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I have a web application that will apply a percentage markup to a product, but the percentage will be specified by a user. For example, the user can indicate they want to mark up a product 5%, 9%, 23%, etc. My problem is, the product price will change as well, and in doing so, end up giving ugly values ($1462.72)
As a result, my users are hoping that we can round the value to the nearest 5\0 value. So if my marked up product price is $1462.72, it would round up to $1465. $1798.02 on the other hand would round up to an even $1800.
Using VB\C#, how can I go about rounding these values?
Thanks!
To round to an arbitrary modulus you can create a function like:
public decimal Round(decimal source, decimal modulus)
{
return (Math.Round(source / modulus) * modulus);
}
and use it in this way:
decimal rounded = Round(1798.02m , 5.0m); // yields 1800.0
decimal rounded = Round(1462.72m , 5.0m); // yields 1465.0
decimal rounded = Round(2481.23m , 5.0m); // yields 2480.0
Note that Math.Round by default rounds midpoint values to the closest even number (e.g. 1.5 and 2.5 would both "round" to 2. In your case, the effect is that any numbers that are exactly between a 0 and 5 number (i.e. 2.5, 7.5) would be rounded to the closest 10:
decimal rounded = Round(1697.50m , 5.0m); // yields 1700.0
decimal rounded = Round(1702.50m , 5.0m); // yields 1700.0
If you want to always round UP on the midpoint just specify that in Round:
return (Math.Round(source / modulus, MidpointRounding.AwayFromZero) * modulus);
You can use the modulus operator to calculate the adjustment needed.
decimal price = 1798.02;
decimal adjustment = price % 5.0;
if(adjustment != 0) //so we don't round up already round numbers
{
price = (price - adjustment) + 5;
}
This will bring it up to the next multiple of 5.
Am calculating small values in decimal, Expected result is = 345.00 but Result getting is = 345.0000
my code is this decimal temp= 6900 * ( 5 / 100);
temp = 345.0000 how to round this ? I want to show 345.00 ? whats my mistake here ?
I tried this code Math.Round(tempdiscountprice, 2); also but it wont work here. Help me
I don't think you're showing your real code. The following will give a result of 0, because (5 / 100) is evaluated using integer arithmetic:
decimal temp = 6900 * (5 / 100);
Console.WriteLine(temp.ToString()); // result is 0
If you use decimal, you will get the following:
decimal temp = 6900 * (5M / 100);
Console.WriteLine(temp.ToString()); // result is 345.00
However the decimal type does preserve trailing zeroes:
decimal temp = 6900 * (5.0000M / 100);
Console.WriteLine(temp.ToString()); // result is 345.0000
Preserving trailing zeroes like this was introduced in .NET 1.1 for more strict conformance with the ECMA CLI specification.
But there's nothing to stop you formatting the output with two decimals, e.g.:
decimal temp = 6900 * (5.0000M / 100);
Console.WriteLine(temp.ToString("N2")); // result is 345.00
As user2864740 has pointed out above, the problem is not with your rounding, but with displaying the value as a string. Try:
decimal temp = 345.0000m;
String.Format("{0:0.00}", temp)
I think you showing us wrong code because this doesn't get 345.00 or 345.0000 as a result. It always gets 0.
Your temp should be 0 because since you doing integer division, your 5 / 100 is always 0. That's why your temp will 6900 * 0 then it will be 0.
If you show us the right code, your problem looks like showing problem instead of rounding problem.
If you want to format your number as a string, take a look at;
Custom Numeric Format Strings
I have decimal currency amounts (from SQl2k* tables as 32,8) and I would like to round them to Millions with 2 decimal places.
The system will be displaying 1000s of rows with multiple amounts in them, and I need to summarise the totals for various categories into a "quick view".
e.g.
123,456,789.12345678
goes to
123.46
6,655,443,332.2110099
goes to
6,655.44
etc
I know there are issues with rounding and decimal/floating point math:
45,454,454.454545
goes to
45.46 OR 45.45 ?
so also any advice on what's best to use would also be very much appreciated.
Two more examples (of what I'd EXPECT...not what might actually be mathematically correct!)
