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How to convert a double into a floating-point string representation without scientific notation in the .NET Framework?
"Small" samples (effective numbers may be of any size, such as 1.5E200 or 1e-200) :
3248971234698200000000000000000000000000000000
0.00000000000000000000000000000000000023897356978234562
None of the standard number formats are like this, and a custom format also doesn't seem to allow having an open number of digits after the decimal separator.
This is not a duplicate of How to convert double to string without the power to 10 representation (E-05) because the answers given there do not solve the issue at hand. The accepted solution in this question was to use a fixed point (such as 20 digits), which is not what I want. A fixed point formatting and trimming the redundant 0 doesn't solve the issue either because the max width for fixed width is 99 characters.
Note: the solution has to deal correctly with custom number formats (e.g. other decimal separator, depending on culture information).
Edit: The question is really only about displaing aforementioned numbers. I'm aware of how floating point numbers work and what numbers can be used and computed with them.
For a general-purpose¹ solution you need to preserve 339 places:
doubleValue.ToString("0." + new string('#', 339))
The maximum number of non-zero decimal digits is 16. 15 are on the right side of the decimal point. The exponent can move those 15 digits a maximum of 324 places to the right. (See the range and precision.)
It works for double.Epsilon, double.MinValue, double.MaxValue, and anything in between.
The performance will be much greater than the regex/string manipulation solutions since all formatting and string work is done in one pass by unmanaged CLR code. Also, the code is much simpler to prove correct.
For ease of use and even better performance, make it a constant:
public static class FormatStrings
{
public const string DoubleFixedPoint = "0.###################################################################################################################################################################################################################################################################################################################################################";
}
¹ Update: I mistakenly said that this was also a lossless solution. In fact it is not, since ToString does its normal display rounding for all formats except r. Live example. Thanks, #Loathing! Please see Lothing’s answer if you need the ability to roundtrip in fixed point notation (i.e, if you’re using .ToString("r") today).
I had a similar problem and this worked for me:
doubleValue.ToString("F99").TrimEnd('0')
F99 may be overkill, but you get the idea.
This is a string parsing solution where the source number (double) is converted into a string and parsed into its constituent components. It is then reassembled by rules into the full-length numeric representation. It also accounts for locale as requested.
Update: The tests of the conversions only include single-digit whole numbers, which is the norm, but the algorithm also works for something like: 239483.340901e-20
using System;
using System.Text;
using System.Globalization;
using System.Threading;
public class MyClass
{
public static void Main()
{
Console.WriteLine(ToLongString(1.23e-2));
Console.WriteLine(ToLongString(1.234e-5)); // 0.00010234
Console.WriteLine(ToLongString(1.2345E-10)); // 0.00000001002345
Console.WriteLine(ToLongString(1.23456E-20)); // 0.00000000000000000100023456
Console.WriteLine(ToLongString(5E-20));
Console.WriteLine("");
Console.WriteLine(ToLongString(1.23E+2)); // 123
Console.WriteLine(ToLongString(1.234e5)); // 1023400
Console.WriteLine(ToLongString(1.2345E10)); // 1002345000000
Console.WriteLine(ToLongString(-7.576E-05)); // -0.00007576
Console.WriteLine(ToLongString(1.23456e20));
Console.WriteLine(ToLongString(5e+20));
Console.WriteLine("");
Console.WriteLine(ToLongString(9.1093822E-31)); // mass of an electron
Console.WriteLine(ToLongString(5.9736e24)); // mass of the earth
Console.ReadLine();
}
private static string ToLongString(double input)
{
string strOrig = input.ToString();
string str = strOrig.ToUpper();
// if string representation was collapsed from scientific notation, just return it:
if (!str.Contains("E")) return strOrig;
bool negativeNumber = false;
if (str[0] == '-')
{
str = str.Remove(0, 1);
negativeNumber = true;
}
string sep = Thread.CurrentThread.CurrentCulture.NumberFormat.NumberDecimalSeparator;
char decSeparator = sep.ToCharArray()[0];
string[] exponentParts = str.Split('E');
string[] decimalParts = exponentParts[0].Split(decSeparator);
// fix missing decimal point:
if (decimalParts.Length==1) decimalParts = new string[]{exponentParts[0],"0"};
int exponentValue = int.Parse(exponentParts[1]);
string newNumber = decimalParts[0] + decimalParts[1];
string result;
if (exponentValue > 0)
{
result =
newNumber +
GetZeros(exponentValue - decimalParts[1].Length);
}
else // negative exponent
{
result =
"0" +
decSeparator +
GetZeros(exponentValue + decimalParts[0].Length) +
newNumber;
result = result.TrimEnd('0');
}
if (negativeNumber)
result = "-" + result;
return result;
}
private static string GetZeros(int zeroCount)
{
if (zeroCount < 0)
zeroCount = Math.Abs(zeroCount);
StringBuilder sb = new StringBuilder();
for (int i = 0; i < zeroCount; i++) sb.Append("0");
return sb.ToString();
}
}
You could cast the double to decimal and then do ToString().
(0.000000005).ToString() // 5E-09
((decimal)(0.000000005)).ToString() // 0,000000005
I haven't done performance testing which is faster, casting from 64-bit double to 128-bit decimal or a format string of over 300 chars. Oh, and there might possibly be overflow errors during conversion, but if your values fit a decimal this should work fine.
Update: The casting seems to be a lot faster. Using a prepared format string as given in the other answer, formatting a million times takes 2.3 seconds and casting only 0.19 seconds. Repeatable. That's 10x faster. Now it's only about the value range.
