Behavior of "unique" with mixed complex numbers

Question

Joel Miller 2020-12-4

2
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/678683-behavior-of-unique-with-mixed-complex-numbers

评论： Bruno Luong 2020-12-11

I have a large set of complex number that I want to map to an integer grid on the complex plane, removing redundant values. A small subset of the data looks like:

Atest=[-12.9753 - 0.8003i;
     -12.9938 - 0.4003i;
     -13.0000 + 0.0000i;
     -12.9938 + 0.4003i;
      12.9938 + 0.4003i;
      13.0000 + 0.0000i;
      12.9938 - 0.4003i;];

I use round to map to the grid:

Around=round(Atest)
Around =
 -13.0000 - 1.0000i
 -13.0000 + 0.0000i
 -13.0000 + 0.0000i
 -13.0000 + 0.0000i
  13.0000 + 0.0000i
  13.0000 + 0.0000i
  13.0000 + 0.0000i

I then use unique to remove the redundant values:

  Aunique=unique(Around,'stable')
Aunique =
 -13.0000 - 1.0000i
 -13.0000 + 0.0000i
 -13.0000 + 0.0000i
  13.0000 + 0.0000i

Clearly, the result is not unique: the value -13 appears twice. If I apply unique to just the las 6 values of Around (the real values) I obtain the expected results:

  Aunique=unique(Around(2:end),'stable')
Aunique =
   -13
    13

I would appreciate any suggestions for getting the unique values of an array with mixed real and complex values.

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Paul 2020-12-5

在 MATLAB Online 中打开

Could this have something to do with how the round function leaves the sign bit? I'm not sure, but I did notice that:

>> Around
Around =
 -13.0000 - 1.0000i
 -13.0000 + 0.0000i
 -13.0000 + 0.0000i
 -13.0000 + 0.0000i
  13.0000 + 0.0000i
  13.0000 + 0.0000i
  13.0000 + 0.0000i
>> format hex
>> Around
Around =
   c02a000000000000   bff0000000000000i
   c02a000000000000   8000000000000000i
   c02a000000000000   0000000000000000i
   c02a000000000000   0000000000000000i
   402a000000000000   0000000000000000i
   402a000000000000   0000000000000000i
   402a000000000000   8000000000000000i
>> Aunique
Aunique =
   c02a000000000000   bff0000000000000i
   c02a000000000000   8000000000000000i
   c02a000000000000   0000000000000000i
   402a000000000000   0000000000000000i
>> format short
>> Aunique
Aunique =
 -13.0000 - 1.0000i
 -13.0000 + 0.0000i
 -13.0000 + 0.0000i
  13.0000 + 0.0000i

So it appears that the Around(2) and Around(3) might not be unique because of the sign bit on the imaginary part?

But, looking at Marco's answer I'm not sure that's the case:

>> [real(Around) imag(Around)]
ans =
   c02a000000000000   bff0000000000000
   c02a000000000000   8000000000000000
   c02a000000000000   0000000000000000
   c02a000000000000   0000000000000000
   402a000000000000   0000000000000000
   402a000000000000   0000000000000000
   402a000000000000   8000000000000000

So imag preserves the sign bit on the imaginary part of Around(2), but unique doesn't seem to care about that using Marco's answer:

>> unique([real(Around) imag(Around)],'rows')
ans =
   -13    -1
   -13     0
    13     0
>> format hex
>> ans
ans =
   c02a000000000000   bff0000000000000
   c02a000000000000   8000000000000000
   402a000000000000   0000000000000000

Maybe unique works differently for complex and real inputs wrt to how it treats the sign bit on a value of zero?

Paul 2020-12-5

在 MATLAB Online 中打开

I'm confused as to why Around(2) and Around(3:4) are not unique, but Around(7) and Around(5:6) are unique? Shouldn't the same rule apply?

Around =
   c02a000000000000   bff0000000000000i
   c02a000000000000   8000000000000000i
   c02a000000000000   0000000000000000i
   c02a000000000000   0000000000000000i
   402a000000000000   0000000000000000i
   402a000000000000   0000000000000000i
   402a000000000000   8000000000000000i

Maybe this is one other thing that you already noticed, but flipping the matrix not only changes which entries are selected as unique, but also changes the number of unique entries?

>> Aunique=unique(Around,'stable')
Aunique =
   c02a000000000000   bff0000000000000i
   c02a000000000000   8000000000000000i
   c02a000000000000   0000000000000000i
   402a000000000000   0000000000000000i
>> Aunique=unique(flipud(Around),'stable')
Aunique =
   402a000000000000   8000000000000000i
   c02a000000000000   0000000000000000i
   c02a000000000000   bff0000000000000i

Very troubling that the numel of the result depends on the order of input.

Will you post back here when you get a response from tech support?

Bruno Luong 2020-12-7

Here is the reply from tech support

"Thank you for your inquiry and provided clear reproduction examples. It does appear as though this is a bug with the unique function executed on rounded complex arrays. The Development team has been informed of this bug and will investigate further.

Given the information above, I will close this case as 'Bug/Enhancement Submitted'. From a Technical Support perspective the case is now closed. From a Development standpoint, on the other hand, the matter is still open. It is being investigated and will be considered for a future Release or Update. Should this bug be fixed, you will be personally notified via email. Then, the status of this service request will be automatically changed to 'Bug Fixed/Enhancement Complete'. "

So obviously it's abug and they will fix it in future release.

Bruno Luong 2020-12-11

在 MATLAB Online 中打开

Another mail from tech:

"I just wanted to offer you a follow up email regarding a workaround for your use case. If you add complex(0)to 'Around', then call unique, it will give the right answer. For example:"

A = [-12.9753 - 0.8003i; -12.9938 - 0.4003i; -12.9938 + 0.4003i; 12.9938 + 0.4003i];
B = round(A) + complex(0);
C = unique(B);

But we already knew this solution from the cross-thread of atan2 and sign bit.

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Answer 1

Marco Riani 2020-12-4

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/678683-behavior-of-unique-with-mixed-complex-numbers#answer_565643

在 MATLAB Online 中打开

Please let me know if I understood correctly your qestion

    Atest=[-12.9753 - 0.8003i;
         -12.9938 - 0.4003i;
         -13.0000 + 0.0000i;
         -12.9938 + 0.4003i;
          12.9938 + 0.4003i;
          13.0000 + 0.0000i;
          12.9938 - 0.4003i;];
      
      Around=round(Atest);
      A=[real(Around) imag(Around)];
      unique(A,'rows')

ans =

-13 -1

-13 0

13 0

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Joel Miller 2020-12-4

Marco, thank you, this works as I had hoped. I would be interested to know why the round function appears to be outputting integers for the real and imaginary components, but they are treated as floating point when combined into a complex number.

Marco Riani 2020-12-4

Dear Joel,

please see

round function on complex numbers - MATLAB Answers - MATLAB Central (mathworks.com)

Best

Marco

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