555444444.444444 => 555.44
555444444.444445 => 555.45
I would expect the final "5" of the second example to cascade up to the decimal place?
decimal d = Math.Round(inValue/1000000m, 2);
double v = Math.Round(45454454.454545, 2);
Will return 45454454.45
So solution would be:
double factor = 1000000;
double v1 = Math.Round(123456789.12345678 / factor, 2); //// 123.46
double v2 = Math.Round(6655443332.2110099 / factor, 2); //// to 6,655.44
IMHO You should use Bankers Rounding instead of Your 'cascade' rounding
I want to do this using the Math.Round function
Here's some examples:
decimal a = 1.994444M;
Math.Round(a, 2); //returns 1.99
decimal b = 1.995555M;
Math.Round(b, 2); //returns 2.00
You might also want to look at bankers rounding / round-to-even with the following overload:
Math.Round(a, 2, MidpointRounding.ToEven);
There's more information on it here.
Try this:
twoDec = Math.Round(val, 2)
If you'd like a string
> (1.7289).ToString("#.##")
"1.73"
Or a decimal
> Math.Round((Decimal)x, 2)
1.73m
But remember! Rounding is not distributive, ie. round(x*y) != round(x) * round(y). So don't do any rounding until the very end of a calculation, else you'll lose accuracy.
Personally I never round anything. Keep it as resolute as possible, since rounding is a bit of a red herring in CS anyway. But you do want to format data for your users, and to that end, I find that string.Format("{0:0.00}", number) is a good approach.
Wikipedia has a nice page on rounding in general.
All .NET (managed) languages can use any of the common language run time's (the CLR) rounding mechanisms. For example, the Math.Round() (as mentioned above) method allows the developer to specify the type of rounding (Round-to-even or Away-from-zero). The Convert.ToInt32() method and its variations use round-to-even. The Ceiling() and Floor() methods are related.
You can round with custom numeric formatting as well.
Note that Decimal.Round() uses a different method than Math.Round();
Here is a useful post on the banker's rounding algorithm.
See one of Raymond's humorous posts here about rounding...
// convert upto two decimal places
String.Format("{0:0.00}", 140.6767554); // "140.67"
String.Format("{0:0.00}", 140.1); // "140.10"
String.Format("{0:0.00}", 140); // "140.00"
Double d = 140.6767554;
Double dc = Math.Round((Double)d, 2); // 140.67
decimal d = 140.6767554M;
decimal dc = Math.Round(d, 2); // 140.67
=========
// just two decimal places
String.Format("{0:0.##}", 123.4567); // "123.46"
String.Format("{0:0.##}", 123.4); // "123.4"
String.Format("{0:0.##}", 123.0); // "123"
can also combine "0" with "#".
String.Format("{0:0.0#}", 123.4567) // "123.46"
String.Format("{0:0.0#}", 123.4) // "123.4"
String.Format("{0:0.0#}", 123.0) // "123.0"
If you want to round a number, you can obtain different results depending on: how you use the Math.Round() function (if for a round-up or round-down), you're working with doubles and/or floats numbers, and you apply the midpoint rounding. Especially, when using with operations inside of it or the variable to round comes from an operation. Let's say, you want to multiply these two numbers: 0.75 * 0.95 = 0.7125. Right? Not in C#
Let's see what happens if you want to round to the 3rd decimal:
double result = 0.75d * 0.95d; // result = 0.71249999999999991
double result = 0.75f * 0.95f; // result = 0.71249997615814209
result = Math.Round(result, 3, MidpointRounding.ToEven); // result = 0.712. Ok
result = Math.Round(result, 3, MidpointRounding.AwayFromZero); // result = 0.712. Should be 0.713
As you see, the first Round() is correct if you want to round down the midpoint. But the second Round() it's wrong if you want to round up.
