# Find duplicate elements and remove the rows that has similar values in one column

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Hamid on 18 Oct 2022
Commented: Jan on 19 Oct 2022
Dear Matlab experts,
I am using the following function to find the rows that has similar value in their 9th column. The speed of calculation is very slow as the data is big. Any suggestions for modifying my code to increase the speed or any other suggestions to achieve that purpose?
function in1=dup_remove(out2)
b=[];
for i=1:size(out2,1)
[r,c]=find(out2(:,9)==out2(i,9));
if(length(r)==1)
b=[b;out2(i,:)];
end
end
if (~isempty(b))
in1=b;
end
end
Jan on 19 Oct 2022
@KSSV: How? I've tried it without success. The only way with standard Matlab functions I've found, uses unique to get a list of occurring values and histcounts to identify the elements, which occur once only. This was much slower than sorting the input, comparing neighbors by diff , remove the duplicates and reproducing the original order.

Jan on 18 Oct 2022
Edited: Jan on 19 Oct 2022
Avoid iteratively growing arrays, because they are extremly expensive. See:
x = [];
for k = 1:1e6
x(k) = rand;
end
This creates a new vector x in each iteration and copies the former contents of the vector to the new one, so Matlab reserves and copies sum(1:1e6)*8 Bytes, which is more than 4 TB!
Pre-allocation solves the problem:
x = zeros(1, 1e6);
for k = 1:1e6
x(k) = rand;
end
Tis reserves 8 MB only and copies just the scalar elements.
function y = dup_remove(x)
x9 = x(:, 9); % Slightly faster than indexing each time
n = size(x,1);
match = false(n, 1);
for i = 1:n
[r, c] = find(x9 == x9(i));
match(i) = (numel(r) == 1);
end
y = x(match, :);
end
It is too strange, to call the input "out2" and the output "in1".
A smarter method:
function y = dup_remove(x)
x9 = x(:, 9); % Slightly faster than indexing each time
T = true(numel(x9), 1);
[S, idx] = sort(x9(:).');
m = [true, diff(S) ~= 0];
ini = strfind(m, [true, false]);
m(ini) = false; % Mark 1st occurence in addition
T(idx) = m; % Restore original order
y = x(T, :);
end
The sorting avoids to compare each element with all others, but only one comparison with the neighbor is required.
##### 2 CommentsShowHide 1 older comment
Hamid on 19 Oct 2022
Edited: Hamid on 19 Oct 2022
@Jan Thank you so much. I learned a lot studying your answer.

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