How do I tell if a decimal or double value is an integer?
For example:
decimal d = 5.0; // Would be true
decimal f = 5.5; // Would be false
or
double d = 5.0; // Would be true
double f = 5.5; // Would be false
The reason I would like to know this is so that I can determine programmatically if I want to output the value using .ToString("N0") or .ToString("N2"). If there is no decimal point value, then I don't want to show that.
For floating point numbers, n % 1 == 0 is typically the way to check if there is anything past the decimal point.
public static void Main (string[] args)
{
decimal d = 3.1M;
Console.WriteLine((d % 1) == 0);
d = 3.0M;
Console.WriteLine((d % 1) == 0);
}
Output:
False
True
Update: As #Adrian Lopez mentioned below, comparison with a small value epsilon will discard floating-point computation mis-calculations. Since the question is about double values, below will be a more floating-point calculation proof answer:
Math.Abs(d % 1) <= (Double.Epsilon * 100)
There are any number of ways to do this. For example:
double d = 5.0;
bool isInt = d == (int)d;
You can also use modulo.
double d = 5.0;
bool isInt = d % 1 == 0;
How about this?
public static bool IsInteger(double number) {
return number == Math.Truncate(number);
}
Same code for decimal.
Mark Byers made a good point, actually: this may not be what you really want. If what you really care about is whether a number rounded to the nearest two decimal places is an integer, you could do this instead:
public static bool IsNearlyInteger(double number) {
return Math.Round(number, 2) == Math.Round(number);
}
Whilst the solutions proposed appear to work for simple examples, doing this in general is a bad idea. A number might not be exactly an integer but when you try to format it, it's close enough to an integer that you get 1.000000. This can happen if you do a calculation that in theory should give exactly 1, but in practice gives a number very close to but not exactly equal to one due to rounding errors.
Instead, format it first and if your string ends in a period followed by zeros then strip them. There are also some formats that you can use that strip trailing zeros automatically. This might be good enough for your purpose.
double d = 1.0002;
Console.WriteLine(d.ToString("0.##"));
d = 1.02;
Console.WriteLine(d.ToString("0.##"));
Output:
1
1.02
bool IsInteger(double num) {
if (ceil(num) == num && floor(num) == num)
return true;
else
return false;
}
Problemo solvo.
Edit: Pwned by Mark Rushakoff.
static bool IsWholeNumber(double x)
{
return Math.Abs(x % 1) < double.Epsilon;
}
Mark Rushakoff's answer may be simpler, but the following also work and may be more efficient since there is no implicit division operation:
bool isInteger = (double)((int)f) == f ;
and
bool isInteger = (decimal)((int)d) == d ;
If you want a single expression for both types, perhaps
bool isInteger = (double)((int)val) == (double)val ;
.NET 7 now has built-in methods for this:
decimal.IsInteger: https://learn.microsoft.com/en-us/dotnet/api/system.decimal.isinteger?view=net-7.0
double.IsInteger: https://learn.microsoft.com/en-us/dotnet/api/system.double.isinteger?view=net-7.0
You can check out the source code at:
https://github.com/dotnet/runtime/blob/main/src/libraries/System.Private.CoreLib/src/System/Decimal.cs
https://github.com/dotnet/runtime/blob/main/src/libraries/System.Private.CoreLib/src/System/Double.cs
If upper and lower bound of Int32 matters:
public bool IsInt32(double value)
{
return value >= int.MinValue && value <= int.MaxValue && value == (int)value;
}
You can use String formatting for the double type. Here is an example:
double val = 58.6547;
String.Format("{0:0.##}", val);
//Output: "58.65"
double val = 58.6;
String.Format("{0:0.##}", val);
//Output: "58.6"
double val = 58.0;
String.Format("{0:0.##}", val);
//Output: "58"
Let me know if this doesn't help.
public static bool isInteger(decimal n)
{
return n - (Int64)n == 0;
}
I faced a similar situation, but where the value is a string. The user types in a value that's supposed to be a dollar amount, so I want to validate that it's numeric and has at most two decimal places.
Here's my code to return true if the string "s" represents a numeric with at most two decimal places, and false otherwise. It avoids any problems that would result from the imprecision of floating-point values.
try
{
// must be numeric value
double d = double.Parse(s);
// max of two decimal places
if (s.IndexOf(".") >= 0)
{
if (s.Length > s.IndexOf(".") + 3)
return false;
}
return true;
catch
{
return false;
}
I discuss this in more detail at http://progblog10.blogspot.com/2011/04/determining-whether-numeric-value-has.html.