This applies to negative numbers:
double result = -0.75 * 0.95; //result = -0.71249999999999991
result = Math.Round(result, 3, MidpointRounding.ToEven); // result = -0.712. Ok
result = Math.Round(result, 3, MidpointRounding.AwayFromZero); // result = -0.712. Should be -0.713
So, IMHO, you should create your own wrap function for Math.Round() that fit your requirements. I created a function in which, the parameter 'roundUp=true' means to round to next greater number. That is: 0.7125 rounds to 0.713 and -0.7125 rounds to -0.712 (because -0.712 > -0.713). This is the function I created and works for any number of decimals:
double Redondea(double value, int precision, bool roundUp = true)
{
if ((decimal)value == 0.0m)
return 0.0;
double corrector = 1 / Math.Pow(10, precision + 2);
if ((decimal)value < 0.0m)
{
if (roundUp)
return Math.Round(value, precision, MidpointRounding.ToEven);
else
return Math.Round(value - corrector, precision, MidpointRounding.AwayFromZero);
}
else
{
if (roundUp)
return Math.Round(value + corrector, precision, MidpointRounding.AwayFromZero);
else
return Math.Round(value, precision, MidpointRounding.ToEven);
}
}
The variable 'corrector' is for fixing the inaccuracy of operating with floating or double numbers.
This is for rounding to 2 decimal places in C#:
label8.Text = valor_cuota .ToString("N2") ;
In VB.NET:
Imports System.Math
round(label8.text,2)
I know its an old question but please note for the following differences between Math round and String format round:
decimal d1 = (decimal)1.125;
Math.Round(d1, 2).Dump(); // returns 1.12
d1.ToString("#.##").Dump(); // returns "1.13"
decimal d2 = (decimal)1.1251;
Math.Round(d2, 2).Dump(); // returns 1.13
d2.ToString("#.##").Dump(); // returns "1.13"
Had a weird situation where I had a decimal variable, when serializing 55.50 it always sets default value mathematically as 55.5. But whereas, our client system is seriously expecting 55.50 for some reason and they definitely expected decimal. Thats when I had write the below helper, which always converts any decimal value padded to 2 digits with zeros instead of sending a string.
public static class DecimalExtensions
{
public static decimal WithTwoDecimalPoints(this decimal val)
{
return decimal.Parse(val.ToString("0.00"));
}
}
Usage should be
var sampleDecimalValueV1 = 2.5m;
Console.WriteLine(sampleDecimalValueV1.WithTwoDecimalPoints());
decimal sampleDecimalValueV1 = 2;
Console.WriteLine(sampleDecimalValueV1.WithTwoDecimalPoints());
Output:
2.50
2.00
One thing you may want to check is the Rounding Mechanism of Math.Round:
http://msdn.microsoft.com/en-us/library/system.midpointrounding.aspx
Other than that, I recommend the Math.Round(inputNumer, numberOfPlaces) approach over the *100/100 one because it's cleaner.
You should be able to specify the number of digits you want to round to using Math.Round(YourNumber, 2)
You can read more here.
Math.Floor(123456.646 * 100) / 100
Would return 123456.64
string a = "10.65678";
decimal d = Math.Round(Convert.ToDouble(a.ToString()),2)
public double RoundDown(double number, int decimalPlaces)
{
return Math.Floor(number * Math.Pow(10, decimalPlaces)) / Math.Pow(10, decimalPlaces);
}
When displaying the value of a decimal currently with .ToString(), it's accurate to like 15 decimal places, and since I'm using it to represent dollars and cents, I only want the output to be 2 decimal places.
Do I use a variation of .ToString() for this?
decimalVar.ToString("#.##"); // returns ".5" when decimalVar == 0.5m
or
decimalVar.ToString("0.##"); // returns "0.5" when decimalVar == 0.5m
or
decimalVar.ToString("0.00"); // returns "0.50" when decimalVar == 0.5m
I know this is an old question, but I was surprised to see that no one seemed to post an answer that;
Didn't use bankers rounding
Keeps the value as a decimal.
This is what I would use:
decimal.Round(yourValue, 2, MidpointRounding.AwayFromZero);
http://msdn.microsoft.com/en-us/library/9s0xa85y.aspx
decimalVar.ToString("F");
This will:
Round off to 2 decimal places eg. 23.456 → 23.46
Ensure that there
are always 2 decimal places eg. 23 → 23.00; 12.5 → 12.50
Ideal for displaying currency.
