Extrating the fractional coefficient of a double

Question

A number 170 can be represented as a normalised form with:
1,7 as the fractional coefficient
+2 as the exponent with 10 as the base.

170 = 1,7 × 10+2

Is there a way to extract the "fractional coefficient" in a mathematical way? Without using the string representation.

var testInputs = 
    new double[] {
        1.7E+1,     // 17
        1.7E+2,     // 170
        1.7E+3,     // 1700

        -1.7E+1,    // -17
        -1.7E+2,    // -170
        -1.7E+3,    // -1700

        1.7E-1,     // 0,17
        1.7E-2,     // 0,017
        1.7E-3,     // 0,0017

        -1.7E-1,    // -0,17
        -1.7E-2,    // -0,017
        -1.7E-3,    // -0,0017
    };

The absolute value of the fractional coefficient of those test is : 1,7.

Nb: , is the decimal separator.

Disclaimer not Homework, I was asking my self if there was a bitwise operation that give result better that somethign like `string GetFractional(double input) => input.ToString("E").Substring(0, input.ToString("E").IndexOf("E")).TrimEnd('0');` not tested. — xdtTransform, Mar 04 '19 at 15:07
how about If x > 0 divide by 10 until the value becomes smaller then 0.... — Mong Zhu, Mar 04 '19 at 15:09
Note that your array here isn't what you expect. The commas are separating the numbers into two distinct values. — DavidG, Mar 04 '19 at 15:09
@canton7, "The absolute value of the fractional coefficient of those test is : 1,7." the result is either 1.7 or -1.7 the part without the E — xdtTransform, Mar 04 '19 at 15:09
[This question](https://stackoverflow.com/questions/389993/extracting-mantissa-and-exponent-from-double-in-c-sharp) may help you — John M, Mar 04 '19 at 15:11
@xdtTransform If you not really care about accuracy you can use logarithm `var log = Math.Log10(1700); var res = Math.Pow(10, log - Math.Truncate(log));` — Aleks Andreev, Mar 04 '19 at 15:15
@JohnM, while this seems revelant and it's from Skeet, the function return Nothing and the value of those variable in debug looked wrong. For -1.7E-3, I got -62 and 7839866231326559. — xdtTransform, Mar 04 '19 at 15:31

score 5 · Accepted Answer · edited Mar 04 '19 at 15:51

EDIT: initial version didn't work for (-1<x<1)

If I get your query right...

var X = -1.7; 
var Y = (decimal)X/(decimal)(Math.Pow(10, (int)Math.Log10(Math.Abs(X))-(X<1 && X > -1?1:0)))

Test:

var inputs = new[] { 1700,170,17,1.7,0.17,0.017,0.0017,
                    -1700, -170, -17, -1.7, -0.17, -0.017, -0.0017 };

decimal  GetFractional(decimal X) =>
    (decimal)X/(decimal)(Math.Pow(10, (int)Math.Log10(Math.Abs(X))-(X<1 && X > -1?1:0)));

foreach (var x in inputs) 
{ 
    WriteLine($"{x,7}\t\t{GetFractional(x)}"); 
}

Result :

1700        1,7
 170        1,7
  17        1,7
 1,7        1,7
0,17        1,7
0,017       1,7
0,0017      1,7
-1700       -1,7
-170        -1,7
 -17        -1,7
-1,7        -1,7
-0,17       -1,7
-0,017      -1,7
-0,0017     -1,7

You should explain what this does and why you use `Floor` through that cast to `int` — Panagiotis Kanavos, Mar 04 '19 at 15:29

score 1 · Answer 2 · answered Mar 04 '19 at 15:21

Here would be an algorithmic approach to solve this:

public double ExtractFraction(double value)
{
    if (value == 0) return value;

    if (value > 10 || value < -10)
    {
        while (value > 10 || value < -10)
        {
            value /= 10.0;
        }
    }
    else if (value < 1 || value > -1)
    {
        while (value < 1 && value > -1)
        {
            value *= 10.0;
        }
    }
    return value;
}

Explanation: If the value is outside the range ]10 -10[ you need to divide by 10 until you slide inside this range

if the value is inside the range ]1 -1[ you need to multiply by 10 until you slide out of this range

Like it. Add 0 Remove 0. It loop a bit but the math is simple. — xdtTransform, Mar 04 '19 at 15:35
@xdtTransform yo, you can round it then as far as you like. I like loops :) — Mong Zhu, Mar 04 '19 at 15:38

score 0 · Answer 3 · answered Jan 18 '23 at 10:56

This is how I extract the exponent and coefficient of a large number (valueToConvert) for a project I worked on:

int Exp = (int)Math.Log10(valueToConvert);
double Coef = (valueToConvert / Math.Pow(10,Exp)).ToString("f2");

For example, if valueToConvert = 5.43E+123, I get Coef = 5.43, Exp = 123.

Extrating the fractional coefficient of a double

3 Answers3