# grabbing number after decimal

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##### 1 Comment

Bart McCoy
on 25 Jul 2018

The "fix" function is useful here, since it rounds toward 0. This is important when extracting the fractional part of a NEGATIVE number:

If value = pi, then "value - fix(value)" = 0.141592653589793

If value = -pi, then "value - fix(value)" = -0.141592653589793

### Accepted Answer

Star Strider
on 20 Aug 2014

If you know you will only have one number after the decimal, just use the rem function:

x = 1234.5;

d = 10*rem(x,1);

produces:

d =

5

##### 1 Comment

Elena
on 22 Feb 2022

### More Answers (5)

David Sanchez
on 20 Aug 2014

N = 1234.5;

R = 10*(N-floor(N))

R =

5

##### 3 Comments

Patrik Ek
on 20 Aug 2014

@Hikaru It is true that R becomes 50 when there is only 1 decimal digit in the number. This is not a bug though, since the second digit indeed would be 0 if N is exactly 1234.5. If the accuracy of N is one digit (or if you only want the first decimal digit), it is still not a problem. Then you explicitely need to state that you do only want one digit and not 2 by setting nDigits=1. This is not an unreasonable approach, since this decision should really be the user's.

And about using round instead of floor I do not buy it. You solve a problem that is unlikely to happen, by creating a problem that will occur in 50% of all the cases. Assume that I would use round in my case; This would give the result R = 54, since round(53.7) is 54. This could possibly be a common case. However, since you do the operation N-floor(N), you are unlikely to get floating point errors. Focus on the real problem instead of obscure errors that will probably not occur anyway. An alternative could be you uploading a solution that is safe from floating point errors. To use round is not really a solution to the problem. It is just a workaround and in this case a workaround that does not work as intended.

Hikaru
on 20 Aug 2014

Edited: Hikaru
on 20 Aug 2014

One way to do this

specialNumber = [1234.5; 9087.3];

d = abs(specialNumber - fix(specialNumber));

answer = d*10; %assuming that your data only has one decimal place

final = round(answer)

Note: This is not really the best solution since subtracting floating points can introduce error.

##### 1 Comment

Patrik Ek
on 20 Aug 2014

Patrik Ek
on 20 Aug 2014

Since there have been a discussion about floating point errors here (which IMO seems to be overkill to care about for this operation) I will provide an integer casting solution. This should be free from floating point issues since no addition or subtraction is done with floating points.

N = 1234.537;

nDigits = 2;

R = cast(floor(10^nDigits*N),'int32') - cast(10^nDigits*floor(N),'int32');

R = cast(R,'double') % If double is important then cast it back

The floor in the first term is there because of matlabs solution of casting, which rounds the double to closest int.

##### 0 Comments

Image Analyst
on 20 Aug 2014

neesha, what if there are 3 or 4 numbers after the decimal point? Do you want only the first one, or all of them? What if you did

fractions = t.specialNumber - floor(t.specialNumber);

where t is the table and specialNumber is the column in the table. So for example 73.8234 would give 0.8234 as a result. Would that work for you? Or do you only want the 8 as 8 and not as 0.8 or 0.8234?

##### 0 Comments

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