Using int.TryParse will yield these results:
var shouldBeInt = 3;
var shouldntBeInt = 3.1415;
var iDontWantThisToBeInt = 3.000f;
Console.WriteLine(int.TryParse(shouldBeInt.ToString(), out int parser)); // true
Console.WriteLine(int.TryParse(shouldntBeInt.ToString(), out parser)); // false
Console.WriteLine(int.TryParse(iDontWantThisToBeInt.ToString(), out parser)); // true, even if I don't want this to be int
Console.WriteLine(int.TryParse("3.1415", out parser)); // false
Console.WriteLine(int.TryParse("3.0000", out parser)); // false
Console.WriteLine(int.TryParse("3", out parser)); // true
Console.ReadKey();
You can simply compare the double against the int cast of the same value.
double d = 5.0m;
if (d == (int)d)
{
....
}
This is my solution to this problem. Maybe someone will useful.
public static bool IsInt(object number, int? decimalPlaces = null)
{
bool isInt;
var splinted = number.ToString().Split(',');
if (splinted.Length == 1)
isInt = true;
else
{
var charsAfterComma = decimalPlaces != null ? splinted[1].Substring(0, (int) decimalPlaces) : splinted[1];
isInt = charsAfterComma.First().ToString() == "0" && charsAfterComma.Replace("0", "") == "";
}
return isInt;
}
Try this:
number == Convert.ToInt16(number);
Perhaps not the most elegant solution but it works if you are not too picky!
bool IsInteger(double num) {
return !num.ToString("0.################").Contains(".");
}
Related
I need to found out if a given String is a pure Integer number, that means it is no double and no float .
What I have tried is a regular expression:
#"^\d+$"
if (Regex.IsMatch(oStreamDataValues[i].ToString(), #"^\d+$") == false)
That works fine but only until a String contains a number like 0.00 or any other digit value.
You can use the TryParse method:
int value;
if(int.TryParse(myString, out value)) {
// it is an integer
}
Why not just try to parse it using int.TryParse:
string s = "-12";
int i;
bool isInt = int.TryParse(s, out i);
If you also want to parse 0.00:
bool isInt;
decimal d;
if (decimal.TryParse(s, out d))
{
if (d % 1 == 0)
{
isInt = true;
}
else
{
isInt = false;
}
}
else
{
isInt = false;
}
As I can see, some floating point values are positive examples in your case
4.00 // <- just 4
1e3 // <- is 1000
1e100 // <- 10...0 (100 zeros)
So I suggest parsing it into Double, and then test if fractional part is 0:
String s = "123.0";
Double v;
if (Double.TryParse(s, NumberStyles.Any, CultureInfo.InvariantCulture, out v))
if ((v - Math.Truncate(v)) < Math.Abs(v) / 1.0e20) //TODO: Adjast tolerance here
Console.Write("this is an integer value");
I personnaly would use Int32 static methods to parse the String.
There is TryParse and Parse (depends if you want to use a try/catch or not).
int returnedNumber = 0;
if(!Int32.TryParse(oStreamDataValues[i].ToString(), out returnedNumber)) {
//enter code here
}
or
bool isInt = true;
int returnedNumber = 0;
try {
returnedNumber = Int32.Parse(oStreamDataValues[i].ToString());
}
catch(FormatException) {
isInt = false;
}
if(!isInt) {
//enter code here
}
Parse method uses the following format :
The s parameter contains a number of the form:
[ws][sign]digits[ws]
Items in square brackets ([ and ]) are optional. The following table describes each element.
ws Optional white space.
sign An optional sign.
digits A sequence of digits ranging from 0 to 9.
References :
Parse : https://msdn.microsoft.com/en-us/library/b3h1hf19%28v=vs.110%29.aspx
TryParse : https://msdn.microsoft.com/en-us/library/f02979c7%28v=vs.110%29.aspx
I have an input type integer that represents a number that needs to be converted to double between 1-100, and the rest is decimal precision.
Example: 1562 -> 15.62 ; 198912 -> 19.8912
Right now, I tried a conversion to string, count the number of characters, take 2 to check how many decimals I have and depending of the result "create" a composite string to get a valid double...
Any idea of there is a better way of resolving convert-precision on runtime.
What about this:
int value = 1562;
decimal d = value;
while (d > 100) {
d /= 10;
}
You can use LINQ Skip and Take like:
string str = "198912";
string newStr = string.Format("{0}.{1}", new string(str.Take(2).ToArray()), new string(str.Skip(2).ToArray()));
double d = double.Parse(newStr, CultureInfo.InvariantCulture);
You can add the checks for length on original string, and also use double.TryParse to see if you get valid values.
If you have an int to begin with then you can use decimal, which would provide you more accurate conversion. Like:
int number = 1562123123;
decimal decimalNumber = number;
while (decimalNumber > 100)
{
decimalNumber /= 10;
}
Here is a mathematical solution. The line lg = Math.Max(lg, 0); changes "2" to return "2.0" instead of "20.0" but I guess that depends on your needs for single digit numbers.
static double ToDoubleBetween1And100(int num)
{
var lg = Math.Floor(Math.Log10(num)) - 1;
lg = Math.Max(lg, 0);
return ((double)num) / Math.Pow(10, lg);
}
I have a variable of decimal type and I want to check the number of digits before decimal point in it.
