Get the decimal part from a double

Question

I want to receive the number after the decimal dot in the form of an integer. For example, only 05 from 1.05 or from 2.50 only 50 not 0.50

Amando

2.50 is just 2.5. I assume you're talking of strings. – Boger

Are you always returning the first two digits after the decimal place? Is the input a decimal, float, string, ...? – Marcellmarcella

An example of usages goes a long way toward getting quality answers, for next time. – Blocker

yes only towo digits ,the input is decimal – Amando

Why not use int y = value.Split('.')[1];? – There

@SearchForKnowledge, . is different in some countries. – Transversal

@There also performing split would take more time than mathematical expressions – Tiffanytiffi

The C# development team should be embarrassed (and take action) that this very common operation has no single Math.function such as Math.DecimalPart(double x) or something. There is way too much "did you try this, did you think about that" for something many of us need to do often. – Hillell

Meta observation: Not getting a "0.50" but getting a number of "50" from "2.50", or getting a number of "05" is one of the most ill-specified requirements I heard or read today.. not mentioning misleading title of the question. That being said, it's interesting how many of the answers focus on the title and focus on how to calculate the non-integral part of a floating number, completely ignoring that the actual question includes quite specific formatting requirements :) Congrats to orad for answering in full. – Bayer

Answer 1

the best of the best way is:

var floatNumber = 12.5523;

var x = floatNumber - Math.Truncate(floatNumber);

result you can convert however you like

Goggles

This should be the accepted answer, a Math solution to a Math question, without all that tedious mucking about in string manipulation – Billowy

I think this solution have a "problem". If the floatNumber was "10.2", the variable x will be something like "0.19999999999999929". Even if this works as expected, the result would be "0.2" and not "2". – Hardunn

@Hardunn Float and Double values always have small errors under subtraction in the last few decimal places. You should not be considering those as significant. If you actually need that kind of precision, you should be using Decimal type. If you can't switch to Decimal (or just don't want to), then you ought to keep track of the expected precision somehow and account for that in comparison operations. I usually use 4 decimal places, but there may be exceptions to that. – Hillell

From msdn.microsoft.com/en-us/library/… : "Because some numbers cannot be represented exactly as fractional binary values, floating-point numbers can only approximate real numbers" – Hillell

Final comment: Using Math.Truncate was the intent of the author's of the Math library for this kind of situation. @Matterai 's answer should be the accepted one. – Hillell

Answer 2

var decPlaces = (int)(((decimal)number % 1) * 100);

This presumes your number only has two decimal places.

Blocker

(318.40d % 1) * 100 outputs 39.9999999999977, you can use casting to get around the rounding error: var decPlaces = (int)(((decimal)number % 1) * 100); – Ayer

@Ayer Good call there - floating point numbers always have these little inaccuracies, so casting it is pretty prudent to ensure consistent behavior. I'll update my answer, though Matterai's is still more technically correct. – Blocker

Answer 3

There is a cleaner and ways faster solution than the 'Math.Truncate' approach:

double frac = value % 1;

Forfeit

a note: for negative values, it is negative: -12.34 % 1 => -0.33999999999999986 – Fra

Answer 4

Solution without rounding problem:

double number = 10.20;
var first2DecimalPlaces = (int)(((decimal)number % 1) * 100);
Console.Write("{0:00}", first2DecimalPlaces);

Outputs: 20

Note if we did not cast to decimal, it would output 19.

Also:

• for 318.40 outputs: 40 (instead of 39)
• for 47.612345 outputs: 61 (instead of 612345)
• for 3.01 outputs: 01 (instead of 1)

If you are working with financial numbers, for example if in this case you are trying to get the cents part of a transaction amount, always use the decimal data type.

Update:

The following will also work if processing it as a string (building on @SearchForKnowledge's answer).

10.2d.ToString("0.00", CultureInfo.InvariantCulture).Split('.')[1]

You can then use Int32.Parse to convert it to int.

Ayer

Thumbs up for mod ones – Stefansson

I used this...10.2d.ToString("0.00", CultureInfo.InvariantCulture).Split('.')[1] – Mejia

Answer 5

Better Way -

        double value = 10.567;
        int result = (int)((value - (int)value) * 100);
        Console.WriteLine(result);

Output -

56

Envious

Did not test, but this suffers from float inaccuracy, no? – Teeny

For 10.20 it will output 19. See my answer. – Ayer

Answer 6

The simplest variant is possibly with Math.truncate()

double value = 1.761
double decPart = value - Math.truncate(value)

Ejection

does not work for value below zero like 0.25 – Fiber

Answer 7

I guess this thread is getting old but I can't believe nobody has mentioned Math.Floor

//will always be .02 cents
(10.02m - System.Math.Floor(10.02m))

Deration

For the super lazy: double fract(double x) { return x - System.Math.Floor(x); } – Bedazzle

Answer 8

var result = number.ToString().Split(System.Globalization.NumberDecimalSeparator)[2]

Returns it as a string (but you can always cast that back to an int), and assumes the number does have a "." somewhere.

