How to seperate fractional and decimal part in a real number
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DSP Masters
am 16 Nov. 2011
Kommentiert: Les Beckham
am 25 Jan. 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 Kommentare
Jeremy Wood
am 5 Jul. 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
am 25 Jul. 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);
Akzeptierte Antwort
Walter Roberson
am 14 Feb. 2016
number = -1.23
integ = fix(number)
frac = mod(abs(number),1)
2 Kommentare
CS MATLAB
am 19 Sep. 2016
What if the number is unknown and you want to compare decimal value with something..
Walter Roberson
am 19 Sep. 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)
Weitere Antworten (5)
Naz
am 16 Nov. 2011
number=1.23;
integ=floor(number);
fract=number-integ;
1 Kommentar
Walter Roberson
am 16 Nov. 2011
That fails on negative numbers. For negative numbers, you need fract=number-ceil(number)
Revant Adlakha
am 24 Feb. 2021
Bearbeitet: Revant Adlakha
am 24 Feb. 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
am 26 Dez. 2012
Bearbeitet: Walter Roberson
am 24 Feb. 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 Kommentare
Les Beckham
am 25 Jan. 2023
Did you try it?
x = [0.2, 1.2 1.0]
integ = fix(x)
fract = abs(x - integ)
Are Mjaavatten
am 9 Feb. 2016
Bearbeitet: Are Mjaavatten
am 9 Feb. 2016
mod(number,1)
5 Kommentare
Are Mjaavatten
am 13 Feb. 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
am 13 Feb. 2016
What about rem instead of mod?
abs(rem(-0.123, 1)) % => 0.123
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