Count all unique elements in a 3d matrix

2013 8 8
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1 个投票

Hi all,
I have created a 3-d matrix randworld and would like to list the unique elements, count the number of unique elements and count the size of the unique elements for randworld(:,:,1), randworld(:,:,2) and randomworld(:,:,3). e.g. 4, 7, 13 is one unique element in this matrix with size 3.
randworld(:,:,1) =
4 4 3 11 11
4 9 3 9 10
2 7 9 3 9
7 6 6 9 3
4 1 15 15 13
randworld(:,:,2) =
7 7 6 3 3
7 11 6 6 10
11 4 11 9 6
4 15 15 6 9
1 8 5 5 5
randworld(:,:,3) =
13 13 9 4 4
13 11 9 3 6
13 4 11 12 3
4 1 1 3 12
11 5 15 15 15
Any help will be appreciated.
Thanks, Vishal

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Azzi Abdelmalek
Azzi Abdelmalek 2013-8-8
why 4,7,13 is unique element, what about other numbers?
Jan
Jan 2013-8-8
I do not understand: "4, 7, 13 is one unique element in this matrix with size 3." I see 9 elements of the value 4.
Could you provide a real example for the outputs for a [3x3x3] array?
Vishal
Vishal 2013-8-8
Sorry, I'll try and make it my question more clear. In the 3-d matrix randworld as listed below, I am trying to find the following:
randworld(:,:,1) =
4 4 3
4 9 3
2 7 9
randworld(:,:,2) =
7 7 6
7 11 6
11 4 11
randworld(:,:,3) =
13 13 9
13 11 9
13 4 11
1. list the unique elements-
4,7,13
2,11,13
9,11,11
7,4,4
3,6,9
9,11,11
2. count the number of unique elements
4,7,13 (count is 3) & 2,11,13 (count is 1) & 9,11,11 (count is 1) & 7,4,4 (count is 1) & 3,6,9 (count is 2)& 9,11,11 (count is 1)

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Azzi Abdelmalek
Azzi Abdelmalek 2013-8-8

1 个投票

a1=randworld(:,:,1)
a2=randworld(:,:,2)
a3=randworld(:,:,3)
v=cell2mat(arrayfun(@(x1,x2,x3) [x1 x2 x3],a1(:),a2(:),a3(:),'un',0));
[a,b,c]=unique(v,'rows','stable')
idx=histc(c,(1:size(a,1))')
% a represent unique vectors
% idx represent the repetition of each vector

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Vishal
Vishal 2013-8-8
works fine. thanks a ton
Evan
Evan 2013-8-8
If that answer solved your problem, be sure to accept it! ;)
Azzi Abdelmalek
Azzi Abdelmalek 2013-8-8
Or simply
a1=randworld(:,:,1);
a2=randworld(:,:,2);
a3=randworld(:,:,3);
v1=[a1(:) a2(:) a3(:)];
[a,b,c]=unique(v,'rows','stable')
idx=histc(c,(1:size(a,1))')
Vishal
Vishal 2013-8-12
Yes, that works fine as well. Just a small typo. In line 4 replace v with v1. Thanks mate.
a1=randworld(:,:,1); a2=randworld(:,:,2); a3=randworld(:,:,3); v1=[a1(:) a2(:) a3(:)]; [a,b,c]=unique(v1,'rows','stable') idx=histc(c,(1:size(a,1))')

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更多回答（2 个）

Jan
Jan 2013-8-8

0 个投票

It sounds like a basic task for unique and histc. But currently I do not understand the needs exactly.

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Azzi Abdelmalek
Azzi Abdelmalek 2013-8-8
I think, he wants to form vectors from each chanel:
vector1= [a(1,1,1) a(1,1,2) a(1,1,3)]
vector2=[a(1,2,1) a(1,2,2) a(1,2,3)]
and so on

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dpb
dpb 2013-8-8

0 个投票

Permutation on above...
rpr=permute(r,[1 3 2]);
r2d=rpr(:,:,1);for i=2:size(rpr,3),r2d=[r2d; rpr(:,:,i)];end
u=unique(r2d,'rows','stable');

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Vishal
Vishal 2013-8-12
This solutions works fine if the task is to just list the unique rows. Thanks
dpb
dpb 2013-8-12
Well, one presumes one would use the u as in the previous to determine the rest having found them. Didn't see any point in repeating that.

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