This is what I've got so far, seems to work, but maybe someone has a better solution:
private static readonly Regex rxScientific = new Regex(#"^(?<sign>-?)(?<head>\d+)(\.(?<tail>\d*?)0*)?E(?<exponent>[+\-]\d+)$", RegexOptions.IgnoreCase|RegexOptions.ExplicitCapture|RegexOptions.CultureInvariant);
public static string ToFloatingPointString(double value) {
return ToFloatingPointString(value, NumberFormatInfo.CurrentInfo);
}
public static string ToFloatingPointString(double value, NumberFormatInfo formatInfo) {
string result = value.ToString("r", NumberFormatInfo.InvariantInfo);
Match match = rxScientific.Match(result);
if (match.Success) {
Debug.WriteLine("Found scientific format: {0} => [{1}] [{2}] [{3}] [{4}]", result, match.Groups["sign"], match.Groups["head"], match.Groups["tail"], match.Groups["exponent"]);
int exponent = int.Parse(match.Groups["exponent"].Value, NumberStyles.Integer, NumberFormatInfo.InvariantInfo);
StringBuilder builder = new StringBuilder(result.Length+Math.Abs(exponent));
builder.Append(match.Groups["sign"].Value);
if (exponent >= 0) {
builder.Append(match.Groups["head"].Value);
string tail = match.Groups["tail"].Value;
if (exponent < tail.Length) {
builder.Append(tail, 0, exponent);
builder.Append(formatInfo.NumberDecimalSeparator);
builder.Append(tail, exponent, tail.Length-exponent);
} else {
builder.Append(tail);
builder.Append('0', exponent-tail.Length);
}
} else {
builder.Append('0');
builder.Append(formatInfo.NumberDecimalSeparator);
builder.Append('0', (-exponent)-1);
builder.Append(match.Groups["head"].Value);
builder.Append(match.Groups["tail"].Value);
}
result = builder.ToString();
}
return result;
}
// test code
double x = 1.0;
for (int i = 0; i < 200; i++) {
x /= 10;
}
Console.WriteLine(x);
Console.WriteLine(ToFloatingPointString(x));
The problem using #.###...### or F99 is that it doesn't preserve precision at the ending decimal places, e.g:
String t1 = (0.0001/7).ToString("0." + new string('#', 339)); // 0.0000142857142857143
String t2 = (0.0001/7).ToString("r"); // 1.4285714285714287E-05
The problem with DecimalConverter.cs is that it is slow. This code is the same idea as Sasik's answer, but twice as fast. Unit test method at bottom.
public static class RoundTrip {
private static String[] zeros = new String[1000];
static RoundTrip() {
for (int i = 0; i < zeros.Length; i++) {
zeros[i] = new String('0', i);
}
}
private static String ToRoundTrip(double value) {
String str = value.ToString("r");
int x = str.IndexOf('E');
if (x < 0) return str;
int x1 = x + 1;
String exp = str.Substring(x1, str.Length - x1);
int e = int.Parse(exp);
String s = null;
int numDecimals = 0;
if (value < 0) {
int len = x - 3;
if (e >= 0) {
if (len > 0) {
s = str.Substring(0, 2) + str.Substring(3, len);
numDecimals = len;
}
else
s = str.Substring(0, 2);
}
else {
// remove the leading minus sign
if (len > 0) {
s = str.Substring(1, 1) + str.Substring(3, len);
numDecimals = len;
}
else
s = str.Substring(1, 1);
}
}
else {
int len = x - 2;
if (len > 0) {
s = str[0] + str.Substring(2, len);
numDecimals = len;
}
else
s = str[0].ToString();
}
if (e >= 0) {
e = e - numDecimals;
String z = (e < zeros.Length ? zeros[e] : new String('0', e));
s = s + z;
}
else {
e = (-e - 1);
String z = (e < zeros.Length ? zeros[e] : new String('0', e));
if (value < 0)
s = "-0." + z + s;
else
s = "0." + z + s;
}
return s;
}
private static void RoundTripUnitTest() {
StringBuilder sb33 = new StringBuilder();
double[] values = new [] { 123450000000000000.0, 1.0 / 7, 10000000000.0/7, 100000000000000000.0/7, 0.001/7, 0.0001/7, 100000000000000000.0, 0.00000000001,
1.23e-2, 1.234e-5, 1.2345E-10, 1.23456E-20, 5E-20, 1.23E+2, 1.234e5, 1.2345E10, -7.576E-05, 1.23456e20, 5e+20, 9.1093822E-31, 5.9736e24, double.Epsilon };
foreach (int sign in new [] { 1, -1 }) {
foreach (double val in values) {
double val2 = sign * val;
String s1 = val2.ToString("r");
String s2 = ToRoundTrip(val2);
double val2_ = double.Parse(s2);
double diff = Math.Abs(val2 - val2_);
if (diff != 0) {
throw new Exception("Value {0} did not pass ToRoundTrip.".Format2(val.ToString("r")));
}
sb33.AppendLine(s1);
sb33.AppendLine(s2);
sb33.AppendLine();
}
}
}
}
The obligatory Logarithm-based solution. Note that this solution, because it involves doing math, may reduce the accuracy of your number a little bit. Not heavily tested.
private static string DoubleToLongString(double x)
{
int shift = (int)Math.Log10(x);
if (Math.Abs(shift) <= 2)
{
return x.ToString();
}
if (shift < 0)
{
double y = x * Math.Pow(10, -shift);
return "0.".PadRight(-shift + 2, '0') + y.ToString().Substring(2);
}
else
{
double y = x * Math.Pow(10, 2 - shift);
return y + "".PadRight(shift - 2, '0');
}
}
Edit: If the decimal point crosses non-zero part of the number, this algorithm will fail miserably. I tried for simple and went too far.
In the old days when we had to write our own formatters, we'd isolate the mantissa and exponent and format them separately.
