Splitting a matrix in subsets??
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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 Commenti
Geoff Hayes
il 7 Ago 2014
Modificato: Geoff Hayes
il 7 Ago 2014
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
il 11 Ago 2014
Risposte (2)
Iain
il 7 Ago 2014
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 Commenti
Iain
il 7 Ago 2014
Change the 1:2:end-1 etc. with the appropriate list.
Usman Ali
il 7 Ago 2014
Iain
il 7 Ago 2014
To match your example:
The first list (for M1) is: [1 3 5 6 8]
The second list (for M2) is: [2 4 5 6 8]
Why that matches your example, I do not know.
Usman Ali
il 7 Ago 2014
Iain
il 11 Ago 2014
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);
Adam
il 11 Ago 2014
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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