What should I do? For example, 467.45 should return 3.
Solution without converting to string (which can be dangerous in case of exotic cultures):
static int GetNumberOfDigits(decimal d)
{
decimal abs = Math.Abs(d);
return abs < 1 ? 0 : (int)(Math.Log10(decimal.ToDouble(abs)) + 1);
}
Note, that this solution is valid for all decimal values
UPDATE
In fact this solution does not work with some big values, for example: 999999999999998, 999999999999999, 9999999999999939...
Obviously, the mathematical operations with double are not accurate enough for this task.
While searching wrong values I tend to use string-based alternatives proposed in this topic. As for me, that is the evidence that they are more reliable and easy-to-use (but be aware of cultures). Loop-based solutions can be faster though.
Thanks to commentators, shame on me, lesson to you.
Instead of converting to string, you can also divide the number by 10 until it equals 0. Interesting is, that the mathematical operations on decimals are much slower than converting the decimal to a string and returning the length (see benchmarks below).
This solution does not use the Math-methods that take a double as input; so all operations are done on decimals and no casting is involved.
using System;
public class Test
{
public static void Main()
{
decimal dec = -12345678912345678912345678912.456m;
int digits = GetDigits(dec);
Console.WriteLine(digits.ToString());
}
static int GetDigits(decimal dec)
{
decimal d = decimal.Floor(dec < 0 ? decimal.Negate(dec) : dec);
// As stated in the comments of the question,
// 0.xyz should return 0, therefore a special case
if (d == 0m)
return 0;
int cnt = 1;
while ((d = decimal.Floor(d / 10m)) != 0m)
cnt++;
return cnt;
}
}
Output is 29. To run this sample, visit this link.
Side note: some benchmarks show surprising results (10k runs):
while ((d = decimal.Floor(d / 10m)) != 0m): 25ms
while ((d = d / 10m) > 1m): 32ms
ToString with Math-double-operations: 3ms
ToString with decimal-operations: 3ms
BigInt (see answer of #Heinzi): 2ms
Also using random numbers instead of always the same value (to avoid possible caching of the decimal to string conversion) showed that the string-based methods are much faster.
I would try this:
Math.Truncate(467.45).ToString().Length
If you want to be sure not having some weird results for different cultures and with negative decimals, you better use this:
var myDecimal = 467.45m;
Math.Truncate(Math.Abs(myDecimal)).ToString(CultureInfo.InvariantCulture).Length
I would prefer the following instead of casting to int to ensure that you can also handle big numbers (e.g. decimal.MaxValue):
Math.Truncate ( Math.Abs ( decValue ) ).ToString( "####" ).Length
decimal d = 467.45M;
int i = (int)d;
Console.WriteLine(i.ToString(CultureInfo.InvariantCulture).Length); //3
As a method;
public static int GetDigitsLength(decimal d)
{
int i = int(d);
return i.ToString(CultureInfo.InvariantCulture).Length;
}
Note: Of course you should check first your decimals full part is bigger than Int32.MaxValue or not. Because if it is, you get an OverflowException.
Is such a case, using long instead of int can a better approach. However even a long (System.Int64) is not big enough to hold every possible decimal value.
As Rawling mentioned, your full part can hold the thousands separator and my code will be broken in such a case. Because in this way, it totally ignores my number contains NumberFormatInfo.NumberGroupSeparator or not.
That's why getting numbers only is a better approach. Like;
i.ToString().Where(c => Char.IsDigit(c)).ToArray()
Here's a recursive example (mostly for fun).
void Main()
{
digitCount(0.123M); //0
digitCount(493854289.543354345M); //10
digitCount(4937854345454545435549.543354345M); //22
digitCount(-4937854345454545435549.543354345M); //22
digitCount(1.0M); //1
//approximately the biggest number you can pass to the function that works.
digitCount(Decimal.MaxValue + 0.4999999M); //29
}
int digitCount(decimal num, int count = 0)
{
//divided down to last digit, return how many times that happened
if(Math.Abs(num) < 1)
return count;
return digitCount(num/10, ++count); //increment the count and divide by 10 to 'remove' a digit
}
Math.Floor(Math.Log10((double)n) + 1); is the way to go.
Converting to int is BAD because decimal may be bigger than int:
Decimal.MaxValue = 79,228,162,514,264,337,593,543,950,335;
Int32.MaxValue = 2,147,483,647; //that is, hexadecimal 0x7FFFFFFF;
Math.Floor(n).ToString().Count(); is bad because it may include thousands seperators.
