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I have the following values that I run through the atan2 function:

a(77:78,:)

ans =

-0.250000000000000 0 -0.950000000000000

-0.260000000000000 0 -0.950000000000000

when I put the above array through the atan2 function, the first value is positive and the second is negative,

atan2(a(77:78,2),a(77:78,3))

ans =

3.141592653589793

-3.141592653589793

If I type in each value of the second row of data by hand, the result is differnt for the atan2 function (it is positive):

atan2(0,-0.95)

ans =

3.141592653589793

If I build a test matrix, manually inputing the values, the atan2 results are both positive.

test = [-0.25, 0 , -0.95; -0.26, 0 , -0.95]

test =

-0.250000000000000 0 -0.950000000000000

-0.260000000000000 0 -0.950000000000000

>> atan2(test(1,2),test(1,3))

ans =

3.141592653589793

>> atan2(test(2,2),test(2,3))

ans =

3.141592653589793

Question: Why are the values of atan2 function positive and negative in the first case but then both positive when I type out the values?

Aside: If done separately:

atan2(a(77,2),a(77,3))

ans =

3.141592653589793

>> atan2(a(78,2),a(78,3))

ans =

-3.141592653589793

Thank you

Stephen Cobeldick
on 25 Nov 2020

Edited: Stephen Cobeldick
on 25 Nov 2020

The reason is a little arcane: negative zero:

The other answers and comments are wrong: this has nothing to do with floating point error or "almost zero" values.

The second column values are exactly zero, and we know that the value is exactly zero because in both long and short format only the exact value zero is displayed as one single digit "0", just as all of your examples show.

So what is the catch?

The catch is simply the IEEE 754 defines two zeros, a positive zero and a negative zero. Same value, different signs. Some (but not all) operations retain the sign when operating on them.

One of your zeros is negative and one is positive, but unfortunately MATLAB** displays them without the sign:

Y = [0;-0] % positive and negative zero!

X = [-0.95;-0.95]; % these values are not really relevant

atan2(Y,X)

Here is a useful trick to detect if a zero is positive or negative (simply displaying it doesn't work):

Y % they look the same ... but are they?

sign(1./Y) % no!

Negative zeros can be generated by some arithmetic operations, or perhaps by converting data from text. If you want to remove negative zeros from an array without changing the sign of the other data, just add zero:

Z = [-0,-2,+0,3]

sign(1./Z)

Z = 0+Z % remove negative zeros

sign(1./Z)

** for what it's worth, Octave does display the sign.

Paul
on 6 Dec 2020

Upstream in this thread it was stated:

"If you want to remove negative zeros from an array without changing the sign of the other data, just add zero"

Just to be clear, adding zero only impacts the real part of a complex number

>> x = [0+1i*3; -0.1-0.1*1i];

>> x = round(x)

x =

0000000000000000 4008000000000000i

8000000000000000 8000000000000000i

>> x + 0

ans =

0000000000000000 4008000000000000i

0000000000000000 8000000000000000i

Unclear if there is a way to simultaneously "fix" the real and imaginary parts other than:

>> z = complex(real(x)+0,imag(x)+0)

z =

4014000000000000 4008000000000000i

0000000000000000 0000000000000000i

or other similar methods that break out the real and imaginary parts, process them, and then recombine.

Bruno Luong
on 6 Dec 2020

>> format hex

>> x = round([0+1i*3; -0.1-0.1*1i])

x =

0000000000000000 4008000000000000i

8000000000000000 8000000000000000i

>> x+complex(0)

ans =

0000000000000000 4008000000000000i

0000000000000000 0000000000000000i

Beside that ATAN2 accepts only real arguments. So not sure this observation is relevant here, and I could't locate the upstream quote.

Paul
on 6 Dec 2020

The upstream quote is in Stephen's original answer.

Wasn't making the observation in the context of ATAN2 alone; was just interested in how to get rid of the negative zero in general.

Adding complex(0) is a much better solution. Thanks for pointing that out.

John D'Errico
on 24 Nov 2020

What you don't understand is that just because MATLAB displays a number represented to 15 digits, does not mean it is EXACTLY the same thing as what you typed in.

In fact, typically when MATLAB shows a number as -0.950000000000000, you should expect there is more to that number than meets the eye. Anyway, a double precision number cannot represent the number you have written exactly. Just as you cannot represent the fraction 2/3 exactly in any finite number of decimal digits, you cannot represent the decimal number 0.95 as an exact binary representation. This number would be represented by the infinitely repeating string...

'11110011001100110011001100110011001100110011001100110...'

So when you typed in 0.95, you certainly were typing in a DIFFERENT number than you havd stored in the array a. And that is why you get a differnt result. It is not atan2, but the numbers you passed to atan2.

Walter Roberson
on 25 Nov 2020

Look at

format long

a(144:145,2)

I suspect the values are not exact 0, and that fact is being hidden by your use of format short

Stephen Cobeldick
on 25 Nov 2020

"I suspect the values are not exact 0, and that fact is being hidden by your use of format short"

@Walter Roberson: can you please show a one single example of a non-zero value that is displayed as the single digit "0" (as all of Richard Bag's second-column examples in this thread show). I can't find any:

>> format short

>> [0,eps(0),-eps(0),-0.95]

ans =

0 0.0000 -0.0000 -0.9500

>> format long

>> [0,eps(0),-eps(0),-0.95]

ans =

0 0.000000000000000 -0.000000000000000 -0.950000000000000

Bruno Luong
on 25 Nov 2020

Edited: Bruno Luong
on 25 Nov 2020

Your result

atan2(a(77:78,2),a(77:78,3))

ans =

3.141592653589793

-3.141592653589793

Type single value

a(78,2)

you will see it's actually negative (not exactly 0).

Yes atan2(y,x) has discontinuity at the semi axes { y = 0 } for x < 0.

Stephen Cobeldick
on 25 Nov 2020

"you will see it's actually negative (not exactly 0)."

It is exactly zero. Negative zero in fact.

MATLAB does not display non-zero numbers as the single digit "0", as all of Richard Bag's examples show.

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