In this article by Jon Skeet (https://csharpindepth.com/articles/FloatingPoint) he provides a link to his DoubleConverter.cs routine that should do exactly what you want. Skeet also refers to this at extracting mantissa and exponent from double in c#.
I have just improvised on the code above to make it work for negative exponential values.
using System;
using System.Text.RegularExpressions;
using System.IO;
using System.Text;
using System.Threading;
namespace ConvertNumbersInScientificNotationToPlainNumbers
{
class Program
{
private static string ToLongString(double input)
{
string str = input.ToString(System.Globalization.CultureInfo.InvariantCulture);
// if string representation was collapsed from scientific notation, just return it:
if (!str.Contains("E")) return str;
var positive = true;
if (input < 0)
{
positive = false;
}
string sep = Thread.CurrentThread.CurrentCulture.NumberFormat.NumberDecimalSeparator;
char decSeparator = sep.ToCharArray()[0];
string[] exponentParts = str.Split('E');
string[] decimalParts = exponentParts[0].Split(decSeparator);
// fix missing decimal point:
if (decimalParts.Length == 1) decimalParts = new string[] { exponentParts[0], "0" };
int exponentValue = int.Parse(exponentParts[1]);
string newNumber = decimalParts[0].Replace("-", "").
Replace("+", "") + decimalParts[1];
string result;
if (exponentValue > 0)
{
if (positive)
result =
newNumber +
GetZeros(exponentValue - decimalParts[1].Length);
else
result = "-" +
newNumber +
GetZeros(exponentValue - decimalParts[1].Length);
}
else // negative exponent
{
if (positive)
result =
"0" +
decSeparator +
GetZeros(exponentValue + decimalParts[0].Replace("-", "").
Replace("+", "").Length) + newNumber;
else
result =
"-0" +
decSeparator +
GetZeros(exponentValue + decimalParts[0].Replace("-", "").
Replace("+", "").Length) + newNumber;
result = result.TrimEnd('0');
}
float temp = 0.00F;
if (float.TryParse(result, out temp))
{
return result;
}
throw new Exception();
}
private static string GetZeros(int zeroCount)
{
if (zeroCount < 0)
zeroCount = Math.Abs(zeroCount);
StringBuilder sb = new StringBuilder();
for (int i = 0; i < zeroCount; i++) sb.Append("0");
return sb.ToString();
}
public static void Main(string[] args)
{
//Get Input Directory.
Console.WriteLine(#"Enter the Input Directory");
var readLine = Console.ReadLine();
if (readLine == null)
{
Console.WriteLine(#"Enter the input path properly.");
return;
}
var pathToInputDirectory = readLine.Trim();
//Get Output Directory.
Console.WriteLine(#"Enter the Output Directory");
readLine = Console.ReadLine();
if (readLine == null)
{
Console.WriteLine(#"Enter the output path properly.");
return;
}
var pathToOutputDirectory = readLine.Trim();
//Get Delimiter.
Console.WriteLine("Enter the delimiter;");
var columnDelimiter = (char)Console.Read();
//Loop over all files in the directory.
foreach (var inputFileName in Directory.GetFiles(pathToInputDirectory))
{
var outputFileWithouthNumbersInScientificNotation = string.Empty;
Console.WriteLine("Started operation on File : " + inputFileName);
if (File.Exists(inputFileName))
{
// Read the file
using (var file = new StreamReader(inputFileName))
{
string line;
while ((line = file.ReadLine()) != null)
{
String[] columns = line.Split(columnDelimiter);
var duplicateLine = string.Empty;
int lengthOfColumns = columns.Length;
int counter = 1;
foreach (var column in columns)
{
var columnDuplicate = column;
try
{
if (Regex.IsMatch(columnDuplicate.Trim(),
#"^[+-]?[0-9]+(\.[0-9]+)?[E]([+-]?[0-9]+)$",
RegexOptions.IgnoreCase))
{
Console.WriteLine("Regular expression matched for this :" + column);
columnDuplicate = ToLongString(Double.Parse
(column,
System.Globalization.NumberStyles.Float));
Console.WriteLine("Converted this no in scientific notation " +
"" + column + " to this number " +
columnDuplicate);
}
}
catch (Exception)
{
}
duplicateLine = duplicateLine + columnDuplicate;
if (counter != lengthOfColumns)
{
duplicateLine = duplicateLine + columnDelimiter.ToString();
}
counter++;
}
duplicateLine = duplicateLine + Environment.NewLine;
outputFileWithouthNumbersInScientificNotation = outputFileWithouthNumbersInScientificNotation + duplicateLine;
}
file.Close();
}
var outputFilePathWithoutNumbersInScientificNotation
= Path.Combine(pathToOutputDirectory, Path.GetFileName(inputFileName));
//Create Directory If it does not exist.
if (!Directory.Exists(pathToOutputDirectory))
Directory.CreateDirectory(pathToOutputDirectory);
using (var outputFile =
new StreamWriter(outputFilePathWithoutNumbersInScientificNotation))
{
outputFile.Write(outputFileWithouthNumbersInScientificNotation);
outputFile.Close();
}
Console.WriteLine("The transformed file is here :" +
outputFilePathWithoutNumbersInScientificNotation);
}
}
}
}
}
This code takes an input directory and based on the delimiter converts all values in scientific notation to numeric format.