If you have a bias towards smaller numbers, you can use something more simple like this.
It is split up into two methods, so the first method is smaller and can be inlined.
Performance is about the same as the solution with the Log10, but without the rounding errors. The Method using Log10, is still the fastest (a bit) specially for numbers > 1 million.
public static int CountNrOfDigitsIfs(decimal d) {
var absD = Math.Abs(d);
// 1
if (absD < 10M) return 1;
// 2
if (absD < 100M) return 2;
// 3
if (absD < 1000M) return 3;
// 4
if (absD < 10000M) return 4;
return CountNrOfDigitsIfsLarge(d);
}
private static int CountNrOfDigitsIfsLarge(decimal d) {
// 5
if (d < 100000M) return 5;
// 6
if (d < 1000000M) return 6;
// 7
if (d < 10000000M) return 7;
// 8
if (d < 100000000M) return 8;
// 9
if (d < 1000000000M) return 9;
// 10
if (d < 10000000000M) return 10;
// 11
if (d < 100000000000M) return 11;
// 12
if (d < 1000000000000M) return 12;
// 13
if (d < 10000000000000M) return 13;
// 14
if (d < 100000000000000M) return 14;
// 15
if (d < 1000000000000000M) return 15;
// 16
if (d < 10000000000000000M) return 16;
// 17
if (d < 100000000000000000M) return 17;
// 18
if (d < 1000000000000000000M) return 18;
// 19
if (d < 10000000000000000000M) return 19;
// 20
if (d < 100000000000000000000M) return 20;
// 21
if (d < 1000000000000000000000M) return 21;
// 22
if (d < 10000000000000000000000M) return 22;
// 23
if (d < 100000000000000000000000M) return 23;
// 24
if (d < 1000000000000000000000000M) return 24;
// 25
if (d < 10000000000000000000000000M) return 25;
// 26
if (d < 100000000000000000000000000M) return 26;
// 27
if (d < 1000000000000000000000000000M) return 27;
// 28
if (d < 10000000000000000000000000000M) return 28;
return 29; // Max nr of digits in decimal
}
This code is generated using the following T4 template:
<#
const int SIGNIFICANT_DECIMALS = 29;
const int SPLIT = 5;
#>
namespace Study.NrOfDigits {
static partial class DigitCounter {
public static int CountNrOfDigitsIfs(decimal d) {
var absD = Math.Abs(d);
<#
for (int i = 1; i < SPLIT; i++) { // Only 29 significant digits
var zeroes = new String('0', i);
#>
// <#= i #>
if (absD < 1<#= zeroes #>M) return <#= i #>;
<#
}
#>
return CountNrOfDigitsIfsLarge(d);
}
private static int CountNrOfDigitsIfsLarge(decimal d) {
<#
for (int i = SPLIT; i < SIGNIFICANT_DECIMALS; i++) { // Only 29 significant digits
var zeroes = new String('0', i);
#>
// <#= i #>
if (d < 1<#= zeroes #>M) return <#= i #>;
<#
}
#>
return <#= SIGNIFICANT_DECIMALS #>; // Max nr of digits in decimal
}
}
}
So, I've run into this before, and solved it with this code:
SqlDecimal d = new SqlDecimal(467.45M);
int digits = d.Precision - d.Scale;
SqlDecimal is part of the System.Data.SqlTypes namespace. "Precision" is the total number of significant digits, while "Scale" is the number of digits after the decimal point.
Now, I know one objection to going this route is that SqlDecimal is part of the SQL Server-specific code. It's a valid point, but I would also point out that it's a part of the .NET framework itself, and has been since at least version 1.1, so it seems like it would be still be applicable no matter what the code around it is doing.
I looked under the hood with a decompiler (JetBrains' dotPeek in this instance), to see if maybe the code for calculating precision and scale could be easily extracted and just used, without pulling in SqlDecimal. The code to calculate scale is very simple, but the method to calculate precision is non-trivial, so if it were me, I'd just go through SqlDecimal.
This will do if you really don't want to use the Log method (which IMO is the best way). It's about the clearest way I can think of of doing this using ToString():
Math.Abs(val).ToString("f0", CultureInfo.InvariantCulture).Length
Or alternatively, if you don't want to count 0.123M as having one digit:
Math.Abs(val).ToString("#", CultureInfo.InvariantCulture).Length
You could use ToString function with a custom format.
Decimal value = 467.45m;
int count = Math.Abs(value).ToString("#", System.Globalization.CultureInfo.InvariantCulture).Length;
The # specify you only want the characters before the .
The System.Globalization.CultureInfo.InvariantCulture ensure you won't get any formating from the Region Option.
