Effacer les filtres
Effacer les filtres

How to find the frequency of each row with certain coordiante in a matrix?

3 vues (au cours des 30 derniers jours)
Hello all,
I have an mx2 matrix with its rows (0,1), (-1,0), (0,1), (0,-1), (1,1), (-1,1), (1,-1),(-1,-1); I would like to find the frequency of each of above coordinates. In other words, if I have A=[1 1;0 1;-1 1;1 0;-1 1], I would like to get something like,
number of times that (1,1) has appeared=1; number of times that (0,1) has appeared=1; number of times that (-1,1) has appeared=2; number of times that (1,0) has appeared=1; number of times that (0,-1) has appeared=0; number of times that (-1,-1) has appeared=0; number of times that (-1,0) has appeared=0; number of times that (1,-1) has appeared=0;
when I use "find" command I get an error. Thank you.

Réponse acceptée

Azzi Abdelmalek
Azzi Abdelmalek le 29 Mar 2014
Modifié(e) : Azzi Abdelmalek le 29 Mar 2014
A=[1 1; 0 1; 1 0; -1 1;-1 1;0 1;0 1]
[ii,jj,kk]=unique(A,'rows','stable')
f=histc(kk,1:numel(jj)); % Frequency
result=[ii f]
  3 commentaires
Azzi Abdelmalek
Azzi Abdelmalek le 29 Mar 2014
Modifié(e) : Azzi Abdelmalek le 29 Mar 2014
Ok use this (without stable option), maybe you have an old version of Matlab
A=[1 1; 0 1; 1 0; -1 1;-1 1;0 1;0 1]
[ii,jj,kk]=unique(A,'rows')
f=histc(kk,1:numel(jj)); % Frequency
result=[ii f]
Mnr
Mnr le 29 Mar 2014
Thank you so much! I really appreciate your help!

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Plus de réponses (1)

Aditya Rathore
Aditya Rathore le 1 Fév 2022
There are two things:
  1. Your row length is small (2)
  2. You want to also retrieve the stored values
I suggest you make a matrix for storing frequencies
freqs = zeros(m, 2);
[rows, ~] = size(A);
for idx=1:rows
i = A(idx, 1);
j = A(idx, 2);
freqs(i+2,j+2) = freqs(i+2, j+2) + 1; %Because you have -1 also
end
When retrieving a value, simply do
count = A(i+2, j+2);
This approach is useful as you can normalize the counts and get probability using
freqs = freqs / sum( freqs, 'all');
I tried it for image based task in RGB space and found this to be fastest solution.

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