Dynamic Matrices in Matlab

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MiauMiau
MiauMiau 2012 年 12 月 13 日
Hi!
Yesterday, I had a very similar question, now I want something a bit more advanced. So I have coded already:
data = [234,3,8;457,5,7;611,7,3;234,9,12]
ID = data(:,1);
X = data(:,2);
U = unique(ID)
data(U(i) == data(:,1),3)
display data
Now I want to create a matrix, where the first column would be the first column of the data matrix above (i.e. "id"). And all the following collumns would then store the values in the third column of data belonging to this id. So it would look like:
[234, 8,12; 457, 7,0; --- ]
But, obviously, I could not set the dimensions of this matrix from the beginning on, the number of columns would have to be dynamical. Is that possible?
Thanks
  2 件のコメント
Azzi Abdelmalek
Azzi Abdelmalek 2012 年 12 月 13 日
It's not clear
MiauMiau
MiauMiau 2012 年 12 月 13 日
My question is not clear? So, I would like to have a matrix as follows:
[ 234,8,12; 457,7,0; 611,3,0]
the first column would be the id - and the following columns of the same row would be the third elements in the data matrix belonging to this id. As in the example, it is not clear, how many third column elements each id has - so the number of columns of the matrix would have to be dynamical. Is that possible to programm in matlab?

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回答 (2 件)

Andrei Bobrov
Andrei Bobrov 2012 年 12 月 13 日
編集済み: Andrei Bobrov 2012 年 12 月 13 日
[a,b,c] = unique(data(:,1));
d = accumarray(c,data(:,3),[],@(x){sort(x)});
n = cellfun(@numel,d);
m = numel(a);
out = [a, zeros(m,max(n))];
for j1 = 1:m
out(j1,2:n(j1)+1) = d{j1};
end
  2 件のコメント
Andrei Bobrov
Andrei Bobrov 2012 年 12 月 13 日
corrected
MiauMiau
MiauMiau 2012 年 12 月 13 日
the output I get is:
>> test
data =
234 3 8
457 5 7
611 7 3
234 9 12
so the data matrix itself...

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Walter Roberson
Walter Roberson 2012 年 12 月 13 日
Is the number of matching elements possibly different between the id's? If it is, then you cannot do this with a numeric matrix and need to use a cell array instead.

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