Thanks
try this one:
public static string DoubleToFullString(double value,
NumberFormatInfo formatInfo)
{
string[] valueExpSplit;
string result, decimalSeparator;
int indexOfDecimalSeparator, exp;
valueExpSplit = value.ToString("r", formatInfo)
.ToUpper()
.Split(new char[] { 'E' });
if (valueExpSplit.Length > 1)
{
result = valueExpSplit[0];
exp = int.Parse(valueExpSplit[1]);
decimalSeparator = formatInfo.NumberDecimalSeparator;
if ((indexOfDecimalSeparator
= valueExpSplit[0].IndexOf(decimalSeparator)) > -1)
{
exp -= (result.Length - indexOfDecimalSeparator - 1);
result = result.Replace(decimalSeparator, "");
}
if (exp >= 0) result += new string('0', Math.Abs(exp));
else
{
exp = Math.Abs(exp);
if (exp >= result.Length)
{
result = "0." + new string('0', exp - result.Length)
+ result;
}
else
{
result = result.Insert(result.Length - exp, decimalSeparator);
}
}
}
else result = valueExpSplit[0];
return result;
}
Being millions of programmers world wide, it's always a good practice to try search if someone has bumped into your problem already. Sometimes there's solutions are garbage, which means it's time to write your own, and sometimes there are great, such as the following:
http://www.yoda.arachsys.com/csharp/DoubleConverter.cs
(details: http://www.yoda.arachsys.com/csharp/floatingpoint.html)
string strdScaleFactor = dScaleFactor.ToString(); // where dScaleFactor = 3.531467E-05
decimal decimalScaleFactor = Decimal.Parse(strdScaleFactor, System.Globalization.NumberStyles.Float);
I don't know if my answer to the question can still be helpful. But in this case I suggest the "decomposition of the double variable into decimal places" to store it in an Array / Array of data of type String.
This process of decomposition and storage in parts (number by number) from double to string, would basically work with the use of two loops and an "alternative" (if you thought of workaround, I think you got it), where the first loop will extract the values from double without converting to String, resulting in blessed scientific notation and storing number by number in an Array. And this will be done using MOD - the same method to check a palindrome number, which would be for example:
String[] Array_ = new double[ **here you will put an extreme value of places your DOUBLE can reach, you must have a prediction**];
for (int i = 0, variableDoubleMonstrous > 0, i++){
x = variableDoubleMonstrous %10;
Array_[i] = x;
variableDoubleMonstrous /= 10;
}
And the second loop to invert the Array values (because in this process of checking a palindrome, the values invert from the last place, to the first, from the penultimate to the second and so on. Remember?) to get the original value:
String[] ArrayFinal = new String[the same number of "places" / indices of the other Array / Data array];
int lengthArray = Array_.Length;
for (int i = 0, i < Array_.Length, i++){
FinalArray[i] = Array_[lengthArray - 1];
lengthArray--;
}
***Warning: There's a catch that I didn't pay attention to. In that case there will be no "." (floating point decimal separator or double), so this solution is not generalized. But if it is really important to use decimal separators, unfortunately the only possibility (If done well, it will have a great performance) is:
**Use a routine to get the position of the decimal point of the original value, the one with scientific notation - the important thing is that you know that this floating point is before a number such as the "Length" position x, and after a number such as the y position - extracting each digit using the loops - as shown above - and at the end "export" the data from the last Array to another one, including the decimal place divider (the comma, or the period , if variable decimal, double or float) in the imaginary position that was in the original variable, in the "real" position of that matrix.
*** The concept of position is, find out how many numbers occur before the decimal point, so with this information you will be able to store in the String Array the point in the real position.
NEEDS THAT CAN BE MADE:
But then you ask:
But what about when I'm going to convert String to a floating point value?
My answer is that you use the second matrix of this entire process (the one that receives the inversion of the first matrix that obtains the numbers by the palindrome method) and use it for the conversion, but always making sure, when necessary, of the position of the decimal place in future situations, in case this conversion (Double -> String) is needed again.
But what if the problem is to use the value of the converted Double (Array of Strings) in a calculation. Then in this case you went around in circles. Well, the original variable will work anyway even with scientific notation. The only difference between floating point and decimal variable types is in the rounding of values, which depending on the purpose, it will only be necessary to change the type of data used, but it is dangerous to have a significant loss of information, look here
I could be wrong, but isn't it like this?
data.ToString("n");
http://msdn.microsoft.com/en-us/library/dwhawy9k.aspx
i think you need only to use IFormat with
ToString(doubleVar, System.Globalization.NumberStyles.Number)
example:
double d = double.MaxValue;
string s = d.ToString(d, System.Globalization.NumberStyles.Number);
My solution was using the custom formats.
try this:
double d;
d = 1234.12341234;
d.ToString("#########0.#########");
Just to build on what jcasso said what you can do is to adjust your double value by changing the exponent so that your favorite format would do it for you, apply the format, and than pad the result with zeros to compensate for the adjustment.
This works fine for me...
double number = 1.5E+200;
string s = number.ToString("#");
//Output: "150000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
I need to convert decimal number a to string b folowing:
b must be haven '.' character. Eg:
a = 12 -> b = "12.0"
a = 1.2 -> b = "1.2"
a = 1.234 -> b = "1.234"
How can I do that with 1 command?
b must be haven exactly 10 character. Eg:
a = 101 -> b = "101.000000"
a = 1.234 -> b = "1.23400000"
a = 1.234567891 -> b = "1.23456789"
(Same question with 1)
decimal a = 12;
var b = a.ToString("N1"); // 12.0
a = 1.2m;
b = a.ToString(); // 1.2
a = 101m;
b = a.ToString("N10"); // 101.0000000000
a = 1.234m;
b = a.ToString("N10"); // 1.2340000000
For the second part of your question - where you want a total length of 10 then:
decimal a = 1.234567891m;
int numberOfDigits = ((int)a).ToString().Length;
var b = a.ToString($"N{9 - numberOfDigits}"); //1.23456789
//Or before C# 6.0
var b = a.ToString("N" + (9 - numberOfDigits)); //1.23456789
Basically ((int)number).ToString().Length gives you the amount of digits before the . (converting to int will remove the fractions) and then reducing that from the number of digits after the . (and also -1 for the decimal point itself)
You can use .ToString() to do this task:
decimal aDecimalVal = 10.234m;
string decimalString = aDecimalVal.ToString("#.000"); // "10.234"
aDecimalVal.ToString("#.00"); // "10.23"
aDecimalVal.ToString("#.0000"); // "10.2340"
The number of 0 after the . in the format string will decide the number of places in the decimal string.
