How can I compute all the possible differences of the vectors in a matrix?
15 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Gianluca Ligresti
am 28 Nov. 2018
Kommentiert: Gianluca Ligresti
am 29 Nov. 2018
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 Kommentare
John D'Errico
am 28 Nov. 2018
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.
Akzeptierte Antwort
Guillaume
am 28 Nov. 2018
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 Kommentare
Guillaume
am 29 Nov. 2018
"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.
Weitere Antworten (2)
Honglei Chen
am 28 Nov. 2018
If you have Statistics Toolbox, you can try
idx = flipud(combnk(1:4,2))
A(idx(:,1),:)-A(idx(:,2),:)
HTH
2 Kommentare
Honglei Chen
am 28 Nov. 2018
Then you can do this
idx = combnk(1:4,2)
idx = sortrows([idx;fliplr(idx)])
A(idx(:,1),:)-A(idx(:,2),:)
HTH
Bruno Luong
am 29 Nov. 2018
Bearbeitet: Bruno Luong
am 29 Nov. 2018
>> 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
0 Kommentare
Siehe auch
Kategorien
Mehr zu Logical finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!