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

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?

Thank you in advance.

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

[KSSV]

Thias can be achieved using unique and ismember.

[Fatih]

I guess you may use

rowIndex = find(ismember(A(:,9),B,'rows'))

to find the row index of the searching vector.

In this code B matrix is searched in all 9th columns of A matrix. If there is a lot of B in A, then rowIndex will be a column vector. If the B is changing, you may set different values in every loop as you wish.

Hope it works for you.

[Hamid]

@Fatih Thanks a lot.

[Hamid]

@KSSV Thank you for your suggestion and the support.

[Jan]

@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.

Answer 1

[Jan]

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.

In your case:

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.

[Jan]

Some timings:

x = randi([0, 65535], 1e4, 9);
n = 10;  % Repeat loops for accurate timings
tic
for k = 1:n
    y0 = dup_remove(x);
end
toc  % Original:

Elapsed time is 2.469073 seconds.

tic
for k = 1:n
    y1 = dup_remove1(x);
end
toc  % Avoid iterative growing:

Elapsed time is 1.181532 seconds.

tic
for k = 1:n
    y11 = dup_remove11(x);
end
toc  % Without FIND:

Elapsed time is 0.974451 seconds.

tic
for k = 1:n
    y2 = dup_remove2(x);
end
toc  % Using SORT and comparison of neighbors:

Elapsed time is 0.011396 seconds.

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

function y = dup_remove1(x)
x9 = x(:, 9);   % Slightly faster than indexing each time
n  = size(x,1);
m  = false(n, 1);
for i = 1:n
    [r, c] = find(x9 == x9(i));
    m(i)   = (numel(r) == 1);
end
y = x(m, :);
end

function y = dup_remove11(x)
x9 = x(:, 9);   % Slightly faster than indexing each time
n  = size(x,1);
m  = false(n, 1);
for i = 1:n
    m(i) = (sum(x9 == x9(i)) == 1);
end
y = x(m, :);
end

function y = dup_remove2(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

[Hamid]

@Jan Thank you so much. I learned a lot studying your answer.