I have a double and need to convert it to a decimal. If the value is too large/small the result should be decimal.MaxValue/decimal.MinValue. What is the best way to do this?
I assume you realize this is a lossy conversion, and should be avoided if possible? If you really need it:
private static readonly double MaxDecimalAsDouble = (double) decimal.MaxValue;
private static readonly double MinDecimalAsDouble = (double) decimal.MinValue;
...
public decimal ConvertWithCap(double input)
{
return input >= MaxDecimalAsDouble ? decimal.MaxValue
: input <= MinDecimalAsDouble ? decimal.MinValue
: (decimal) input;
}
This way the cap is checked in terms of the double value, but then return the capped value hard-coded to be the decimal value. This avoids the problem where double can't accurately represent the max/min values of decimal.
(Even at the largest scales of decimal, consecutive integers can be represented - but that's not the case for double.)
What would you consider as large and small values for the double??
I feel a simple if else statement will do the job
public decimal getDecimal(double value,double large,double small)
{
if(value <= large && value >= small)
return (decimal) value;
else if (value>large)
return decimal.MaxValue;
else
return decimal.MinValue;
}
This is just typed into the site, so please forgive syntax and spelling
Related
This question already has answers here:
Round a double to x significant figures
(17 answers)
Closed 4 years ago.
This is not a duplicate question. There is an answer posted in the question. Hope it can help.
There are two doubles with the same value with decimals.
(Sorry, this is not a good case. because it will return false sometimes, but I can't find the case. If you try this case, it may not have any problem. So don't waste time to test it.)
double a = 0.70448;
double b = 0.70441;
I want to compare them with only 4 decimals.
I have this helper function to round them down to 4 decimals first.
public static double RoundDown(this double value, int decimals)
{
var multiplier = Math.Pow(10, decimals);
return Math.Floor(value * multiplier) / multiplier;
}
And then I want to check if a is larger than b like this:
RoundDown(a, 4) > RoundDown(b, 4)
Sometimes, for some cases, it will return true even they look equal. I understand very well this is floating issue, so I would like to know if there any elegant solution to compare them.
Updates:
I have tried to multiply it and compare them in integer. However, for this solution, I need to handle double infinity and NAN.
private static CompareResult Compare(double a, double b, double decimals = 0)
{
var multiplier = Math.Pow(10, decimals);
var aInt = Convert.ToInt32(a * multiplier);
var bInt = Convert.ToInt32(b * multiplier);
return aInt > bInt ? CompareResult.Greater : aInt < bInt ? CompareResult.Less : CompareResult.Equal;
}
private enum CompareResult
{
Greater,
Less,
Equal
}
System.OverflowException is thrown if one of the double is larger than int max or infinity. Also, this is not an elegant way to compare double.
Importants:
I am not going to round down with x significant figures. I have already provide this solution in my question, my question is: Even round down to x significant figures, it will return true when comparing them.
Again
I am not finding a way to round down or truncate the doubles to x significant digits. I have no problem on this part.
Answer
Thanks for #m88 answer. But it still cannot solve my problem.
I finally solve this issue using sigma. (Reference: http://forums.codeguru.com/showthread.php?506300-float-double-value-comparison-significant-figures.)
Thanks to some people misunderstand the problem and vote it as a duplicated question. I can't post my answer for others facing the same problem. So I post the answer in my question. I hope it can help others.
public static int CompareTo(this double value1, double value2, int decimals)
{
var diff = value1 - value2;
var sigma = Math.Pow(10, -decimals - 1);
return Math.Abs(diff) < sigma ? 0 : diff > 0 ? 1 : -1;
}
If you use the Math.Round method to round a and b to 4 decimals, a (0.7045) will always be greater than b (0.7044):
const double a = 0.70448;
const double b = 0.70441;
if (Math.Round(a, 4) > Math.Round(b, 4))
...