Updates: So you want to find the number of digits after the decimal points, So the changes in the code will be like the following:
decimal aDecimalVal = 10.2343455m;
int count = BitConverter.GetBytes(decimal.GetBits(aDecimalVal)[3])[2];
string formatString = String.Format("N{0}",count.ToString());
string decimalString = aDecimalVal.ToString(formatString); // "10.2343455"
I manage to do it using double. Is this what you need?
I don't quite get the second part of your question.
double a = 12;
string b = a.ToString("0.0000000000######");
Console.WriteLine(b);
if you want the result as a string, just parse it and format it to one decimal places:
string strTemp = 12;
double temp = Double.Parse(strTemp, CultureInfo.InvariantCulture);
string result = temp.ToString("N1", CultureInfo.InvariantCulture);
Round off to 2 decimal places eg. 27.658 => 27.66
Ensure that there are always 2 decimal places eg. 12 => 12.00, 12.5 => 12.50
Good fit for currency amounts.
strTemp.ToString("F");
For the first one, if you don't know how many the digits the variable could be, you can have a extension method:
public static string ToSpecificFormat(this decimal value)
{
var count = BitConverter.GetBytes(decimal.GetBits(value)[3])[2];
return value.ToString(count == 0 ? "N1" : "N" + count);
}
This will make sure there is at least 1 digit in the output.
For the second one, the selected answer will fail if the number > 1000000000. This one should work:
public static string ToFixedLength(this decimal value)
{
if (value >= 1000000000m) return value.ToString("N0");
var format = 9 - Math.Floor(value).ToString().Length;
return value.ToString("N" + format);
}
output:
1.234m.ToFixedLength(); // 1.23400000
101m.ToFixedLength(); // 101.000000
123456789123m.ToFixedLength(); // 123,456,789,123
I have some fields returned by a collection as
2.4200
2.0044
2.0000
I want results like
2.42
2.0044
2
I tried with String.Format, but it returns 2.0000 and setting it to N0 rounds the other values as well.
I ran into the same problem but in a case where I do not have control of the output to string, which was taken care of by a library. After looking into details in the implementation of the Decimal type (see http://msdn.microsoft.com/en-us/library/system.decimal.getbits.aspx),
I came up with a neat trick (here as an extension method):
public static decimal Normalize(this decimal value)
{
return value/1.000000000000000000000000000000000m;
}
The exponent part of the decimal is reduced to just what is needed. Calling ToString() on the output decimal will write the number without any trailing 0. E.g.
1.200m.Normalize().ToString();
Is it not as simple as this, if the input IS a string? You can use one of these:
string.Format("{0:G29}", decimal.Parse("2.0044"))
decimal.Parse("2.0044").ToString("G29")
2.0m.ToString("G29")
This should work for all input.
Update Check out the Standard Numeric Formats I've had to explicitly set the precision specifier to 29 as the docs clearly state:
However, if the number is a Decimal and the precision specifier is omitted, fixed-point notation is always used and trailing zeros are preserved
Update Konrad pointed out in the comments:
Watch out for values like 0.000001. G29 format will present them in the shortest possible way so it will switch to the exponential notation. string.Format("{0:G29}", decimal.Parse("0.00000001",System.Globalization.CultureInfo.GetCultureInfo("en-US"))) will give "1E-08" as the result.
In my opinion its safer to use Custom Numeric Format Strings.
decimal d = 0.00000000000010000000000m;
string custom = d.ToString("0.#########################");
// gives: 0,0000000000001
string general = d.ToString("G29");
// gives: 1E-13
I use this code to avoid "G29" scientific notation:
public static string DecimalToString(this decimal dec)
{
string strdec = dec.ToString(CultureInfo.InvariantCulture);
return strdec.Contains(".") ? strdec.TrimEnd('0').TrimEnd('.') : strdec;
}
EDIT: using system CultureInfo.NumberFormat.NumberDecimalSeparator :
public static string DecimalToString(this decimal dec)
{
string sep = CultureInfo.CurrentCulture.NumberFormat.NumberDecimalSeparator;
string strdec = dec.ToString(CultureInfo.CurrentCulture);
return strdec.Contains(sep) ? strdec.TrimEnd('0').TrimEnd(sep.ToCharArray()) : strdec;
}
Use the hash (#) symbol to only display trailing 0's when necessary. See the tests below.
decimal num1 = 13.1534545765;
decimal num2 = 49.100145;
decimal num3 = 30.000235;
num1.ToString("0.##"); //13.15%
num2.ToString("0.##"); //49.1%
num3.ToString("0.##"); //30%
I found an elegant solution from http://dobrzanski.net/2009/05/14/c-decimaltostring-and-how-to-get-rid-of-trailing-zeros/
Basically
decimal v=2.4200M;
v.ToString("#.######"); // Will return 2.42. The number of # is how many decimal digits you support.
