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Splitting a matrix in subsets??

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Usman  Ali
Usman Ali 2014 年 8 月 7 日
コメント済み: Iain 2014 年 8 月 11 日
Dear experts, I have a scenario matrix from which I would like to generate a binary tree, the binary tree structure matrix is as follow:
if true
% code
M= [8 25 25 25 25 25;
8 25 36 36 36 36;
8 25 25 15 15 15;
8 25 36 50 50 50;
8 25 25 10 10 10;
8 25 25 6 6 6;
8 25 36 70 70 70;
8 25 25 3 3 3];
end
where 1st and 2nd row are mother scenarios and the rest are the branching scenarios. I would like to seperate this matrix to these two scenario that which rows is a subset of which mother row. so that the end result should be
if true
%
M1 = [8 25 25 25 25 25;
8 25 25 15 15 15;
8 25 25 10 10 10;
8 25 25 6 6 6;
8 25 25 3 3 3];
%
M2 = [8 25 36 36 36 36;
8 25 36 50 50 50;
8 25 36 70 70 70];
end
  2 件のコメント
Geoff Hayes
Geoff Hayes 2014 年 8 月 7 日
編集済み: Geoff Hayes 2014 年 8 月 7 日
Usman - in your example, M1 and M2 have three rows in common. How are you determining which row belongs to which mother row?
Usman  Ali
Usman Ali 2014 年 8 月 11 日
Ohhhh my Bad :( , let me correct it. Actually the M1 includes all the branches of [8 25 25 25 25 25] and M2 includes all the branches connected to [8 25 36 36 36 36].
if true
% code
M1 = [8 25 25 25 25 25; 8 25 25 15 15 15;
8 25 25 10 10 10; 8 25 25 6 6 6;
8 25 25 3 3 3];
%
M2 = [8 25 36 36 36 36; 8 25 36 50 50 50;
8 25 36 70 70 70];
end

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回答 (2 件)

Iain
Iain 2014 年 8 月 7 日
M = [8 25 25 25 25 25; 8 25 36 36 36 36; 8 25 25 15 15 15;
8 25 36 50 50 50; 8 25 25 10 10 10; 8 25 25 6 6 6;
8 25 36 70 70 70; 8 25 25 3 3 3];
M1 = M(1:2:(end-1),:);
M2 = M(2:2:(end),:);
  5 件のコメント
Usman  Ali
Usman Ali 2014 年 8 月 7 日
I guess your ans is some what manual. the posted matrix is just an example, in real i have a very big matrix of [7000 x 50]. what if the order of rows is changed that's why I need some automatic solution that should compare the selected rows with mother row and splitt them into different subsets.
Iain
Iain 2014 年 8 月 11 日
You need some logic that fills in the lists.
If the first list is defined by rows that have 25 in the 3rd column, then the first list is:
list1 = find(M(:,3) == 25);

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Adam
Adam 2014 年 8 月 11 日
M1 = M( M(:,3) == 25,:)
M2 = M( M(:,3) == 36,:)
would seem to do the job in that case though that assumes that column 3 contains the value that distinguishes the two branching paths, irrespective of the values in the other columns.

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