Using find to compare matrices
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Hi
I have a matrix where the first column contains id's and the second some sort of information about thes id's, as an example such as below. What I want to do now, is to be able to say which information belongs to which id, for instance, for the example below, the id 1 has the "values" 5 and 9. So I would like to write code such that I could return for each id it's values. Now I have written that:
Y = [1,2,3,3,1; 5,6,7,8,9]'
U = unique(Y(:,1))
for i= 1:length(U)
[A B] = find(U(i)==Y(:,1))
end
Where the column with the 1,2,3,3,1 is storing these id's. Does that make any sense? What do I actually get returned in the [A B]?
Thanks a lot
采纳的回答
You do not need find
Y = [1,2,3,3,1; 5,6,7,8,9]'
Y((Y(:,1)==1),2) % id =1
Y((Y(:,1)==2),2) % id=2
Y((Y(:,1)==3),2) % id=3
4 个评论
MiauMiau
2012-12-12
Thomas
2012-12-12
This should get all the values
Y = [1001,1010,1011,1011,1001; 5,6,7,8,9]'
U=unique(Y(:,1))
for ii=1:numel(U)
Y((Y(:,1)==U(ii)),2)
end
MiauMiau
2012-12-12
In your case it give the row and column number as answers to the find
eg for id=1
you get
A =
1
5
B =
1
1
i.e. 1 is present in y in row 1 column 1 and row 5 column 1
the generic form of find is
[row,col] = find(X, ...)
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