A very low level approach, but I belive this would be the most performant way by only using fast integer calculations (and no slow string parsing and culture sensitive methods):
public static decimal Normalize(this decimal d)
{
int[] bits = decimal.GetBits(d);
int sign = bits[3] & (1 << 31);
int exp = (bits[3] >> 16) & 0x1f;
uint a = (uint)bits[2]; // Top bits
uint b = (uint)bits[1]; // Middle bits
uint c = (uint)bits[0]; // Bottom bits
while (exp > 0 && ((a % 5) * 6 + (b % 5) * 6 + c) % 10 == 0)
{
uint r;
a = DivideBy10((uint)0, a, out r);
b = DivideBy10(r, b, out r);
c = DivideBy10(r, c, out r);
exp--;
}
bits[0] = (int)c;
bits[1] = (int)b;
bits[2] = (int)a;
bits[3] = (exp << 16) | sign;
return new decimal(bits);
}
private static uint DivideBy10(uint highBits, uint lowBits, out uint remainder)
{
ulong total = highBits;
total <<= 32;
total = total | (ulong)lowBits;
remainder = (uint)(total % 10L);
return (uint)(total / 10L);
}
This is simple.
decimal decNumber = Convert.ToDecimal(value);
return decNumber.ToString("0.####");
Tested.
Cheers :)
Depends on what your number represents and how you want to manage the values: is it a currency, do you need rounding or truncation, do you need this rounding only for display?
If for display consider formatting the numbers are x.ToString("")
http://msdn.microsoft.com/en-us/library/dwhawy9k.aspx and
http://msdn.microsoft.com/en-us/library/0c899ak8.aspx
If it is just rounding, use Math.Round overload that requires a MidPointRounding overload
http://msdn.microsoft.com/en-us/library/ms131274.aspx)
If you get your value from a database consider casting instead of conversion:
double value = (decimal)myRecord["columnName"];
This will work:
decimal source = 2.4200m;
string output = ((double)source).ToString();
Or if your initial value is string:
string source = "2.4200";
string output = double.Parse(source).ToString();
Pay attention to this comment.
Trying to do more friendly solution of DecimalToString (https://stackoverflow.com/a/34486763/3852139):
private static decimal Trim(this decimal value)
{
var s = value.ToString(CultureInfo.InvariantCulture);
return s.Contains(CultureInfo.InvariantCulture.NumberFormat.NumberDecimalSeparator)
? Decimal.Parse(s.TrimEnd('0'), CultureInfo.InvariantCulture)
: value;
}
private static decimal? Trim(this decimal? value)
{
return value.HasValue ? (decimal?) value.Value.Trim() : null;
}
private static void Main(string[] args)
{
Console.WriteLine("=>{0}", 1.0000m.Trim());
Console.WriteLine("=>{0}", 1.000000023000m.Trim());
Console.WriteLine("=>{0}", ((decimal?) 1.000000023000m).Trim());
Console.WriteLine("=>{0}", ((decimal?) null).Trim());
}
Output:
=>1
=>1.000000023
=>1.000000023
=>
how about this:
public static string TrimEnd(this decimal d)
{
string str = d.ToString();
if (str.IndexOf(".") > 0)
{
str = System.Text.RegularExpressions.Regex.Replace(str.Trim(), "0+?$", " ");
str = System.Text.RegularExpressions.Regex.Replace(str.Trim(), "[.]$", " ");
}
return str;
}
You can just set as:
decimal decNumber = 23.45600000m;
Console.WriteLine(decNumber.ToString("0.##"));
The following code could be used to not use the string type:
int decimalResult = 789.500
while (decimalResult>0 && decimalResult % 10 == 0)
{
decimalResult = decimalResult / 10;
}
return decimalResult;
Returns 789.5
In case you want to keep decimal number, try following example:
number = Math.Floor(number * 100000000) / 100000000;
Here is an Extention method I wrote, it also removes dot or comma if it`s the last character (after the zeros were removed):
public static string RemoveZeroTail(this decimal num)
{
var result = num.ToString().TrimEnd(new char[] { '0' });
if (result[result.Length - 1].ToString() == "." || result[result.Length - 1].ToString() == ",")
{
return result.Substring(0, result.Length - 1);
}
else
{
return result;
}
}
To remove trailing zero's from a string variable dateTicks, Use
return new String(dateTicks.Take(dateTicks.LastIndexOf(dateTicks.Last(v => v != '0')) + 1).ToArray());
Additional Answer:
In a WPF Application using XAML you could use
{Binding yourDecimal, StringFormat='#,0.00#######################'}
The above answer will preserve the zero in some situations so you could still return 2.00 for example
{Binding yourDecimal, StringFormat='#,0.#########################'}
If you want to remove ALL trailing zeros, adjust accordingly.
The following code will be able to remove the trailing 0's. I know it's the hard way but it works.
private static string RemoveTrailingZeros(string input)
{
for (int i = input.Length - 1; i > 0; i-- )
{
if (!input.Contains(".")) break;
if (input[i].Equals('0'))
{
input= input.Remove(i);
}
else break;
}
return input;
}
string.Format("{0:G29}", decimal.Parse("2.00"))
string.Format("{0:G29}", decimal.Parse(Your_Variable))
try this code:
string value = "100";
value = value.Contains(".") ? value.TrimStart('0').TrimEnd('0').TrimEnd('.') : value.TrimStart('0');
Very simple answer is to use TrimEnd(). Here is the result,
double value = 1.00;
string output = value.ToString().TrimEnd('0');
Output is 1
If my value is 1.01 then my output will be 1.01
try like this
string s = "2.4200";
s = s.TrimStart("0").TrimEnd("0", ".");
and then convert that to float
I have some fields returned by a collection as
2.4200
2.0044
2.0000
I want results like
2.42
2.0044
2
I tried with String.Format, but it returns 2.0000 and setting it to N0 rounds the other values as well.
I ran into the same problem but in a case where I do not have control of the output to string, which was taken care of by a library. After looking into details in the implementation of the Decimal type (see http://msdn.microsoft.com/en-us/library/system.decimal.getbits.aspx),
I came up with a neat trick (here as an extension method):
public static decimal Normalize(this decimal value)
{
return value/1.000000000000000000000000000000000m;
}
The exponent part of the decimal is reduced to just what is needed. Calling ToString() on the output decimal will write the number without any trailing 0. E.g.
