## combining matrix rows with repeated elements

さんによって質問されました 2014 年 7 月 31 日

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]}

#### 5 件のコメント

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

### dpb (view profile)

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"... :)

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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## 1 件の回答

### Jian Wei (view profile)

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.
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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