This answer is pretty much lifted from Calculate System.Decimal Precision and Scale but with a minor change to fit the question asked.
class Program
{
static void Main()
{
decimal dec = 467.45m;
Console.WriteLine(dec.GetNumberOfDigitsBeforeDecimalPlace());
}
}
public static class DecimalEx
{
public static int GetNumberOfDigitsBeforeDecimalPlace(this decimal dec)
{
var x = new System.Data.SqlTypes.SqlDecimal(dec);
return x.Precision - x.Scale;
}
}
Also if you want to do it without using the SqlDecimal class check out Jon Skeet's answer for the same question.
var sep = Convert.ToChar(CultureInfo.CurrentCulture.NumberFormat.NumberDecimalSeparator);
var count = d.ToString().TakeWhile(c => c != sep).Count();
The mathematical way of doing this (and probably the fastest) is to get a logarytm of base 10 of a absolute value of this number and round it
up.
Math.Floor(Math.Log10(Math.Abs(val)) + 1);
TLDR all the other answers. I wrote this in PHP, and the math would be the same. (If I knew C# I'd have written in that language.)
$input=21689584.999;
$input=abs($input);
$exp=0;
do{
$test=pow(10,$exp);
if($test > $input){
$digits=$exp;
}
if($test == $input){
$digits=$exp+1;
}
$exp++;
}while(!$digits);
if($input < 1){$digits=0;}
echo $digits;
I don't doubt there's a better way, but I wanted to throw in my $.02
EDIT:
I php-ized the code I mentioned in my comments, but did away with the int conversion.
function digitCount($input){
$digits=0;
$input=abs($input);
while ($input >= 1) {
$digits++;
$input=$input/10;
//echo $input."<br>";
}
return $digits;
}
$big=(float)(PHP_INT_MAX * 1.1);
echo digitCount($big);
Use modulo, i'm not a C# programmer, but I'm pretty sure this solution work:
double i = 1;
int numberOfDecimals = 0;
while (varDouble % i != varDouble)
{
numberOfDecimals++;
i*=10;
}
return numberOfDecimals;
This would be the Java solution
public class test {
public static void main(String args[]) {
float f = 1.123f;
int a = (int) f;
int digits = 0;
while (a > 0) {
digits++;
a=a/10;
}
System.out.println("No Of digits before decimal="+digits);
}
}
If you treat zeros or lack of zeroes as 1 number, this is OK. If you want zero to return zero or lack of zero to return zero, then there are a few edge cases to work out which shouldn't be too hard to add. Also, should Absolute value to handle negative numbers. Added that test case as well.
const decimal d = 123.45m;
const decimal d1 = 0.123m;
const decimal d2 = .567m;
const decimal d3 = .333m;
const decimal d4 = -123.45m;
NumberFormatInfo currentProvider = NumberFormatInfo.InvariantInfo;
var newProvider = (NumberFormatInfo) currentProvider.Clone();
newProvider.NumberDecimalDigits = 0;
string number = d.ToString("N", newProvider); //returns 123 = .Length = 3
string number1 = d1.ToString("N", newProvider); //returns 0 = .Length = 1
string number2 = d2.ToString("N", newProvider); //returns 1 = .Length = 1
string number3 = d3.ToString("N", newProvider); //returns 0 = .Length = 1
string number4 = Math.Abs(d4).ToString("N", newProvider); //returns 123 = .Length = 3
Here's a somewhat final solution, if you find a test case that doesn't work, let me know. It should return 3,0,0,0,3 for the test cases provided.
public static int NumbersInFrontOfDecimal(decimal input)
{
NumberFormatInfo currentProvider = NumberFormatInfo.InvariantInfo;
var newProvider = (NumberFormatInfo)currentProvider.Clone();
newProvider.NumberDecimalDigits = 0;
var absInput = Math.Abs(input);
var numbers = absInput.ToString("N", newProvider);
//Handle Zero and < 1
if (numbers.Length == 1 && input < 1.0m)
{
return 0;
}
return numbers.Length;
}
Here is my optimized version of the code inspired by Gray's answer:
static int GetNumOfDigits(decimal dTest)
{
int nAnswer = 0;
dTest = Math.Abs(dTest);
//For loop version
for (nAnswer = 0; nAnswer < 29 && dTest > 1; ++nAnswer)
{
dTest *= 0.1M;
}
//While loop version
//while (dTest > 1)
//{
// nAnswer++;
// dTest *= 0.1M;
//}
return (nAnswer);
}
If you don't want the Math.Abs to be called inside this function then be sure to use it
outside the function on the parameter before calling GetNumOfDigits.
I decided to remove the other codes to reduce clutter in my answer, even though they helped me get to this point...
If there is any improvements needed, then let me know and I will update it :).
In order to get an accurate and culturally agnostic answer I do the following:
Use System.Numerics.BigInteger, whose constructor accepts a decimal and doesn't seem to produce any rounding errors.
