How can I compute all the possible differences of the vectors in a matrix?
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Gianluca Ligresti
2018 年 11 月 28 日
コメント済み: Gianluca Ligresti
2018 年 11 月 29 日
Hi all,
I have a problem: given a matrix A such as
A = [1,2,3;
4,0,1;
2,3,5;
6,0,1];
is there any matlab function that returns a matrix with all the possible differences of rows? I mean: a matrix in which I have in the first row (1,2,3)-(4,0,1)=(-3,2,2), in the second row (1,2,3)-(2,3,5)=(-1,-1,-2) and so forth for all the rows? In other words I want to compute all the possible combinations of differences.
Thank you very much
2 件のコメント
John D'Errico
2018 年 11 月 28 日
Is row1 - row2 different from row2 - row1?
Do you want to compute the difference row1-row1?
These are important factors which you must answer for a useful solution.
採用された回答
Guillaume
2018 年 11 月 28 日
result = A - permute(A, [3 2 1])
would return a size(A, 1) x size(A, 2) x size(A, 1) array where result(i, :, j) is the difference between A(i, :) and A(j, :)
5 件のコメント
Guillaume
2018 年 11 月 29 日
"A matrix A that is very very big (20.000 rows x 300 columns) so with this code I obtain an 'OUT OF MEMORY' error."
Regardless of the method used to generate all the combinations, you will end up with 20,000 x 19,999 combinations = 399,980,000 combinations (with 300 columns each). This will requires around 890 GB of memory to store. Even if you could store that, whatever processing you want to do with these combinations would probably take a long time.
"Is there a way to avoid this problem?"
The only way to avoid this problem is to change your algorithm so that you don't require that many combinations.
その他の回答 (2 件)
Honglei Chen
2018 年 11 月 28 日
If you have Statistics Toolbox, you can try
idx = flipud(combnk(1:4,2))
A(idx(:,1),:)-A(idx(:,2),:)
HTH
2 件のコメント
Honglei Chen
2018 年 11 月 28 日
Then you can do this
idx = combnk(1:4,2)
idx = sortrows([idx;fliplr(idx)])
A(idx(:,1),:)-A(idx(:,2),:)
HTH
Bruno Luong
2018 年 11 月 29 日
編集済み: Bruno Luong
2018 年 11 月 29 日
>> A = [1,2,3;
4,0,1;
2,3,5;
6,0,1];
>> r2 = nchoosek(1:size(A,1),2)
r2 =
1 2
1 3
1 4
2 3
2 4
3 4
>> A(r2(:,1),:)-A(r2(:,2),:)
ans =
-3 2 2
-1 -1 -2
-5 2 2
2 -3 -4
-2 0 0
-4 3 4
If you want to difference of 2 rows AND the opposite
>> R2 = sortrows([r2;fliplr(r2)])
R2 =
1 2
1 3
1 4
2 1
2 3
2 4
3 1
3 2
3 4
4 1
4 2
4 3
>> A(R2(:,1),:)-A(R2(:,2),:)
ans =
-3 2 2
-1 -1 -2
-5 2 2
3 -2 -2
2 -3 -4
-2 0 0
1 1 2
-2 3 4
-4 3 4
5 -2 -2
2 0 0
4 -3 -4
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