I need to parse the string "1.2345E-02" (a number expressed in exponential notation) to a decimal data type, but Decimal.Parse("1.2345E-02") simply throws an error
It is a floating point number, you have to tell it that:
decimal d = Decimal.Parse("1.2345E-02", System.Globalization.NumberStyles.Float);
It works if you specify NumberStyles.Float:
decimal x = decimal.Parse("1.2345E-02", NumberStyles.Float);
Console.WriteLine(x); // Prints 0.012345
I'm not entirely sure why this isn't supported by default - the default is to use NumberStyles.Number, which uses the AllowLeadingWhite, AllowTrailingWhite, AllowLeadingSign, AllowTrailingSign, AllowDecimalPoint, and AllowThousands styles. Possibly it's performance-related; specifying an exponent is relatively rare, I suppose.
In addition to specifying the NumberStyles I would recommend that you use the decimal.TryParse function such as:
decimal result;
if( !decimal.TryParse("1.2345E-02", NumberStyles.Any, CultureInfo.InvariantCulture, out result) )
{
// do something in case it fails?
}
As an alternative to NumberStyles.Any you could use a specific set if you're certain of your formats. e.g:
NumberStyles.AllowExponent | NumberStyles.Float
decimal d = Decimal.Parse("1.2345E-02", System.Globalization.NumberStyles.Float);
Be cautious about the selected answer: there is a subtility specifying System.Globalization.NumberStyles.Float in Decimal.Parse which could lead to a System.FormatException because your system might be awaiting a number formated with ',' instead of '.'
For instance, in french notation, "1.2345E-02" is invalid, you have to convert it to "1,2345E-02" first.
In conclusion, use something along the lines of:
Decimal.Parse(valueString.Replace('.',','), System.Globalization.NumberStyles.Float);
The default NumberStyle for decimal.Parse(String) is NumberStyles.Number, so if you just want to add the functionality to allow exponents, then you can do a bitwise OR to include NumberStyles.AllowExponent.
decimal d = decimal
.Parse("1.2345E-02", NumberStyles.Number | NumberStyles.AllowExponent);
I've found that passing in NumberStyles.Float, in some cases, changes the rules by which the string is processed and results in a different output from NumberStyles.Number (the default rules used by decimal.Parse).
For example, the following code will generate a FormatException in my machine:
CultureInfo culture = new CultureInfo("");
culture.NumberFormat.NumberDecimalDigits = 2;
culture.NumberFormat.NumberDecimalSeparator = ".";
culture.NumberFormat.NumberGroupSeparator = ",";
Decimal.Parse("1,234.5", NumberStyles.Float, culture); // FormatException thrown here
I'd recommend using the input NumberStyles.Number | NumberStyles.AllowExponent, as this will allow exponential numbers and will still process the string under the decimal rules.
CultureInfo culture = new CultureInfo("");
culture.NumberFormat.NumberDecimalDigits = 2;
culture.NumberFormat.NumberDecimalSeparator = ".";
culture.NumberFormat.NumberGroupSeparator = ",";
Decimal.Parse("1,234.5",NumberStyles.Number | NumberStyles.AllowExponent, culture); // Does not generate a FormatException
To answer the poster's question, the right answer should instead be:
decimal x = decimal.Parse("1.2345E-02", NumberStyles.Number | NumberStyles.AllowExponent);
Console.WriteLine(x);
Warning about using NumberStyles.Any:
"6.33E+03" converts to 6330 as expected. In German, decimal points are represented by commas, but 6,33E+03 converts to 633000! This is a problem for my customers, as the culture that generates the data is not known and may be different than the culture that is operating on the data. In my case, I always have scientific notation, so I can always replace comma to decimal point before parsing, but if you are working with arbitrary numbers, like pretty-formatted numbers like 1,234,567 then that approach doesn't work.
You don't need to replace the dots (respectively the commas) just specify the input IFormatProvider:
float d = Single.Parse("1.27315", System.Globalization.NumberStyles.Float, new CultureInfo("en-US"));
float d = Single.Parse("1,27315", System.Globalization.NumberStyles.Float, new CultureInfo("de-DE"));
If you want to check and convert the exponent value use this
string val = "1.2345E-02";
double dummy;
bool hasExponential = (val.Contains("E") || val.Contains("e")) && double.TryParse(val, out dummy);
if (hasExponential)
{
decimal d = decimal.Parse(val, NumberStyles.Float);
}
Hope this helps someone.
Related
post.Min.ToString("0.00").Replace(",", ".").Replace(".00", string.Empty)
post.Min is a double such as 12,34 or 12,00. Expected output is 12.34 or 12.
I basically want to replace the comma by a point, and cut the .00 part if any.
I am asking because I couldn't find anything, or because I don't exactly know what to search. This has an high change of being a duplicate, I simply can't find it. Please let me know.
The simplest solution would appear to be to use CultureInfo.InvariantCulture, and I reject the suggestion that this is any more complicated than using a series of replaces as you demonstrated in your question.
post.Min.ToString("0.##", CultureInfo.InvariantCulture);
# is the digit placeholder, described as the docs like this:
Replaces the "#" symbol with the corresponding digit if one is present; otherwise, no digit appears in the result string.
