Filtering a matrix to some small matrices

1 次查看(过去 30 天)
Hi everyone,
Suppose I have a matrix:
a = [12,7;12,5;2,7;23,3;23,43;23,12;3,5;76,21;76,31;4,3;4,7;4,9];
Then I want to have some sub-divide matrix like following matrixes:
A = [2,7;3,5;4,9]
B = [12,7;12,5;76,21;76,31]
C = [23,3;23,43;23,12;4,3;4,7;4,9]
If first row in matrix a repeated once (e.g. 2-3-4), then that row goes to matrix A If first row in matrix a repeated twice (e.g. 12-76), then that row goes to matrix B If first row in matrix a repeated triple (e.g. 23-4), then that row goes to matrix C ...
Repeation frequency is from 1 to 6 times.

采纳的回答

Azzi Abdelmalek
Azzi Abdelmalek 2014-7-14
编辑:Azzi Abdelmalek 2014-7-14
a = [12,7;12,5;2,7;23,3;23,43;23,12;3,5;76,21;76,31;4,3;4,7;4,9]
c1=a(:,1)
[ii,jj,kk]=unique(c1,'stable')
aa=histc(kk,1:numel(jj))
for k=1:max(aa)
jdx=ii(ismember(aa,k))
jj=ismember(c1,jdx)
out{k}=a(jj,:)
end
celldisp(out)

更多回答(1 个)

Matz Johansson Bergström
编辑:Matz Johansson Bergström 2014-7-14
I am sure there is a function in statistics that does this for you, but I quickly wrote a very dirty version using cells, just to handle the matrices in a nice way. You can replace the cells if you like. I assume that the highest frequency is 6, as you state.
Tally = {}; %the matrices
for i = 1:6 %assume frequency is at most 6
Tally{i} = []; %"allocate"
end
[un, in] = unique(a(:, 1));
%Instead of fixing the number of matrices, we store them in a cell and
%count the number of repeats
for i = 1:size(un, 1)
num = un(i); %the number we are counting on at the moment
inds = find( a(:, 1) == num ); %the indices where 'num' is
freq = length(inds);
Tally{freq} = [Tally{freq}; a(inds, :)];
%append the new coordinate in index 'freq'
end
So, a very quick and dirty solution. If I run it on your sample and print my cell Tally, we get
>Tally{:}
ans =
2 7
3 5
ans =
12 7
12 5
76 21
76 31
ans =
4 3
4 7
4 9
23 3
23 43
23 12
ans =
[]
ans =
[]
ans =
[]
I also noticed that [4,9] should not be in A. Hope this helps.

类别

Help CenterFile Exchange 中查找有关 Matrices and Arrays 的更多信息

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by