1.200m.Normalize().ToString();
Is it not as simple as this, if the input IS a string? You can use one of these:
string.Format("{0:G29}", decimal.Parse("2.0044"))
decimal.Parse("2.0044").ToString("G29")
2.0m.ToString("G29")
This should work for all input.
Update Check out the Standard Numeric Formats I've had to explicitly set the precision specifier to 29 as the docs clearly state:
However, if the number is a Decimal and the precision specifier is omitted, fixed-point notation is always used and trailing zeros are preserved
Update Konrad pointed out in the comments:
Watch out for values like 0.000001. G29 format will present them in the shortest possible way so it will switch to the exponential notation. string.Format("{0:G29}", decimal.Parse("0.00000001",System.Globalization.CultureInfo.GetCultureInfo("en-US"))) will give "1E-08" as the result.
In my opinion its safer to use Custom Numeric Format Strings.
decimal d = 0.00000000000010000000000m;
string custom = d.ToString("0.#########################");
// gives: 0,0000000000001
string general = d.ToString("G29");
// gives: 1E-13
I use this code to avoid "G29" scientific notation:
public static string DecimalToString(this decimal dec)
{
string strdec = dec.ToString(CultureInfo.InvariantCulture);
return strdec.Contains(".") ? strdec.TrimEnd('0').TrimEnd('.') : strdec;
}
EDIT: using system CultureInfo.NumberFormat.NumberDecimalSeparator :
public static string DecimalToString(this decimal dec)
{
string sep = CultureInfo.CurrentCulture.NumberFormat.NumberDecimalSeparator;
string strdec = dec.ToString(CultureInfo.CurrentCulture);
return strdec.Contains(sep) ? strdec.TrimEnd('0').TrimEnd(sep.ToCharArray()) : strdec;
}
Use the hash (#) symbol to only display trailing 0's when necessary. See the tests below.
decimal num1 = 13.1534545765;
decimal num2 = 49.100145;
decimal num3 = 30.000235;
num1.ToString("0.##"); //13.15%
num2.ToString("0.##"); //49.1%
num3.ToString("0.##"); //30%
I found an elegant solution from http://dobrzanski.net/2009/05/14/c-decimaltostring-and-how-to-get-rid-of-trailing-zeros/
Basically
decimal v=2.4200M;
v.ToString("#.######"); // Will return 2.42. The number of # is how many decimal digits you support.
A very low level approach, but I belive this would be the most performant way by only using fast integer calculations (and no slow string parsing and culture sensitive methods):
public static decimal Normalize(this decimal d)
{
int[] bits = decimal.GetBits(d);
int sign = bits[3] & (1 << 31);
int exp = (bits[3] >> 16) & 0x1f;
uint a = (uint)bits[2]; // Top bits
uint b = (uint)bits[1]; // Middle bits
uint c = (uint)bits[0]; // Bottom bits
while (exp > 0 && ((a % 5) * 6 + (b % 5) * 6 + c) % 10 == 0)
{
uint r;
a = DivideBy10((uint)0, a, out r);
b = DivideBy10(r, b, out r);
c = DivideBy10(r, c, out r);
exp--;
}
bits[0] = (int)c;
bits[1] = (int)b;
bits[2] = (int)a;
bits[3] = (exp << 16) | sign;
return new decimal(bits);
}
private static uint DivideBy10(uint highBits, uint lowBits, out uint remainder)
{
ulong total = highBits;
total <<= 32;
total = total | (ulong)lowBits;
remainder = (uint)(total % 10L);
return (uint)(total / 10L);
}
This is simple.
decimal decNumber = Convert.ToDecimal(value);
return decNumber.ToString("0.####");
Tested.
Cheers :)
Depends on what your number represents and how you want to manage the values: is it a currency, do you need rounding or truncation, do you need this rounding only for display?
If for display consider formatting the numbers are x.ToString("")
http://msdn.microsoft.com/en-us/library/dwhawy9k.aspx and
http://msdn.microsoft.com/en-us/library/0c899ak8.aspx
If it is just rounding, use Math.Round overload that requires a MidPointRounding overload
http://msdn.microsoft.com/en-us/library/ms131274.aspx)
If you get your value from a database consider casting instead of conversion:
double value = (decimal)myRecord["columnName"];
This will work:
decimal source = 2.4200m;
string output = ((double)source).ToString();
Or if your initial value is string:
string source = "2.4200";
string output = double.Parse(source).ToString();
Pay attention to this comment.
