Asked by DSP Masters
on 16 Nov 2011

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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Answer by Naz
on 16 Nov 2011

Accepted answer

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

Walter Roberson
on 16 Nov 2011

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

Answer by Resam Makvandi
on 26 Dec 2012

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.

Answer by Are Mjaavatten
on 9 Feb 2016

Edited by Are Mjaavatten
on 9 Feb 2016

mod(number,1)

Show 2 older comments

Walter Roberson
on 10 Feb 2016

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
on 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 Simon
on 13 Feb 2016

What about `rem` instead of `mod`?

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

Answer by Walter Roberson
on 14 Feb 2016

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

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## 3 Comments

## Jan Simon (view profile)

Direct link to this comment:http://www.mathworks.com/matlabcentral/answers/21339#comment_46396

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

## Jerry Gregoire (view profile)

Direct link to this comment:http://www.mathworks.com/matlabcentral/answers/21339#comment_103137

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 Simon (view profile)

Direct link to this comment:http://www.mathworks.com/matlabcentral/answers/21339#comment_342446

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.