How to seperate fractional and decimal part in a real number

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..

[Jan]

Are you talking of numbers or strings? The quotes in '1.23' might be misleading.

[Jerry Gregoire]

Jan Its my pet peeve when a poster poses a question and it is responded to with another unnecessary question. Yes, in Matlab syntax, '0.23' indicates a string, but it is really obvious that he meant 0.23. I guess my wish to responders is simply, 'Just answer the question already' !!

[Jan]

Some years later: @Jerry: Many questions in this forum are based on the inaccurate knowledge about the classes of variables. I tend to ask for a clarification instead of speculating of what seems obvious.

[Jeremy Wood]

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]

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);

Answer 1

[Walter Roberson]

number = -1.23
integ = fix(number)
frac = mod(abs(number),1)

[CS MATLAB]

What if the number is unknown and you want to compare decimal value with something..

[Walter Roberson]

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)

Answer 2

[Naz]

number=1.23;
integ=floor(number);
fract=number-integ;

[Walter Roberson]

That fails on negative numbers. For negative numbers, you need fract=number-ceil(number)

Answer 3

[Revant Adlakha]

How about this?

sign(x)*(abs(x) - floor(abs(x)))

% Number -> x = -1.23
% Answer -> -0.23

% Number -> x = 1.23
% Answer -> 0.23

[Setsuna Yuuki.]

Thanks!

Answer 4

[Resam Makvandi]

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.

[KOMAL VERMA]

what if there is array

like x=[0.2, 1.2 1.0]

[Les Beckham]

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

Answer 5

[Are Mjaavatten]

 mod(number,1)

[Walter Roberson]

>> mod(-0.123,1)

ans =
                     0.877

However, 0.877 is neither the whole number nor the fraction of -0.123

[Are Mjaavatten]

Walter is right of course. To work for both positive and negative numbers my solution must be mofified to

 mod(abs(number),1)*sign(number)

or just

 mod(abs(number),1)

depending on you definition of the fraction part. I prefer these to the accepted answer because it does not require intermediate variables, but this is a matter of taste.

[Walter Roberson]

The accepted answer by Naz does not use any intermediate variables. The task is to return each of the parts. Naz's solution happens to calculate one part and use it to calculate the other as well, but that does not make either one an intermediate variable.

[Are Mjaavatten]

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]

What about rem instead of mod?

abs(rem(-0.123, 1))   % =>  0.123

Answer 6

[Kh.Ehsanur Rahman]

what if the number is -1.23.