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Removing adjacent duplicate numbers in rows of a matrix

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AA on 21 Jun 2019
Commented: infinity on 23 Jun 2019
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.

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Answers (2)

infinity on 21 Jun 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

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Rik on 22 Jun 2019
I would suggest using the code below instead. With a NaN you avoid confusion if the last value happens to be a 0.
clear
xx=[1 2 3 3 4 4; 1 1 2 2 3 3 ; 5 5 5 3 3 2]
%yy = zeros(size(xx));
yy=NaN(size(xx))
for row = 1:size(xx,1)
temp = unique(xx(row,:),'stable');
yy(row,1:length(temp)) = temp;
end
AA on 22 Jun 2019
Hi,
somehow I see an error when I modify the xx and increase the set of numbers. Adjacent numbers can repeat like 3 or 2 or 5 but they are not displayed in the output.
clear
xx=[1 2 3 3 4 4 3; 1 1 2 2 3 3 2; 5 5 5 3 3 2 5]
%yy = zeros(size(xx));
yy=NaN(size(xx))
for row = 1:size(xx,1)
temp = unique(xx(row,:),'stable');
yy(row,1:length(temp)) = temp;
end
The last digit 3 or 2 or 5 is not shown in the output. Instead i get
1 2 3 4 NaN NaN NaN
1 2 3 NaN NaN NaN NaN
5 3 2 NaN NaN NaN NaN
but I should get
1 2 3 4 3 NaN NaN
1 2 3 2 NaN NaN NaN
5 3 2 5 NaN NaN NaN
How can I modify the formula?
infinity on 23 Jun 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

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Rik on 22 Jun 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)

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