Removing adjacent duplicate numbers in rows of a matrix
1 visualización (últimos 30 días)
Mostrar comentarios más antiguos
Hi,
assume the following matrix format:
xx=[1 2 3 3 4 4; 1 1 2 2 3 3 ; 5 5 5 3 3 2]
I want to remove the adjacent duplicate numbers in each row of this matrix so I get the following output:
result=[1 2 3 4; 1 2 3 ; 5 3 2]
Help would be very much appreciated with a formula. Thanks.
0 comentarios
Respuestas (2)
infinity
el 21 de Jun. de 2019
Hello,
In your problem, you may get error since your result is not a matrix (number of colume of the first row is different with other rows).
So, you result should be a cell like this
clear
xx=[1 2 3 3 4 4; 1 1 2 2 3 3 ; 5 5 5 3 3 2]
for i = 1:size(xx,1)
yy{i} = unique(xx(i,:));
end
12 comentarios
infinity
el 23 de Jun. de 2019
Hello,
You can modify the code like this
clear
% xx=[1 2 3 3 4 4; 1 1 2 2 3 3 ; 5 5 5 3 3 2]
% xx=[1 2 3 3 4 4; 1 1 2 2 3 3 ; 5 4 5 3 3 2]
xx=[1 2 3 3 4 4 3; 1 1 2 2 3 3 2; 5 5 5 3 3 2 5]
% xx=[1 2 3 3 4 4 3 3; 1 1 2 2 3 3 2 3; 5 5 5 3 3 2 5 5]
n = size(xx,2);
% yy = 0*xx;
yy=NaN(size(xx));
for i = 1:size(xx,1)
[temp,ia,ib] = unique(xx(i,:),'stable');
f = diff(ib)~=0;
idx = [find(f);length(ib)];
% yy(i,1:length(temp)) = temp;
yy(i,1:length(idx)) = xx(i,idx);
end
yy
Rik
el 22 de Jun. de 2019
You can use diff to find any repeats and remove them. The compare to eps is to avoid strange things happening once you put in decimal numbers, which can lead to rounding errors, which is why exp(log(3))==3 returns false.
xx=[1 2 3 3 4 4 3; 1 1 2 2 3 3 2; 5 5 5 3 3 2 5];
yy=NaN(size(xx));
for row = 1:size(xx,1)
temp = xx(row,:);
%keep values that are different from the one before
%L=[true diff(temp)~=0];%line below is safe for float rounding
L=[true abs(diff(temp))>=(2*eps)];
yy(row,1:sum(L)) = temp(L);
end
%remove trailing NaN cols:
while all(isnan(yy(:,end)))
yy(:,end)=[];
end
disp(yy)
0 comentarios
Ver también
Categorías
Más información sobre Loops and Conditional Statements en Help Center y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!