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How to delete/cancel trailing zeros in complex and imaginary numbers?

19 vues (au cours des 30 derniers jours)
Joshua Ford
Joshua Ford le 17 Mai 2021
Commenté : Rik le 17 Mai 2021
Whenever I use or create imaginary and complex numbers, they are always saved with trailing zeros, for example 1.22000000 + 2.150000000i . I want to mimic the behaviour and structure of fixed-point while still being in floating-point. I want 1.22 + 2.15i. How do I delete or cancel the trailing zeros? I have tried different ways and as soon as the imaginary unit is present, the zeros come back.
  1 commentaire
Stephen23
Stephen23 le 17 Mai 2021
"I want 1.22 + 2.15i."
Why?
What specific operations are you performing that cannot be achieved using normal binary floating point numbers?

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Réponses (2)

Rik
Rik le 17 Mai 2021
The way data is stored and the way it is displayed is not necessarilly connected. You can influence the way data is presented with the format function. If you want more control, you need the fprintf or sprintf functions.
format shortG
1.22 + 2.15i
ans =
1.22 + 2.15i
  3 commentaires
James Tursa
James Tursa le 17 Mai 2021
@Joshua Ford You can't have that with double precision numbers. The storage is binary floating point and you can't change that. In fact neither of your 1.22 or 2.15 example numbers can be represented exactly in double precision floating point, and a decimal conversion will have non-zero digits beyond the ones you display above. If you insist on representing these decimal numbers exactly, then you will have to resort to fixed point or Symbolic Toolbox which will greatly slow down your code.
Rik
Rik le 17 Mai 2021
You already have the values very close to that. You can round to a specific number of decimals, but due to what James wrote you can't get close than this:
round(1.223 + 2.148i,2)
ans = 1.2200 + 2.1500i

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Paul Hoffrichter
Paul Hoffrichter le 17 Mai 2021
Modifié(e) : Paul Hoffrichter le 17 Mai 2021
a = 1.22000000 + 2.150000000i
a = 1.2200 + 2.1500i
fprintf("%g + %gi\n", real(a), imag(a));
1.22 + 2.15i
fprintf("%.2f + %.2fi\n", real(a), imag(a));
1.22 + 2.15i
  3 commentaires
Joshua Ford
Joshua Ford le 17 Mai 2021
Does this store the numbers like this or does it only display the numbers as shown?
Paul Hoffrichter
Paul Hoffrichter le 17 Mai 2021
>> Does this store the numbers like this or does it only display the numbers as shown?
The storage has not changed with the choice of display.
Below shows two different variables set to a value with and without trailing zeros. But the two number numbers are represented internally exactly the same way.
a = 1.22 + 2.15i;
b = 1.22000000 + 2.150000000i;
a == b
ans = logical
1

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