Change or remove duplicate matrix elements

I have the matrix:

value =
  3.1727
  5.2495
  5.2708
  3.3852
  5.6222
  5.2708
  5.1444
  4.9834
  5.7499
  5.7499
  3.4728
  3.4728
  5.3560
  5.7499
  3.4728
  5.7286
  6.1225
  3.6539
  3.0351
  4.3020
  5.2296
  3.8040
  4.6747
  5.4412
  3.6539

and I want to add 0.1 to any duplicate entries, so that all values are unique, and none are removed.

I have tried:

value=sort(value)

for i=1:(length(value)-1)

    if value(i+1)==value(i);

        value(i+1)=(value(i+1)+0.1);
    end

end

But for some reason it has no impact on the matrix..

Your help is greatly appreciated, thanks!

**EDIT: how can i remove all but the first occurance of a duplicate value? unique does not work for me.

[Sean de Wolski]

So what if by adding 0.1 you duplicate a value? You'll have to do a while loop checking to ensure this didn't happen and rerunning the engine again if it did.

[A]

That's true. What if on the other hand I want to completely remove any duplicate entries?

Answer 1

[Sean de Wolski]

X = [1 2 3 3 4 5 2]';
X2 = sort(X,1);
idx = [false;diff(X2)<(10^-6)];  %equal to 10^-6th precision
X2(idx) = X2(idx)+.1;

[A]

Thanks for your response.

This is the output with my matrix:

3.0351

3.1727

3.3852

3.4728*

3.4728*

3.4728*

3.6539

3.7539

3.8040

4.3020

4.6747

4.9834

5.1444

5.2296

5.2495

5.2708

5.3708

5.3560

5.4412

5.6222

5.7286

5.7499*

5.7499*

5.8499

6.1225

As in the response above, it works on some entires but entirely skips others..

[Jan]

Yup. Then these values are *not* equal.

The two 5.7499 can be 5.749900000000001 and 5.74989999999999.

[Sean de Wolski]

you could easily add a tolerance to the diff operator, I'll do it in a second.

[Sean de Wolski]

Try it now, they need to be accurate to 10^-6 places. Tighten or loosen the tolerance as you wish

[A]

Great, I used it twice and all of the values are now unique.

I am also interested to know how I can remove all but the first instance of a repeated value. unique does not work, and I suspect this may have something to do with the tolerence, as you mentioned. I have yet to decide of removing or changing the multiple values is more ideal for my task. Thank you for your help!

[A]

ah, just did X2(idx)=[]

Thanks so much for your help!

[Fangjun Jiang]

@Sean, this won't work. The problem is not the floating point comparison. It's the multiple duplication. If the first round has 3 duplications, then after the round, you still have 2 duplications.

X = [1 2 3 3 4 5 2]';

X2 = sort(X,1);

idx = [false;diff(X2)<(10^-6)]; %equal to 10^-6th precision

X2(idx) = X2(idx)+.1

[Sean de Wolski]

Thst's true, hence the while-loop I mentioned in the comment above.

Answer 2

[Fangjun Jiang]

It works in certain degree but your algorith has a flaw. You'll see it clearly running the following.

value=sort(value);
NewValue=value;

for i=1:(length(NewValue)-1)

      if NewValue(i+1)==NewValue(i);
          NewValue(i+1)=(NewValue(i+1)+0.1);
      end
  end

[value NewValue]

[A]

Thank you for your response!

This is what is returned:

3.0351 3.0351

3.1727 3.1727

3.3852 3.3852

3.4728 3.4728 ***

3.4728 3.4728 ***

3.4728 3.4728 ***

3.6539 3.6539

3.6539 3.7539

3.8040 3.8040

4.3020 4.3020

4.6747 4.6747

4.9834 4.9834

5.1444 5.1444

5.2296 5.2296

5.2495 5.2495

5.2708 5.2708

5.2708 5.3708

5.3560 5.3560

5.4412 5.4412

5.6222 5.6222

5.7286 5.7286

5.7499 5.7499

5.7499 5.7499 ***

5.7499 5.8499

6.1225 6.1225

it worked in some places, but the (***) mark where it didn't work at all.. this is very confusing. Any idea why this is happening?

[A]

Ah sorry thats a bit hard to read, I didn't realize the numbers woudl be so close together. It is [value NewValue] as you put in your code.

[Fangjun Jiang]

The reason is simple. You added 0.1 to the 2nd duplicated value which will change the comparison of your next loop.

Answer 3

[MaVo]

I had a similar issue with a time sequence in 0.005 s steps, but I figured out a solution that worked for me. My problem during debugging was, that it didn't go into the if-condition. I solved this by comparing integers than double values. Maybe this helps someone.

s_length = length(s_time);
ctr = 1;

 for ctr = 2:s_length
    % Checking values for debugging
    new = uint64(s_time(ctr,1)/0.005);
    old = uint64(s_time((ctr-1),1)/0.005);
    before = (ctr-1);

   if new == old
       s_time(ctr:end,1) = s_time(ctr:end,1)+0.005;
   end

   if new > (old + 1)
       s_time(ctr:end) = s_time(ctr:end) - 0.005;
   end 
end