Check out the documentation on ToString("F") (thanks to Jon Schneider).
If you just need this for display use string.Format
String.Format("{0:0.00}", 123.4567m); // "123.46"
http://www.csharp-examples.net/string-format-double/
The "m" is a decimal suffix. About the decimal suffix:
http://msdn.microsoft.com/en-us/library/364x0z75.aspx
Given decimal d=12.345; the expressions d.ToString("C") or String.Format("{0:C}", d) yield $12.35 - note that the current culture's currency settings including the symbol are used.
Note that "C" uses number of digits from current culture. You can always override default to force necessary precision with C{Precision specifier} like String.Format("{0:C2}", 5.123d).
If you want it formatted with commas as well as a decimal point (but no currency symbol), such as 3,456,789.12...
decimalVar.ToString("n2");
There's a very important characteristic of Decimal that isn't obvious:
A Decimal 'knows' how many decimal places it has based upon where it came from
The following may be unexpected :
Decimal.Parse("25").ToString() => "25"
Decimal.Parse("25.").ToString() => "25"
Decimal.Parse("25.0").ToString() => "25.0"
Decimal.Parse("25.0000").ToString() => "25.0000"
25m.ToString() => "25"
25.000m.ToString() => "25.000"
Doing the same operations with Double will result in zero decimal places ("25") for all of the above examples.
If you want a decimal to 2 decimal places there's a high likelyhood it's because it's currency in which case this is probably fine for 95% of the time:
Decimal.Parse("25.0").ToString("c") => "$25.00"
Or in XAML you would use {Binding Price, StringFormat=c}
One case I ran into where I needed a decimal AS a decimal was when sending XML to Amazon's webservice. The service was complaining because a Decimal value (originally from SQL Server) was being sent as 25.1200 and rejected, (25.12 was the expected format).
All I needed to do was Decimal.Round(...) with 2 decimal places to fix the problem regardless of the source of the value.
// generated code by XSD.exe
StandardPrice = new OverrideCurrencyAmount()
{
TypedValue = Decimal.Round(product.StandardPrice, 2),
currency = "USD"
}
TypedValue is of type Decimal so I couldn't just do ToString("N2") and needed to round it and keep it as a decimal.
Here is a little Linqpad program to show different formats:
void Main()
{
FormatDecimal(2345.94742M);
FormatDecimal(43M);
FormatDecimal(0M);
FormatDecimal(0.007M);
}
public void FormatDecimal(decimal val)
{
Console.WriteLine("ToString: {0}", val);
Console.WriteLine("c: {0:c}", val);
Console.WriteLine("0.00: {0:0.00}", val);
Console.WriteLine("0.##: {0:0.##}", val);
Console.WriteLine("===================");
}
Here are the results:
ToString: 2345.94742
c: $2,345.95
0.00: 2345.95
0.##: 2345.95
===================
ToString: 43
c: $43.00
0.00: 43.00
0.##: 43
===================
ToString: 0
c: $0.00
0.00: 0.00
0.##: 0
===================
ToString: 0.007
c: $0.01
0.00: 0.01
0.##: 0.01
===================
Math.Round Method (Decimal, Int32)
Very rarely would you want an empty string if the value is 0.
decimal test = 5.00;
test.ToString("0.00"); //"5.00"
decimal? test2 = 5.05;
test2.ToString("0.00"); //"5.05"
decimal? test3 = 0;
test3.ToString("0.00"); //"0.00"
The top rated answer is incorrect and has wasted 10 minutes of (most) people's time.
Mike M.'s answer was perfect for me on .NET, but .NET Core doesn't have a decimal.Round method at the time of writing.
In .NET Core, I had to use:
decimal roundedValue = Math.Round(rawNumber, 2, MidpointRounding.AwayFromZero);
A hacky method, including conversion to string, is:
public string FormatTo2Dp(decimal myNumber)
{
// Use schoolboy rounding, not bankers.
myNumber = Math.Round(myNumber, 2, MidpointRounding.AwayFromZero);
return string.Format("{0:0.00}", myNumber);
}
None of these did exactly what I needed, to force 2 d.p. and round up as 0.005 -> 0.01
Forcing 2 d.p. requires increasing the precision by 2 d.p. to ensure we have at least 2 d.p.
then rounding to ensure we do not have more than 2 d.p.
