 6;4 7;8 9;10 9]
And want to get,
C = {[1 2 5 6],[3 4 7],[8 9 10]}

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Javad
2014 年 7 月 31 日
Some rows have repeated elements (e.g. 2, 5), and I want to join all these rows which have one same element two by two, without repeating the same elements. For example the first matrix in C is the combination of A(1,:)=(1 2), A(3,:)=(2 5) and A(4,:)=(5 6) that have 2 as the same element in two first ones and 5 in two last ones.
dpb
2014 年 7 月 31 日
Which is what I said...start w/ the unique values you have then look where they're located and build the array therefrom...
>> [u,ia,ib]=unique(A(:),'stable');
>> u'
ans =
1 3 2 5 4 8 10 6 7 9
>> ia'
ans =
1 2 3 4 5 6 7 11 12 13
>> ib'
ans =
1 2 3 4 5 6 7 3 5 4 8 9 10 10
>>
As you can see, you get the list of values and their locations which you can use to select which ones go where...I'll leave that as "exercise for the student"... :)
Javad
2014 年 8 月 8 日
Thanks for your point. I had a look at documentations which say u=A(ia) and A=u(ib), but I could not figure out how I can relate indices to separate them as I want. As a beginner, it's a little difficult for me to do this exercise!

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回答者: Jian Wei
2014 年 7 月 31 日

If I understand it correctly, you are trying to cluster the vertices in a graph into different groups according to whether they are connected or not. For general cases, the problem might be nontrivial. Please try the following code to see if it generates the results you need.
% adjacency matrix
M = zeros(max(max(10)));
for k = 1:size(A,1)
M(A(k,1),A(k,2)) = 1;
M(A(k,2),A(k,1)) = 1;
end
flag = zeros(length(M),1);
C = cell(0);
for k = 1:length(M)
if flag(k)==0
p1 = k;
flag(k) = 1;
p2 = p1;
for ki = 1:length(p1)
node = p1(ki);
p2 = union(p2,find(M(node,:)==1));
end
while length(p2)~=length(p1)
p1 = p2;
p2 = p1;
for ki = 1:length(p1)
node = p1(ki);
p2 = union(p2,find(M(node,:)==1));
end
end
for ki = 1:length(p2)
flag(p2(ki)) = 1;
end
C{end+1} = p2;
end
end
C is the cell which contains different groups of vertices that are connected with each other.

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