How to seperate fractional and decimal part in a real number

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DSP Masters
DSP Masters 2011-11-16
评论： Les Beckham 2023-1-25
Hi, Please help me in seperating fractional and decimal part in a real number. For example: If the value is '1.23', I need to seperate decimal part '1' and 'fractional part '0.23'.
Thanks and regards, soumya..
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Jeremy Wood
Jeremy Wood 2017-7-5
Try using the floor operator to get the greatest integer below your number then subtract out your integer. For example 1.5 - floor(1.5) 0.5. It's trickier with negative numbers though so try using the absolute value of the number then when you get your fractional part multiply it by -1 so for -1.5 you would do -1*(1.5 - floor(1.5))
Bart McCoy
Bart McCoy 2018-7-25
EXTRACTING THE INTEGER PART
Extracting the integer part can be the most tricky part. MATLAB's "fix" function rounds toward zero, which is useful because it extracts the integer part of BOTH positive and negative numbers. It returns doubles and also works on NxM arrays.
By contrast, the "ceil" function always rounds upward, to the next integer in the POSITIVE direction; "floor" always rounds down, to the next integer in the NEGATIVE direction. Use whatever makes sense, but note:
INTEGER EXTRACTION: fix(pi) = 3; fix(-pi) = -3;
ROUNDING UP: ceil(pi) = 4; ceil(-pi) = -3;
ROUNDING DOWN: floor(pi) = 3; floor(-pi)= -4;
EXTRACTING THE FRACTIONAL PART:
fractional_part = value - fix(value);

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采纳的回答

Walter Roberson
Walter Roberson 2016-2-14
number = -1.23
integ = fix(number)
frac = mod(abs(number),1)
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CS MATLAB
CS MATLAB 2016-9-19
What if the number is unknown and you want to compare decimal value with something..
Walter Roberson
Walter Roberson 2016-9-19
Comparing the fraction is risky
http://matlab.wikia.com/wiki/FAQ#Why_is_0.3_-_0.2_-_0.1_.28or_similar.29_not_equal_to_zero.3F
If you want to compare to a certain number of decimal places, N, I recommend comparing round(number*10^N)

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更多回答（5 个）

Naz
Naz 2011-11-16
number=1.23;
integ=floor(number);
fract=number-integ;
Revant Adlakha
Revant Adlakha 2021-2-24
编辑：Revant Adlakha 2021-2-24
How about this?
sign(x)*(abs(x) - floor(abs(x)))
% Number -> x = -1.23
% Answer -> -0.23
% Number -> x = 1.23
% Answer -> 0.23
Resam Makvandi
Resam Makvandi 2012-12-26
编辑：Walter Roberson 2021-2-24
i think the better way is to use:
number = 1.23;
integ = fix(number);
fract = abs(number - integ);
it works for both negative and positive values.
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Les Beckham
Les Beckham 2023-1-25
Ran in:
Did you try it?
x = [0.2, 1.2 1.0]
x = 1×3
0.2000 1.2000 1.0000
integ = fix(x)
integ = 1×3
0 1 1
fract = abs(x - integ)
fract = 1×3
0.2000 0.2000 0

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Are Mjaavatten
Are Mjaavatten 2016-2-9
编辑：Are Mjaavatten 2016-2-9
mod(number,1)
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Are Mjaavatten
Are Mjaavatten 2016-2-13
Point taken. I should be old enough to have learned to read the problem definition. Still, I think it is nice to have a single command for the fractional part.

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Kh.Ehsanur Rahman
Kh.Ehsanur Rahman 2016-2-13
what if the number is -1.23.

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