The easiest way to explain the question is by an example:
>> A = [1 5; 1 5; 1 6; 1 6; 1 6; 1 6; 1 7; 2 5; 2 6; 2 6; 2 6; 2 7; 2 7; 2 8]
A =
1 5
1 5
1 6
1 6
1 6
1 6
1 7
2 5
2 6
2 6
2 6
2 7
2 7
2 8
What I want to have as output is something similar to
result =
1 6 4
1 5 2
2 6 3
2 7 2
Which means top two frequent pairs for each value in first column. So most frequent pairs came with 1 are 6 and 5 and most frequent pairs came with 2 are 6 and 7. 3rd column shows frequency of pairs.
I use matlab 2016 in linux.
You could use unique and histc to count the frequency of occurrence of each row. Then using frequency, you can proceed with your calculations.
[unique_rows,~,ind] = unique(A,'rows');
counts = histc(ind,unique(ind));
Now, you could combine the frequency count and sort them.
arr = [unique_rows,counts];
[sorted,~]=sortrows(arr,[1 -3])
sorted =
1 6 4
1 5 2
1 7 1
2 6 3
2 7 2
2 5 1
2 8 1
Related
This question already has answers here:
MATLAB: how to pass in the diagonal of a matrix as an argument in another matrix?
(2 answers)
Closed 6 years ago.
I have a matrix
A = repmat(1:7,7,1);
I have index vectors
idx1 = [1 3 5];
idx2 = [1 3 5];
I want to access A at the 2d coordinates denoted by idx1(i),idx2(i).
When I do
A(idx1,idx2) = 0;
I get for each element in idx 1, all the elements in idx2 as well.
I want only the corresponding elements to be assigned the zero value.
Again: I get
A =
0 2 0 4 0 6 7
1 2 3 4 5 6 7
0 2 0 4 0 6 7
1 2 3 4 5 6 7
0 2 0 4 0 6 7
1 2 3 4 5 6 7
1 2 3 4 5 6 7
but I want
A =
0 2 3 4 5 6 7
1 2 3 4 5 6 7
1 2 0 4 5 6 7
1 2 3 4 5 6 7
1 2 3 4 0 6 7
1 2 3 4 5 6 7
1 2 3 4 5 6 7
How to achieve this?
Thanks
The easiest way is probably to use sub2ind to generate the linear indices needed to index into A:
linear_ind = sub2ind(size(A),idx1,idx2);
A(linear_ind) = 0;
Ok so I am clustering data into clusters which are then indexed using a column. The data is in the form of motion vectors and so my data will look like this after being clustered:
[index x y x' y']
for example:
[1 3 5 4 6;
1 4 6 5 7;
2 3 5 4 6;
2 8 9 9 3;
3 2 3 2 4]
in above array there are 3 clusters, with clusters 1 and 2 each containing 2 vectors.
My problem is that I sometimes have to delete clusters based on certain criteria, and may be left with:
[2 3 5 4 6;
2 8 9 9 3;
3 2 3 2 4]
I want to be able to correct the index after deletion, so that it starts at 1 and ends with the number of clusters. So in this case replace the 2s with 1s and 3s with 2s.
Im sure there must be a simple way using a for loop but Ive been trying for a while and can't get ti right?
Assuming your matrix is called data, try this:
>> data = [2 3 5 4 6;
2 8 9 9 3;
3 2 3 2 4]
data =
2 3 5 4 6
2 8 9 9 3
3 2 3 2 4
>> data(:,1) = cumsum(diff(data([1 1:end], 1)) ~= 0) + 1
data =
1 3 5 4 6
1 8 9 9 3
2 2 3 2 4
A simple call to unique will help you do that. You can use the third output of it to assign each unique and new ID using the first column of the new data matrix (index vector) to replace its first column. Also, make sure you use the 'stable' flag so that it assigns IDs in order of occurrence from top to bottom:
%// Data setup
A = [1 3 5 4 6;
1 4 6 5 7;
2 3 5 4 6;
2 8 9 9 3;
3 2 3 2 4];
%-----
B = A(3:end,:); %// Remove first two rows
%// Go through the other IDs and reassign to unique IDs from 1 up to whatever
%// is left
[~,~,id] = unique(B(:,1), 'stable');
%// Replace the first column of the new matrix with the new IDs
B(:,1) = id; %// Replace first column with new IDs
We get:
>> B
B =
1 3 5 4 6
1 8 9 9 3
2 2 3 2 4
Given any number. Lets say for example 5, I need to generate a matrix similar to this:
1 2 3 4 5
2 2 3 4 5
3 3 3 4 5
4 4 4 4 5
5 5 5 5 5
How to generate a matrix similar to this using Matlab?