Use BigInteger.Abs() to remove any sign.
Use BigInteger.ToString() with the "#" format to suppress any separators that might occur.
Code
decimal num = 123213123.123123M;
int length = BigInteger.Abs((BigInteger)num).ToString("#").Length;
You could do this by rounding the number, then getting the length of the new number. You could do it like this:
var number = 476.43;
var newNumber = Math.round(number);
//get the length of the rounded number, and subtract 1 if the
//number is negative (remove the negative sign from the count)
int digits = newNumber.ToString().Length - (number < 0 ? 1 : 0);
The other solutions will lose digits if the number is too large.
public int Digits(Decimal i)
{
NumberFormatInfo format = CultureInfo.CurrentCulture.NumberFormat;
var str = Math.Abs(i).ToString().Replace(format.NumberGroupSeparator, "");
var index = str.IndexOf(format.NumberDecimalSeparator);
var digits = index == -1 ? str.Length : index;
}
Algorithm:
Convert |decimal| to String.
If "." exist in the decimal, cut before it, otherwise consider the whole number.
Return string length.
Example:
3.14 --> 3.14 --> "3.14" --> "3.14".Substring(0,1) --> "3".Length --> 1
-1024 --> 1024 --> "1024" --> IndexOf(".") = -1 --> "1024" --> 4
Code:
static int getNumOfDigits (decimal num)
{
string d = Math.Abs(num).ToString();
if (d.IndexOf(".") > -1)
{
d = d.Substring(0, d.IndexOf("."));
}
return d.Length;
}
I haven't tested this but I would keep it straightforward and do:
decimal value = 467.45;
string str = Convert.toString(value); // convert your decimal type to a string
string[] splitStr = str.split('.'); // split it into an array (use comma separator assuming you know your cultural context)
Console.WriteLine(splitStr[0].Count); // get the first element. You can also get the number of figures after the point by indexing the next value in the array.
This does not handle negative numbers. If you care about those then considering taking the absolute value. Furthermore, if you want 0 before the decimal place to not be counted then you can use a simple if statement to check it.
simple :
string value = "467.45";
int count = value.split('.')[0] == "0" ? 0 : value.split('.')[0].ToString().Replace("-","").Count();
In my C# program I have a double obtained from some computation and its value is something like 0,13999 or 0,0079996 but this value has to be presented to a human so it's better displayed as 0,14 or 0,008 respectively.
So I need to round the value, but have no idea to which precision - I just need to "throw away those noise digits".
How could I do that in my code?
To clarify - I need to round the double values to a precision that is unknown at compile time - this needs to be determined at runtime. What would be a good heuristic to achieve this?
You seem to want to output a value which is not very different to the input value, so try increasing numbers of digits until a given error is achieved:
static double Round(double input, double errorDesired)
{
if (input == 0.0)
return 0.0;
for (int decimals = 0; decimals < 17; ++decimals)
{
var output = Math.Round(input, decimals);
var errorAchieved = Math.Abs((output - input) / input);
if (errorAchieved <= errorDesired)
return output;
}
return input;
}
}
static void Main(string[] args)
{
foreach (var input in new[] { 0.13999, 0.0079996, 0.12345 })
{
Console.WriteLine("{0} -> {1} (.1%)", input, Round(input, 0.001));
Console.WriteLine("{0} -> {1} (1%)", input, Round(input, 0.01));
Console.WriteLine("{0} -> {1} (10%)", input, Round(input, 0.1));
}
}
private double PrettyRound(double inp)
{
string d = inp.ToString();
d = d.Remove(0,d.IndexOf(',') + 1);
int decRound = 1;
bool onStartZeroes = true;
for (int c = 1; c < d.Length; c++ )
{
if (!onStartZeroes && d[c] == d[c - 1])
break;
else
decRound++;
if (d[c] != '0')
onStartZeroes = false;
}
inp = Math.Round(inp, decRound);
return inp;
}
Test:
double d1 = 0.13999; //no zeroes
double d2 = 0.0079996; //zeroes
double d3 = 0.00700956; //zeroes within decimal
Response.Write(d1 + "<br/>" + d2 + "<br/>" + d3 + "<br/><br/>");
d1 = PrettyRound(d1);
d2 = PrettyRound(d2);
d3 = PrettyRound(d3);
Response.Write(d1 + "<br/>" + d2 + "<br/>" + d3 +"<br/><br/>");
Prints:
0,13999
0,0079996
0,00700956
0,14
0,008
0,007
Rounds your numbers as you wrote in your example..
I can think of a solution though it isn't very efficient...
My assumption is that you can tell when a number is in the "best" human readable format when extra digits make no difference to how it is rounded.
eg in the example of 0,13999 rounding it to various numbers of decimal places gives:
0
0.1
0.14
0.14
0.14
0.13999
I'd suggest that you could loop through and detect that stable patch and cut off there.