Byplay

Answer 9

int last2digits = num - (int) ((double) (num /  100) * 100);

Wynd

Question is to get first 2 decimal digits. For number 318.401567d your solution outputs 0.401567 where 40 is expected. – Ayer

Answer 10

    public static string FractionPart(this double instance)
    {
        var result = string.Empty;
        var ic = CultureInfo.InvariantCulture;
        var splits = instance.ToString(ic).Split(new[] { ic.NumberFormat.NumberDecimalSeparator }, StringSplitOptions.RemoveEmptyEntries);
        if (splits.Count() > 1)
        {
            result = splits[1];
        }
        return result;
    }

Nynorsk

Answer 11

 string input = "0.55";
    var regex1 = new System.Text.RegularExpressions.Regex("(?<=[\\.])[0-9]+");
    if (regex1.IsMatch(input))
    {
        string dp= regex1.Match(input ).Value;
    }

Zomba

It would be awesome if you could add a description on how the regex helps solving the problem – Ecumenicist

Answer 12

In my tests this was 3-4 times slower than the Math.Truncate answer, but only one function call. Perhaps someone likes it:

var float_number = 12.345;
var x = Math.IEEERemainder(float_number , 1)

Deejay

Answer 13

var d = 1.5;
var decimalPart = Convert.ToInt32(d.ToString().Split('.')[1]);

it gives you 5 from 1.5 :)

Microparasite

This will give you 25 for 0.25 though – Urbannai

@Urbannai as the question wants :) – Microparasite

No, he wants 50 for 0.5 – Urbannai

@Urbannai no, he wants 50 from 2.50 – Microparasite

There is no difference between 2.5 and 2.50 – Urbannai

its from the question: For example, only 05 from 1.05 or from 2.50 only 50 not 0.50 – Microparasite

Answer 14

You may remove the dot . from the double you are trying to get the decimals from using the Remove() function after converting the double to string so that you could do the operations required on it

Consider having a double _Double of value of 0.66781, the following code will only show the numbers after the dot . which are 66781

double _Double = 0.66781; //Declare a new double with a value of 0.66781
string _Decimals = _Double.ToString().Remove(0, _Double.ToString().IndexOf(".") + 1); //Remove everything starting with index 0 and ending at the index of ([the dot .] + 1) 

Another Solution

You may use the class Path as well which performs operations on string instances in a cross-platform manner

double _Double = 0.66781; //Declare a new double with a value of 0.66781
string Output = Path.GetExtension(D.ToString()).Replace(".",""); //Get (the dot and the content after the last dot available and replace the dot with nothing) as a new string object Output
//Do something

Epoch

Answer 15

My answer is based on a suspected use-case behind this question and people coming to this question.

The following example program will display a decimal value in a way that appears like a currency like 34.99 or 1.00 unless it has further precision, in which case it will display the whole precision like, 290.19882 or 128.00001

Please critique. Remember the design is for display.

using System;

public class Program
{
    public static void Main()
    {
        Console.WriteLine(ConvertToString(5.00m));
        Console.WriteLine(ConvertToString(5.01m));
        Console.WriteLine(ConvertToString(5.99m));
        Console.WriteLine(ConvertToString(5.000000191m));
        Console.WriteLine(ConvertToString(5.000000191000m));
        Console.WriteLine(ConvertToString(51028931298373.000000191000m));
    }

    public static string ConvertToString(decimal value)
    {
        decimal whole = Math.Truncate(value);
        decimal fractional = value - whole;
        return ConvertToString(whole, fractional);
    }

    public static string ConvertToString(decimal whole, decimal fractional)
    {
        if (fractional == 0m)
        {
            return $"{whole:F0}.00";
        }
        else
        {
            string fs = fractional.ToString("F28").Substring(2).TrimEnd('0');
            return $"{whole:F0}.{fs}";
        }
    }
}

Results

5.00
5.01
5.99
5.000000191
5.000000191
51028931298373.000000191

Priapitis

Answer 16

Use a regex: Regex.Match("\.(?\d+)") Someone correct me if I'm wrong here

Wheen

Not very popular with Regex.Match but I've received the following error when trying to get the value of the decimal ArgumentException was unhandled: parsing "\.(?\d+)" - Unrecognized grouping construct. – Epoch

Answer 17

It is very simple

       float moveWater =  Mathf.PingPong(theTime * speed, 100) * .015f;
       int    m = (int)(moveWater);
       float decimalPart= moveWater -m ;

       Debug.Log(decimalPart);

Spout

Answer 18

Why not use int y = value.Split('.')[1];?

The Split() function splits the value into separate content and the 1 is outputting the 2nd value after the .

There

This will work: Int32.Parse((12.05123d.ToString("0.00").Split('.')[1])) – Ayer

Because you are casting it to a string to do string manipulation. I don't think you realize just how slow and wasteful this is. – Doyenne