If you want to truncate the values, you need to be aware of the fact that not all fractions can be accurately represented in a double. If you want "exact" truncating, you might consider converting the double value to a string, truncate the string and then convert the truncated string value back to double. Something like this:
private static double Truncate(double d, int decimals)
{
string s = d.ToString(System.Globalization.CultureInfo.InvariantCulture);
int index = s.IndexOf(System.Globalization.CultureInfo.InvariantCulture.NumberFormat.NumberDecimalSeparator);
if (index > -1)
return Convert.ToDouble($"{s.Substring(0, index + 1)}{s.Substring(index + 1, decimals)}", System.Globalization.CultureInfo.InvariantCulture);
return d;
}
Usage:
const double a = 0.70448;
const double b = 0.70441;
if (Truncate(a, 4) >= Truncate(b, 4))
....
Obviously, if you don't want any "floating issues" as you said in the chat, you cannot work with floating point data types.
You want to truncate, not round:
double a = Math.Truncate(100 * 0.70448) / 100;
double b = Math.Truncate(100 * 0.70441) / 100;
if (a > b)
{
// ...
}
Note that fractions cannot be accurately represented in a double, as per #mm8's comment.
Why this code http://ideone.com/YRcICG
void Main()
{
double a = 0.00004;
Int32 castToInt = (Int32)(1.0/a);
Int32 convertToInt = Convert.ToInt32(1.0/a);
Console.WriteLine("{0} {1:F9} {2:F9}", castToInt == convertToInt, castToInt, convertToInt);
Console.WriteLine((((int)(1.0/(1.0/25000))) == 24999));
}
results in
False 24999,000000000 25000,000000000
True
in context of CLR/C# implementation
The trick lies in the way the double is represented so (1.0/a) will be represented in the following way:
(1.0/a) = 24999.99999999999636202119290828704833984375.
When you use cast you get only the first part of this number, while the convert Method works in a different way, here is the code for the Convert method:
public static int ToInt32(double value)
{
if (value >= 0.0)
{
if (value < 2147483647.5)
{
int num = (int)value;
double num2 = value - (double)num;
if (num2 > 0.5 || (num2 == 0.5 && (num & 1) != 0))
{
num++;
}
return num;
}
}
else
{
if (value >= -2147483648.5)
{
int num3 = (int)value;
double num4 = value - (double)num3;
if (num4 < -0.5 || (num4 == -0.5 && (num3 & 1) != 0))
{
num3--;
}
return num3;
}
}
throw new OverflowException(Environment.GetResourceString("Overflow_Int32"));
}
As you can see there is an if statement that checks the difference between casted value and original double, in your example it is:
int num = (int)value;
double num2 = value - (double)num;
24999.99999999999636202119290828704833984375 - 24999 > 0.5
, hence you get the increment.
In your calculation, the answer to 1.0/0.00004 is getting converted to a value very slightly smaller than 2500, because floating-point numbers can't precisely represent all possible values. Given that,
Why do the two integers have different values?
Casting a double to an int truncates the value, so everything after the decimal point is discarded.
Convert.ToInt32 on a double rounds to the nearest integer.
Why are floating point numbers imprecise?
See the excellent article linked in another answer: What Every Computer Scientist Should Know About Floating-Point Arithmetic
How can I represent values precisely?
You could use a decimal type rather than double. There are pros and cons to doing this, see decimal vs double! - Which one should I use and when?
Cast trunk the floating number
(Int32)41.548 == 41
Convert round the number (feature?)
Convert.ToInt32(41.548) == 42
Floating point math is not exact. Simple values like 0.2 cannot be precisely represented using binary floating point numbers, and the limited precision of floating point numbers means that slight changes in the order of operations can change the result. A must read:
What Every Computer Scientist Should Know About Floating-Point Arithmetic
The IEEE standard divides exceptions into 5 classes: overflow, underflow, division by zero, invalid operation and inexact. There is a separate status flag for each class of exception. The meaning of the first three exceptions is self-evident. Invalid operation covers the situations listed in TABLE D-3, and any comparison that involves a NaN.
I have tried to do this - Making any float value to int value. What I tried to achieve is make it 2.00-2.99 goes to 2.
Note: Im not trying to do Approximation (lower than 2.49 goes to 2, and those 2.50+ goes to 3).
