How to seperate fractional and decimal part in a real number
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DSP Masters
el 16 de Nov. de 2011
Comentada: Les Beckham
el 25 de En. de 2023
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..
5 comentarios
Jeremy Wood
el 5 de Jul. de 2017
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
el 25 de Jul. de 2018
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);
Respuesta aceptada
Walter Roberson
el 14 de Feb. de 2016
number = -1.23
integ = fix(number)
frac = mod(abs(number),1)
2 comentarios
CS MATLAB
el 19 de Sept. de 2016
What if the number is unknown and you want to compare decimal value with something..
Walter Roberson
el 19 de Sept. de 2016
Comparing the fraction is risky
If you want to compare to a certain number of decimal places, N, I recommend comparing round(number*10^N)
Más respuestas (5)
Naz
el 16 de Nov. de 2011
number=1.23;
integ=floor(number);
fract=number-integ;
1 comentario
Walter Roberson
el 16 de Nov. de 2011
That fails on negative numbers. For negative numbers, you need fract=number-ceil(number)
Revant Adlakha
el 24 de Feb. de 2021
Editada: Revant Adlakha
el 24 de Feb. de 2021
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
el 26 de Dic. de 2012
Editada: Walter Roberson
el 24 de Feb. de 2021
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.
2 comentarios
Les Beckham
el 25 de En. de 2023
Did you try it?
x = [0.2, 1.2 1.0]
integ = fix(x)
fract = abs(x - integ)
Are Mjaavatten
el 9 de Feb. de 2016
Editada: Are Mjaavatten
el 9 de Feb. de 2016
mod(number,1)
5 comentarios
Are Mjaavatten
el 13 de Feb. de 2016
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.
Jan
el 13 de Feb. de 2016
What about rem instead of mod?
abs(rem(-0.123, 1)) % => 0.123
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