I'd use bsxfun:
n = 5;
matrix = bsxfun(#max, 1:n, (1:n).');
An alternative (probably slower) is to use ndgrid:
n = 5;
[ii, jj] = ndgrid(1:n);
matrix = max(ii, jj);
Nothing will ever beat bsxfun as used by Luis Mendo., but for the sake of reminding people of the existence of Matlab's gallery function, here another approach:
n = 5;
A = gallery('minij',n)
B = n + 1 - A(end:-1:1,end:-1:1)
A =
1 1 1 1 1
1 2 2 2 2
1 2 3 3 3
1 2 3 4 4
1 2 3 4 5
B =
1 2 3 4 5
2 2 3 4 5
3 3 3 4 5
4 4 4 4 5
5 5 5 5 5
I have some matrix :
A = [ 1 2 3 4 5 6;
1 2 3 4 5 6]
B = [ 6 5 4 3 2 1;
6 5 4 3 2 1]
C = [ 1 2 3 4 5 6;
1 2 3 4 5 6]
what is code to make this following matrix:
Result = [1 2 9 9 10 11 5 5 5 6;
1 2 9 9 10 11 5 5 5 6]
Note : Actually the above matrix is sum of 3 matrix above which had been already rearranged like as the following matrix. #sum is sum which is based on column.
1 2 3 4 5 6
1 2 3 4 5 6
6 5 4 3 2 1
6 5 4 3 2 1
1 2 3 4 5 6
1 2 3 4 5 6
And. I sum first row by first row, and second row by second row.
To do what you say above:
Result = zeros(size(A) + [0,4]);
Result(:,1:size(A,2)) = A;
Result(:,3:end-2) = Result(:,3:end-2) + B;
Result(:,5:end) = Result(:, 5:end) + C;
The point is, you can select a subregion of a matrix, and assign another matrix to it. You just have to make sure both sides of the assignment are the same shape.
For a matrix A (4 rows, 1000 columns). I want to group the columns of the matrix A which have the same value for the third line. so I must have sub matrix with a third row that contains the same value.
for example:
if:
A =
1 4 5 2 2 2 2 1 1 5
1 4 5 4 4 2 2 4 5 2
3 3 3 3 4 1 3 5 3 4
4 5 5 5 4 1 5 5 5 5
then
A1 =
1 4 5 2 2 1
1 4 5 4 2 5
3 3 3 3 3 3
4 5 5 5 5 5
A2 =
2 5
4 2
4 4
4 5
A3 =
2
2
1
1
the result can be in the form of a cell.
here's one possible hack (warning: I haven't been able to check this):
A =
1 4 5 2 2 2 2 1 1 5
1 4 5 4 4 2 2 4 5 2
3 3 3 3 4 1 3 5 3 4
4 5 5 5 4 1 5 5 5 5
specialRow=3;
unqCols = unique(A(specialRow,:));
numUnq = length(unqCols);
sepMats{numUnq}=[];
for i=1:numUnq
sepMats{i} = A(:,A(specialRow,:)==unqCols(i));
end
In the example you shown, there are 4 unique elements in the 3rd row, so you should obtain 4 submatrices, but you only show 3 ?
Here is one way:
clear all;
%data
A = [1 4 5 2 2 2 2 1 1 5;
1 4 5 4 4 2 2 4 5 2;
3 3 3 3 4 1 3 5 3 4;
4 5 5 5 4 1 5 5 5 5
]
%engine
row = 3;
b = unique(A(row,:));
r = arrayfun(#(i) A(:,A(row,:)==b(i)),1:length(b), 'UniformOutput',false);
r{:}
You can make the assignment in a single line using ACCUMARRAY:
A = [1 4 5 2 2 2 2 1 1 5;
1 4 5 4 4 2 2 4 5 2;
3 3 3 3 4 1 3 5 3 4;
4 5 5 5 4 1 5 5 5 5
];
out = accumarray(A(3,:)', (1:size(A,2)), [], #(x){A(:,x)} );
With this, out{i} contains all columns of A where the third row of A equals i (and empty in case there is no valid column).
If you want out{i} to contain columns corresponding to the i-th smallest unique value in the third row of A, you can use GRP2IDX from the statistics toolbox first:
[idx,correspondingEntryInA] = grp2idx(A(3,:)'); %'#
out = accumarray(idx, (1:size(A,2)), [], #(x){A(:,x)} );
Here, out{i} contains the columns corresponding to correspondingEntryInA(i).