Math.Round(exactResult * 1.00m, 2, MidpointRounding.AwayFromZero)
6.665m.ToString() -> "6.67"
6.6m.ToString() -> "6.60"
The top-rated answer describes a method for formatting the string representation of the decimal value, and it works.
However, if you actually want to change the precision saved to the actual value, you need to write something like the following:
public static class PrecisionHelper
{
public static decimal TwoDecimalPlaces(this decimal value)
{
// These first lines eliminate all digits past two places.
var timesHundred = (int) (value * 100);
var removeZeroes = timesHundred / 100m;
// In this implementation, I don't want to alter the underlying
// value. As such, if it needs greater precision to stay unaltered,
// I return it.
if (removeZeroes != value)
return value;
// Addition and subtraction can reliably change precision.
// For two decimal values A and B, (A + B) will have at least as
// many digits past the decimal point as A or B.
return removeZeroes + 0.01m - 0.01m;
}
}
An example unit test:
[Test]
public void PrecisionExampleUnitTest()
{
decimal a = 500m;
decimal b = 99.99m;
decimal c = 123.4m;
decimal d = 10101.1000000m;
decimal e = 908.7650m
Assert.That(a.TwoDecimalPlaces().ToString(CultureInfo.InvariantCulture),
Is.EqualTo("500.00"));
Assert.That(b.TwoDecimalPlaces().ToString(CultureInfo.InvariantCulture),
Is.EqualTo("99.99"));
Assert.That(c.TwoDecimalPlaces().ToString(CultureInfo.InvariantCulture),
Is.EqualTo("123.40"));
Assert.That(d.TwoDecimalPlaces().ToString(CultureInfo.InvariantCulture),
Is.EqualTo("10101.10"));
// In this particular implementation, values that can't be expressed in
// two decimal places are unaltered, so this remains as-is.
Assert.That(e.TwoDecimalPlaces().ToString(CultureInfo.InvariantCulture),
Is.EqualTo("908.7650"));
}
You can use system.globalization to format a number in any required format.
For example:
system.globalization.cultureinfo ci = new system.globalization.cultureinfo("en-ca");
If you have a decimal d = 1.2300000 and you need to trim it to 2 decimal places then it can be printed like this d.Tostring("F2",ci); where F2 is string formating to 2 decimal places and ci is the locale or cultureinfo.
for more info check this link
http://msdn.microsoft.com/en-us/library/dwhawy9k.aspx
https://msdn.microsoft.com/en-us/library/dwhawy9k%28v=vs.110%29.aspx
This link explains in detail how you can handle your problem and what you can do if you want to learn more. For simplicity purposes, what you want to do is
double whateverYouWantToChange = whateverYouWantToChange.ToString("F2");
if you want this for a currency, you can make it easier by typing "C2" instead of "F2"
The most applicable solution is
decimalVar.ToString("#.##");
Double Amount = 0;
string amount;
amount=string.Format("{0:F2}", Decimal.Parse(Amount.ToString()));
If you need to keep only 2 decimal places (i.e. cut off all the rest of decimal digits):
decimal val = 3.14789m;
decimal result = Math.Floor(val * 100) / 100; // result = 3.14
If you need to keep only 3 decimal places:
decimal val = 3.14789m;
decimal result = Math.Floor(val * 1000) / 1000; // result = 3.147
var arr = new List<int>() { -4, 3, -9, 0, 4, 1 };
decimal result1 = arr.Where(p => p > 0).Count();
var responseResult1 = result1 / arr.Count();
decimal result2 = arr.Where(p => p < 0).Count();
var responseResult2 = result2 / arr.Count();
decimal result3 = arr.Where(p => p == 0).Count();
var responseResult3 = result3 / arr.Count();
Console.WriteLine(String.Format("{0:#,0.000}", responseResult1));
Console.WriteLine(String.Format("{0:#,0.0000}", responseResult2));
Console.WriteLine(String.Format("{0:#,0.00000}", responseResult3));
you can put as many 0 as you want.