I have done this so far
public int RemoveDecimal(float value)
{
string TempText = (string)value;
TempText.Remove(IndexOf('.'));
return int.parse(TempText);
}
But this will get an error when the float value ends with .00
How can I achieve this?
Thanks for any help
Math.Floor is the function you need.
It:
Returns the largest integer less than or equal to the specified number.
Example:
var val = Math.Floor(value);
Or, you could simply cast to an integer - this will simply ignore the decimal portion, so long as the range of the decimal is within the range of an int (otherwise you will get an exception):
int noDecimals = (int)value;
you can't always convert a float to an int because most of the values in float are out of range for int. you should return a string.
the following will do just that.
public string RemoveDecimal(float value)
{
string TempText = value.ToString("#");
return TempText;
}
Simply casting a floating point number to int does what you want.
i = (int)myFloat;
It truncates the fractional digits, i.e. it always goes in the direction of 0.
(int)2 == 2
(int)1.9 == 1
(int)-1.5 == -1
This will obviously not work correctly if the result is outside the valid range of int. If you want to achieve the same thing, but with a floating point number as result, Math.Truncate is what you need.
I have a double typed variable. This variable stores information that is part of a more complex formula. Importantly, this variable can only include information up to the tenths location, or one decimal position (i.e. 10.1, 100.2, etc). However, when determining this value, it must be calculated such that anything past the tenths location is truncated, not rounded. For instance:
if the value equals 10.44, The variable value should be 10.4.
if the value equals 10.45, The variable value should also be set to 10.4
How do I truncate values in C# with respect to a decimal place?
Using an extension method:
public static double RoundDown(this double value, int digits)
{
int factor = Math.Pow(10,digits);
return Math.Truncate(value * factor) / factor;
}
Then you simply use it like this:
double rounded = number.RoundDown(2);
You have to do that by your own:
public static decimal Truncate(decimal value, int decimals)
{
if ((decimals < 0) || (decimals > 28))
{
throw new ArgumentOutOfRangeException("decimals", "The number of fractional decimals must be between 0 and 28.");
}
decimal integral = Math.Truncate(value);
decimal fractional = value - integral;
decimal shift = (decimal)Math.Pow(10, decimals);
fractional = Math.Truncate(shift * fractional);
fractional = fractional / shift;
return (integral + fractional);
}
System.Math.Truncate (d * 10) / 10
Generally, if you're working with numbers where the precise decimal representation is important, you should use decimal - not double.
With decimal, you can do something like...
decimal d = ...;
d = decimal.Truncate(d*10)/10;
If you use a double value, your truncated number will not generally be precisely representable - you may end up with excess digits or minor rounding errors. For example Math.Truncate((4.1-4.0)*10) is not 1, but 0.
While I would probably use Phillippe's answer, if you wanted to avoid scaling the number up (unlikely to be a problem for 1dp), you could:
public static double RoundDown(this double x, int numPlaces)
{
double output = Math.Round(x, numPlaces, MidpointRounding.AwayFromZero);
return (output > x ? output - Math.Pow(10, -numPlaces) : output);
}
How can i truncate the leading digit of double value in C#,I have tried Math.Round(doublevalue,2) but not giving the require result. and i didn't find any other method in Math class.
For example i have value 12.123456789 and i only need 12.12.
EDIT: It's been pointed out that these approaches round the value instead of truncating. It's hard to genuinely truncate a double value because it's not really in the right base... but truncating a decimal value is more feasible.
You should use an appropriate format string, either custom or standard, e.g.
string x = d.ToString("0.00");
or
string x = d.ToString("F2");
It's worth being aware that a double value itself doesn't "know" how many decimal places it has. It's only when you convert it to a string that it really makes sense to do so. Using Math.Round will get the closest double value to x.xx00000 (if you see what I mean) but it almost certainly won't be the exact value x.xx00000 due to the way binary floating point types work.
If you need this for anything other than string formatting, you should consider using decimal instead. What does the value actually represent?
I have articles on binary floating point and decimal floating point in .NET which you may find useful.
