Help needed with matrix in for-loop and output of matrix

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Karsten
Karsten el 23 de Dic. de 2014
Comentada: Karsten el 24 de Dic. de 2014
Hi, i have a question regarding some tasks to change the content of a matrix. The goal of this operation is to 'clear' a matrix consisting out of 2 columns and a random amount of rows. The operation has to be done using a function-file. The input has to be x = [:,1] and y = [:,2]. After that the first step is to put them together in a matrix A = [x,y] and the rows need to be sorted from lowest to highest value from the first column, B = sortrows(A,1). The problem however is that in the first column the x-values are supposed to be unique with it's matching y-values (2nd column).
So let's say you got a x = [1;3;4;3;2;2;4] and y = [5;1;3;2;9;6;1] which will then transfer into the matrix B = [1,5;2,9;2,6;3,1;3,2;4,3;4,1]. Now as you can see there are several x-values that aren't unique in the first column. The result of this operation should be the next matrix C = [1,5;2,6;3,1;4,1].
  1. If an x-value is unique, then it's y-value is the one to be taking.
  2. If an x-value is not unique, then the y-value should be the lowest value matching with these x-values.
How can i write a code into a for-loop(?) with these 2 conditions so it can be used for any matrix with n x 2 dimension? This is what i had so far, yet i have no idea if this is anywhere near the solution..
function [ new_matrix ] = matrix( x,y )
A = [x,y];
B = sortrows(A,1);
n = size(B,1);
for i = 1:1:n
if B(i,1) == B(i+1,1) && B(i,2) <= B(i+1,2)
B(i+1,:) = [];
elseif B(i,1) == B(i+1,1) && B(i,2) >= B(i+1,2)
B(i,:) = [];
end
end
This is an example of how it's supposed to work, but for any matrix with n x 2 dimensions.
%

Respuesta aceptada

Sara
Sara el 23 de Dic. de 2014
With x,y column vectors:
A = [x,y];
B = sortrows(A,1);
unique_x = unique(B(:,1));
new_matrix = zeros(numel(unique_x),2);
new_matrix(:,1) = unique_x;
for i = 1:numel(unique_x)
k = find(B(:,1) == unique_x(i));
new_matrix(i,2) = min(B(k,2));
end
  1 comentario
Karsten
Karsten el 24 de Dic. de 2014
Well that doe seem to work! Thanks alot Sara, much appreciated!

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