This method seems to do this:
public double CustomRound(double d)
{
double currentRound = 0;
int stability = 0;
int roundLevel = 0;
while (stability < 3)
{
roundLevel++;
double current = Math.Round(d, roundLevel);
if (current == currentRound)
{
stability++;
}
else
{
stability = 1;
currentRound=current;
}
}
return Math.Round(d, roundLevel);
}
This code might be cleanable but it does the job and is a sufficient proof of concept. :)
I should emphasise that that initial assumption (that no change when rounding) is the criteria we are looking at which means that something like 0.3333333333 will not get rounded at all. With the examples given I'm unable to say if this is correct or not but I assume if this is a double issues that the problem is with the very slight variations from the "right" value and the value as a double.
Heres what I tried:
public decimal myRounding(decimal number)
{
double log10 = Math.Log10((double) number);
int precision = (int)(log10 >= 0 ? 0 : Math.Abs(log10)) + (number < 0.01m ? 1 : 2);
return Math.Round(number, precision);
}
test:
Console.WriteLine(myRounding(0.0000019999m)); //0.000002
Console.WriteLine(myRounding(0.0003019999m)); //0.0003
Console.WriteLine(myRounding(2.56777777m)); //2.57
Console.WriteLine(myRounding(0.13999m)); //0.14
Console.WriteLine(myRounding(0.0079996m)); //0.008
You can do it without converting to string. This is what I created fast:
private static double RoundDecimal(double number)
{
double temp2 = number;
int temp, counter = 0;
do
{
temp2 = 10 * temp2;
temp = (int)temp2;
counter++;
} while (temp < 1);
return Math.Round(number, counter < 2 ? 2 : counter);
}
or
private static double RoundDecimal(double number)
{
int counter = 0;
if (number > 0) {
counter = Math.Abs((int) Math.Log10(number)) + 1;
return Math.Round(arv, counter < 2 ? 2 : counter);
}
After giving it another thought I did the following and looks like it does what I want so far.
I iterate over the number of digits and compare Round( value, number ) and Round( value, number + 1 ). Once they are equal (not == of course - I compare the difference against a small number) then number is the number of digits I'm looking for.
Double.ToString() can take a string format as an argument. This will display as many characters as you require, rounding to the decimal place. E.G:
double Value = 1054.32179;
MessageBox.Show(Value.ToString("0.000"));
Will display "1054.322".
Source
Generic formats (i.e, pre-generated)
How to generate custom formats
You can use no of digits with Math.Round Function
Double doubleValue = 4.052102;
Math.Round(doubleValue, 2);
This will return 4.05 as your required answer.
This is tested code, can u explain me how i am wrong. So i need to change.
I have a situation that I cannot change: one database table (table A) accepts 6 decimal places, while a related column in a different table (table B) only has 3 decimal places.
I need to copy from A to B, but if A has more than 3 decimal places the extra data will be lost. I cant change the table definition but I can add a workaround. So I'm trying to find out how to check if a decimal has more than 3 decimal places or not?
eg
Table A
Id, Qty, Unit(=6dp)
1, 1, 0.00025
2, 4000, 0.00025
Table B
Id, TotalQty(=3dp)
I want to be able to find out if Qty * Unit from Table A has more than 3 decimals (row 1 would fail, row 2 would pass):
if (CountDecimalPlaces(tableA.Qty * tableA.Unit) > 3)
{
return false;
}
tableB.TotalQty = tableA.Qty * tableA.Unit;
How would I implement the CountDecimalPlaces(decimal value) {} function?
You could compare the value of the number rounded to 3 decimal places with the original value.
if (Decimal.Round(valueDecimal, 3) != valueDecimal)
{
//Too many decimals
}
This works for 3 decimal places, and it can be adapted for a generic solution:
static bool LessThan3DecimalPlaces(decimal dec)
{
decimal value = dec * 1000;
return value == Math.Floor(value);
}
static void Test()
{
Console.WriteLine(LessThan3DecimalPlaces(1m * 0.00025m));
Console.WriteLine(LessThan3DecimalPlaces(4000m * 0.00025m));
}
For a real generic solution, you'll need to "deconstruct" the decimal value in its parts - take a look at Decimal.GetBits for more information.