Trying to do more friendly solution of DecimalToString (https://stackoverflow.com/a/34486763/3852139):
private static decimal Trim(this decimal value)
{
var s = value.ToString(CultureInfo.InvariantCulture);
return s.Contains(CultureInfo.InvariantCulture.NumberFormat.NumberDecimalSeparator)
? Decimal.Parse(s.TrimEnd('0'), CultureInfo.InvariantCulture)
: value;
}
private static decimal? Trim(this decimal? value)
{
return value.HasValue ? (decimal?) value.Value.Trim() : null;
}
private static void Main(string[] args)
{
Console.WriteLine("=>{0}", 1.0000m.Trim());
Console.WriteLine("=>{0}", 1.000000023000m.Trim());
Console.WriteLine("=>{0}", ((decimal?) 1.000000023000m).Trim());
Console.WriteLine("=>{0}", ((decimal?) null).Trim());
}
Output:
=>1
=>1.000000023
=>1.000000023
=>
how about this:
public static string TrimEnd(this decimal d)
{
string str = d.ToString();
if (str.IndexOf(".") > 0)
{
str = System.Text.RegularExpressions.Regex.Replace(str.Trim(), "0+?$", " ");
str = System.Text.RegularExpressions.Regex.Replace(str.Trim(), "[.]$", " ");
}
return str;
}
You can just set as:
decimal decNumber = 23.45600000m;
Console.WriteLine(decNumber.ToString("0.##"));
The following code could be used to not use the string type:
int decimalResult = 789.500
while (decimalResult>0 && decimalResult % 10 == 0)
{
decimalResult = decimalResult / 10;
}
return decimalResult;
Returns 789.5
In case you want to keep decimal number, try following example:
number = Math.Floor(number * 100000000) / 100000000;
Here is an Extention method I wrote, it also removes dot or comma if it`s the last character (after the zeros were removed):
public static string RemoveZeroTail(this decimal num)
{
var result = num.ToString().TrimEnd(new char[] { '0' });
if (result[result.Length - 1].ToString() == "." || result[result.Length - 1].ToString() == ",")
{
return result.Substring(0, result.Length - 1);
}
else
{
return result;
}
}
To remove trailing zero's from a string variable dateTicks, Use
return new String(dateTicks.Take(dateTicks.LastIndexOf(dateTicks.Last(v => v != '0')) + 1).ToArray());
Additional Answer:
In a WPF Application using XAML you could use
{Binding yourDecimal, StringFormat='#,0.00#######################'}
The above answer will preserve the zero in some situations so you could still return 2.00 for example
{Binding yourDecimal, StringFormat='#,0.#########################'}
If you want to remove ALL trailing zeros, adjust accordingly.
The following code will be able to remove the trailing 0's. I know it's the hard way but it works.
private static string RemoveTrailingZeros(string input)
{
for (int i = input.Length - 1; i > 0; i-- )
{
if (!input.Contains(".")) break;
if (input[i].Equals('0'))
{
input= input.Remove(i);
}
else break;
}
return input;
}
string.Format("{0:G29}", decimal.Parse("2.00"))
string.Format("{0:G29}", decimal.Parse(Your_Variable))
try this code:
string value = "100";
value = value.Contains(".") ? value.TrimStart('0').TrimEnd('0').TrimEnd('.') : value.TrimStart('0');
Very simple answer is to use TrimEnd(). Here is the result,
double value = 1.00;
string output = value.ToString().TrimEnd('0');
Output is 1
If my value is 1.01 then my output will be 1.01
try like this
string s = "2.4200";
s = s.TrimStart("0").TrimEnd("0", ".");
and then convert that to float
I have a compressed string value I'm extracting from an import file. I need to format this into a parcel number, which is formatted as follows: ##-##-##-###-###. So therefore, the string "410151000640" should become "41-01-51-000-640". I can do this with the following code:
String.Format("{0:##-##-##-###-###}", Convert.ToInt64("410151000640"));
However, The string may not be all numbers; it could have a letter or two in there, and thus the conversion to the int will fail. Is there a way to do this on a string so every character, regardless of if it is a number or letter, will fit into the format correctly?
Regex.Replace("410151000640", #"^(.{2})(.{2})(.{2})(.{3})(.{3})$", "$1-$2-$3-$4-$5");
Or the slightly shorter version
Regex.Replace("410151000640", #"^(..)(..)(..)(...)(...)$", "$1-$2-$3-$4-$5");
I would approach this by having your own formatting method, as long as you know that the "Parcel Number" always conforms to a specific rule.
public static string FormatParcelNumber(string input)
{
if(input.length != 12)
throw new FormatException("Invalid parcel number. Must be 12 characters");
return String.Format("{0}-{1}-{2}-{3}-{4}",
input.Substring(0,2),
input.Substring(2,2),
input.Substring(4,2),
input.Substring(6,3),
input.Substring(9,3));
}
This should work in your case:
string value = "410151000640";
for( int i = 2; i < value.Length; i+=3){
value = value.Insert( i, "-");
}
Now value contains the string with dashes inserted.
EDIT
I just now saw that you didn't have dashes between every second number all the way, to this will require a small tweak (and makes it a bit more clumsy also I'm afraid)
string value = "410151000640";
for( int i = 2; i < value.Length-1; i+=3){
if( value.Count( c => c == '-') >= 3) i++;
value = value.Insert( i, "-");
}
If its part of UI you can use MaskedTextProvider in System.ComponentModel
MaskedTextProvider prov = new MaskedTextProvider("aa-aa-aa-aaa-aaa");
prov.Set("41x151000a40");
string result = prov.ToDisplayString();
Here is a simple extension method with some utility:
public static string WithMask(this string s, string mask)
{
var slen = Math.Min(s.Length, mask.Length);
var charArray = new char[mask.Length];
var sPos = s.Length - 1;
for (var i = mask.Length - 1; i >= 0 && sPos >= 0;)
if (mask[i] == '#') charArray[i--] = s[sPos--];
else
charArray[i] = mask[i--];
return new string(charArray);
}
Use it as follows:
var s = "276000017812008";
var mask = "###-##-##-##-###-###";
var dashedS = s.WithMask(mask);
You can use it with any string and any character other than # in the mask will be inserted. The mask will work from right to left. You can tweak it to go the other way if you want.
Have fun.
If i understodd you correctly youre looking for a function that removes all letters from a string, aren't you?
I have created this on the fly, maybe you can convert it into c# if it's what you're looking for:
Dim str As String = "410151000vb640"
str = String.Format("{0:##-##-##-###-###}", Convert.ToInt64(MakeNumber(str)))
Public Function MakeNumber(ByVal stringInt As String) As String
Dim sb As New System.Text.StringBuilder
For i As Int32 = 0 To stringInt.Length - 1
If Char.IsDigit(stringInt(i)) Then
sb.Append(stringInt(i))
End If
Next
Return sb.ToString
End Function