Matlab Array Formation From Data Set

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Calum
Calum 2014 年 9 月 25 日
回答済み: Guillaume 2014 年 10 月 24 日
I have an array of data e.g:
A = [300 10 1; 100 20 2; 100 30 4; 200 30 5;]
I would like to compile the third column data from in a new array as follows:
B = [1 0 0; 0 2 0; 0 4 5;]
where the X and Y 'scales' of array B are known and match the values found in columns 1 and 2 of array A:
X = [300 100 200]
Y = [10 20 30]
Any help is much appreciated,
Regards,
CS
  2 件のコメント
dpb
dpb 2014 年 9 月 25 日
編集済み: dpb 2014 年 9 月 25 日
1)
size(A)==[4,3] but size(B)==[3,3] ???
2)
Show/define the process by which you "scale" to get the values in B
I can make a reasonable fixup for 1) by presuming
A(3,:)=[];
to remove the third row. I can then make a first stab for 2) for
B(:,1)=fix(A(:,1)/X(1));
but I see no such similar relationship to the desired results shown for the other two columns...
Guillaume
Guillaume 2014 年 9 月 25 日
編集済み: Guillaume 2014 年 9 月 25 日
What happens when two rows of A have got the same X and Y?

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回答 (2 件)

Michael Haderlein
Michael Haderlein 2014 年 9 月 25 日
編集済み: Michael Haderlein 2014 年 9 月 25 日
Ha, got it:
X=[300 100 200];Y=[10 20 30];
A=[300 10 1;100 20 2;100 30 4;200 30 5];
[x,y]=meshgrid(X,Y)
[C,ia,ib]=intersect([x(:) y(:)],A(:,1:2),'rows')
B=zeros(length(Y),length(X));
B(ia)=A(ib,3)
B =
1 0 0
0 2 0
0 4 5
Maybe there's a shorter solution, but this one works fine.
  2 件のコメント
dpb
dpb 2014 年 9 月 25 日
Very creative from the input!!! :)
Calum
Calum 2014 年 10 月 24 日
Perfect! Thanks :)

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Guillaume
Guillaume 2014 年 10 月 24 日
The solution I had in mind when I commented on the question was:
X=[300 100 200];Y=[10 20 30];
A=[300 10 1;100 20 2;100 30 4;200 30 5];
[~, col] = ismember(A(:, 1), X);
[~, row] = ismember(A(:, 2), Y);
B=zeros(numel(Y), numel(X));
B(sub2ind(size(B), row, col)) = A(:, 3);
However, my question still stands: 'What happens when two rows of A have got the same X and Y?' The above will simply use the latest value.

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