Update: this is a simple implementation of a generic solution which works for all decimals whose integer part is less than long.MaxValue (you'd need something like a "big integer" for a trully generic function).
static decimal CountDecimalPlaces(decimal dec)
{
Console.Write("{0}: ", dec);
int[] bits = Decimal.GetBits(dec);
ulong lowInt = (uint)bits[0];
ulong midInt = (uint)bits[1];
int exponent = (bits[3] & 0x00FF0000) >> 16;
int result = exponent;
ulong lowDecimal = lowInt | (midInt << 32);
while (result > 0 && (lowDecimal % 10) == 0)
{
result--;
lowDecimal /= 10;
}
return result;
}
static void Foo()
{
Console.WriteLine(CountDecimalPlaces(1.6m));
Console.WriteLine(CountDecimalPlaces(1.600m));
Console.WriteLine(CountDecimalPlaces(decimal.MaxValue));
Console.WriteLine(CountDecimalPlaces(1m * 0.00025m));
Console.WriteLine(CountDecimalPlaces(4000m * 0.00025m));
}
This is a very simple one line code to get count of decimals in a Decimal:
decimal myDecimal = 1.000000021300010000001m;
byte decimals = (byte)((Decimal.GetBits(myDecimal)[3] >> 16) & 0x7F);
Multiplying a number with 3 decimal places by 10 to the power of 3 will give you a number with no decimal places. It's a whole number when the modulus % 1 == 0. So I came up with this...
bool hasMoreThanNDecimals(decimal d, int n)
{
return !(d * (decimal)Math.Pow(10, n) % 1 == 0);
}
Returns true when n is less than (not equal) to the number of decimal places.
The basics is to know how to test if there are decimal places, this is done by comparing the value to its rounding
double number;
bool hasDecimals = number == (int) number;
Then, to count 3 decimal places, you just need to do the same for your number multiplied by 1000:
bool hasMoreThan3decimals = number*1000 != (int) (number * 1000)
All of the solutions proposed so far are not extensible ... fine if you are never going to check a value other than 3, but I prefer this because if the requirement changes the code to handle it is already written. Also this solution wont overflow.
int GetDecimalCount(decimal val)
{
if(val == val*10)
{
return int.MaxValue; // no decimal.Epsilon I don't use this type enough to know why... this will work
}
int decimalCount = 0;
while(val != Math.Floor(val))
{
val = (val - Math.Floor(val)) * 10;
decimalCount++;
}
return decimalCount;
}
carlosfigueira solution will need to check for 0 otherwise "while ((lowDecimal % 10) == 0)" will produce an infinity loop when called with dec = 0
static decimal CountDecimalPlaces(decimal dec)
{
if (dec == 0)
return 0;
int[] bits = Decimal.GetBits(dec);
int exponent = bits[3] >> 16;
int result = exponent;
long lowDecimal = bits[0] | (bits[1] >> 8);
while ((lowDecimal % 10) == 0)
{
result--;
lowDecimal /= 10;
}
return result;
}
Assert.AreEqual(0, DecimalHelper.CountDecimalPlaces(0m));
Assert.AreEqual(1, DecimalHelper.CountDecimalPlaces(0.5m));
Assert.AreEqual(2, DecimalHelper.CountDecimalPlaces(10.51m));
Assert.AreEqual(13, DecimalHelper.CountDecimalPlaces(10.5123456978563m));
One more option based on #RodH257's solution, but reworked as an extension method:
public static bool HasThisManyDecimalPlacesOrLess(this decimal value, int noDecimalPlaces)
{
return (Decimal.Round(value, noDecimalPlaces) == value);
}
You can then call that as:
If !(tableA.Qty * tableA.Unit).HasThisManyDecimalPlacesOrLess(3)) return;
There is probably a more elegant way to do this, but off the top of my head I would try
a = multiply by 1000
b = truncate a
if (b != a) then there is additional precision that has been lost
bool CountDecimalPlaces(decimal input)
{
return input*1000.0 == (int) (input*1000);
}
Here is my version:
public static int CountDecimalPlaces(decimal dec)
{
var a = Math.Abs(dec);
var x = a;
var count = 1;
while (x % 1 != 0)
{
x = a * new decimal(Math.Pow(10, count++));
}
var result = count - 1;
return result;
}
I tried first #carlosfigueira/#Henrik Stenbæk, but their version does not work with 324000.00m
TEST:
Console.WriteLine(CountDecimalPlaces(0m)); //0
Console.WriteLine(CountDecimalPlaces(0.5m)); //1
Console.WriteLine(CountDecimalPlaces(10.51m)); //2
Console.WriteLine(CountDecimalPlaces(10.5123456978563m)); //13
Console.WriteLine(CountDecimalPlaces(324000.0001m)); //4
Console.WriteLine(CountDecimalPlaces(324000.0000m)); //0
could you convert it to a string and just do a len function or would that not cover your situation?
follow up question:
would 300.4 be ok?
Public Function getDecimalCount(decWork As Decimal) As Integer
Dim intDecimalCount As Int32 = 0
Dim strDecAbs As String = decWork.ToString.Trim("0")
intDecimalCount = strDecAbs.Substring(strDecAbs.IndexOf(".")).Length -1
Return intDecimalCount
End Function