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

14 ビュー (過去 30 日間)
Joshua Ford 2021 年 5 月 17 日
コメント済み: Rik 2021 年 5 月 17 日
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.
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Stephen23 2021 年 5 月 17 日
"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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### 回答 (2 件)

Rik 2021 年 5 月 17 日
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
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James Tursa 2021 年 5 月 17 日
@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 2021 年 5 月 17 日
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 2021 年 5 月 17 日

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
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Joshua Ford 2021 年 5 月 17 日
Does this store the numbers like this or does it only display the numbers as shown?
Paul Hoffrichter 2021 年 5 月 17 日
>> 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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