What have you tried? It works as expected for me:
double original = 12.123456789;
double truncated = Math.Truncate(original * 100) / 100;
Console.WriteLine(truncated); // displays 12.12
double original = 12.123456789;
double truncated = Truncate(original, 2);
Console.WriteLine(truncated.ToString());
// or
// Console.WriteLine(truncated.ToString("0.00"));
// or
// Console.WriteLine(Truncate(original, 2).ToString("0.00"));
public static double Truncate(double value, int precision)
{
return Math.Truncate(value * Math.Pow(10, precision)) / Math.Pow(10, precision);
}
How about:
double num = 12.12890;
double truncatedNum = ((int)(num * 100))/100.00;
This could work (although not tested):
public double RoundDown(this double value, int digits)
{
int factor = Math.Pow(10,digits);
return Math.Truncate(value * factor) / factor;
}
Then you simply use it like this:
double rounded = number.RoundDown(2);
This code....
double x = 12.123456789;
Console.WriteLine(x);
x = Math.Round(x, 2);
Console.WriteLine(x);
Returns this....
12.123456789
12.12
What is your desired result that is different?
If you want to keep the value as a double, and just strip of any digits after the second decimal place and not actually round the number then you can simply subtract 0.005 from your number so that round will then work. For example.
double x = 98.7654321;
Console.WriteLine(x);
double y = Math.Round(x - 0.005, 2);
Console.WriteLine(y);
Produces this...
98.7654321
98.76
There are a lot of answers using Math.Truncate(double).
However, the approach using Math.Truncate(double) can lead to incorrect results.
For instance, it will return 5.01 truncating 5.02, because multiplying of double values doesn't work precisely and 5.02*100=501.99999999999994
If you really need this precision, consider, converting to Decimal before truncating.
public static double Truncate(double value, int precision)
{
decimal power = (decimal)Math.Pow(10, precision);
return (double)(Math.Truncate((decimal)value * power) / power);
}
Still, this approach is ~10 times slower.
I'm sure there's something more .netty out there but why not just:-
double truncVal = Math.Truncate(val * 100) / 100;
double remainder = val-truncVal;
If you are looking to have two points after the decimal without rounding the number, the following should work
string doubleString = doublevalue.ToString("0.0000"); //To ensure we have a sufficiently lengthed string to avoid index issues
Console.Writeline(doubleString
.Substring(0, (doubleString.IndexOf(".") +1) +2));
The second parameter of substring is the count, and IndexOf returns to zero-based index, so we have to add one to that before we add the 2 decimal values.
This answer is assuming that the value should NOT be rounded
For vb.net use this extension:
Imports System.Runtime.CompilerServices
Module DoubleExtensions
<Extension()>
Public Function Truncate(dValue As Double, digits As Integer)
Dim factor As Integer
factor = Math.Pow(10, digits)
Return Math.Truncate(dValue * factor) / factor
End Function
End Module
I use a little formatting class that I put together which can add gaps and all sorts.
Here is one of the methods that takes in a decimal and return different amounts of decimal places based on the decimal display setting in the app
public decimal DisplayDecimalFormatting(decimal input, bool valueIsWeightElseMoney)
{
string inputString = input.ToString();
if (valueIsWeightElseMoney)
{
int appDisplayDecimalCount = Program.SettingsGlobal.DisplayDecimalPlacesCount;
if (appDisplayDecimalCount == 3)//0.000
{
inputString = String.Format("{0:#,##0.##0}", input, displayCulture);
}
else if (appDisplayDecimalCount == 2)//0.00
{
inputString = String.Format("{0:#,##0.#0}", input, displayCulture);
}
else if (appDisplayDecimalCount == 1)//0.0
{
inputString = String.Format("{0:#,##0.0}", input, displayCulture);
}
else//appDisplayDecimalCount 0 //0
{
inputString = String.Format("{0:#,##0}", input, displayCulture);
}
}
else
{
inputString = String.Format("{0:#,##0.#0}", input, displayCulture);
}
//Check if worked and return if worked, else return 0
bool itWorked = false;
decimal returnDec = 0.00m;
itWorked = decimal.TryParse(inputString, out returnDec);
if (itWorked)
{
return returnDec;
}
else
{
return 0.00m;
}
}
object number = 12.123345534;
string.Format({"0:00"},number.ToString());