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

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

Answer 2

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

This presumes your number only has two decimal places.

Answer 3

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

double frac = value % 1;

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.

Answer 5

Better Way -

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

Output -

56

Answer 6

The simplest variant is possibly with Math.truncate()

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

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))

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.

Answer 9

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

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;
    }

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;
    }

Answer 12

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

it gives you 5 from 1.5 :)

Answer 13

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

Answer 14

Here's an extension method I wrote for a similar situation. My application would receive numbers in the format of 2.3 or 3.11 where the integer component of the number represented years and the fractional component represented months.

// Sample Usage
int years, months;
double test1 = 2.11;
test1.Split(out years, out months);
// years = 2 and months = 11

public static class DoubleExtensions
{
    public static void Split(this double number, out int years, out int months)
    {
        years = Convert.ToInt32(Math.Truncate(number));

        double tempMonths = Math.Round(number - years, 2);
        while ((tempMonths - Math.Floor(tempMonths)) > 0 && tempMonths != 0) tempMonths *= 10;
        months = Convert.ToInt32(tempMonths);
    }
}

Answer 15

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)

Answer 16

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

Answer 17

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

Answer 18

It is very simple

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

       Debug.Log(decimalPart);

Answer 19

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 .