Try it online
If you use this in a lot of places, and that's why you want to keep it simple, you could make an extension method:
public static class MyExtensions
{
public static string ToHappyString(this double value)
{
return value.ToString("0.##", CultureInfo.InvariantCulture);
}
}
And then you just have to call .ToHappyString() wherever you use it. For example, post.Min.ToHappyString()
You can use .ToString("0.##").
like,
// Considered german culture; May be this is your current culture
CultureInfo culture = new CultureInfo("de");
double number1 = Double.Parse("12,34", culture);
double number2 = Double.Parse("12,00", culture);
Console.WriteLine(number1.ToString("0.##"));
Console.WriteLine(number2.ToString("0.##"));
Output:
12.34
12
.Net fiddle
Checkout the ToString overloads article on MSDN about examples of the N format. This is also covered in the Standard Numeric Format Strings article.
Relevant examples:
// Formatting of 1054.32179:
// N: 1,054.32
// N0: 1,054
// N1: 1,054.3
// N2: 1,054.32
// N3: 1,054.322
For the dot instead of comma to do it properly, in combination with N0 use:
System.Globalization.CultureInfo customCulture = (System.Globalization.CultureInfo)System.Threading.Thread.CurrentThread.CurrentCulture.Clone();
customCulture.NumberFormat.NumberDecimalSeparator = ".";
System.Threading.Thread.CurrentThread.CurrentCulture = customCulture;
double.ToString("0.##") to consider decimal places only if not .00 and you can create your own Number Format without using Culture:
NumberFormatInfo nfi = new NumberFormatInfo();
nfi.NumberDecimalSeparator = ".";
post.Min.ToString("0.##", nfi);
I already searched for my problem but I wasn't successfully and that's the reason I'm here.
All I want to do is reading a string like "3.14" and convert it to double.
Enough said... here is my code:
using System;
namespace GlazerCalcApplication
{
class MainClass
{
public static void Main (string[] args)
{
string heightString;
double height;
heightString = Console.ReadLine();
height = Convert.ToDouble(heightString);
Console.WriteLine(height);
}
}
}
Output:
3.14
314
Press any key to continue...
Why is my double value not 3.14?
Instead of Convert.ToDouble() I also tried it with double.Parse() but I received the same behaviour. Reading strings like 3,14 is no problem.
Maybe I should also mention that I use MonoDevelop and a linux OS.
Thanks in advance.
Try specifying the culture as Invariant:
height = Convert.ToDouble(heightString,CultureInfo.InvariantCulture);
It seems the decimal seperator of your culture is comma instead of dot therefore dot is truncated after conversion.
Convert.ToDouble(string) uses Double.Parse(string, CultureInfo.CurrentCulture) method explicitly.
Here how it's implemented;
public static double ToDouble(String value) {
if (value == null)
return 0;
return Double.Parse(value, CultureInfo.CurrentCulture);
}
It is likely your CurrentCulture's NumberFormatInfo.NumberDecimalSeparator property is not . (dot). That's why you can't parse a string with . as a date seperator.
Example in LINQPad;
CultureInfo c = new CultureInfo("de-DE");
c.NumberFormat.NumberDecimalSeparator.Dump(); // Prints ,
As a solution, you can create a new reference of your CurrentCulture and assing it's NumberDecimalSeparator property to . like;
double height;
CultureInfo c = new CultureInfo("de-DE");
c.NumberFormat.NumberDecimalSeparator = ".";
height = Convert.ToDouble("3.14", c);
Judging by the result I take it you are in a culture zone where comma is the normal decimal separator.
Also, I take it that you want both dot and comma to be used for decimal separation.
If not, the below is not the proper solution.
The fastest solution for using both would be
height = Convert.ToDouble(heightString.Replace('.', ',');
This would mean that both dots and comma's are used as comma and thus parsed as a decimal separator.
If you only want to use a dot as separator, you can use invariantculture or a specific numberformatinfo. Invariant culture is already shown in the other posts. numberformat info example:
var nfi = new NumberFormatInfo { NumberDecimalSeparator = "." };
height = double.Parse(heightString,nfi);
For completeness, the example below shows both using numberformatinfo for setting the dot as decimal separator, as well as replacing comma with dots, so both characters are used for decimals
var nfi = new NumberFormatInfo { NumberDecimalSeparator = "." };
height = double.Parse(heightString.Replace(',', '.'),nfi);
Different .Net cultures (countries) have different decimal separators.
If you expect input values to be in some specific format - either use some particular culture or InvariantCulture. Also consider using double.Parse as it geve more flexibility on parsing the values than generic Convert.ToDouble.
var d = double.Parse(heightString, CultureInfo.InvariantCulture);
If you expect user to enter value in local format - your code is fine, but either your expectation of "local format" is wrong, or "current culture" set incorrectly.
After search in google, using below code still can not be compiled:
decimal h = Convert.ToDecimal("2.09550901805872E-05");
decimal h2 = Decimal.Parse(
"2.09550901805872E-05",
System.Globalization.NumberStyles.AllowExponent);
You have to add NumberStyles.AllowDecimalPoint too:
Decimal.Parse("2.09550901805872E-05", NumberStyles.AllowExponent | NumberStyles.AllowDecimalPoint);
MSDN is clear about that:
Indicates that the numeric string can be in exponential notation. The
AllowExponent flag allows the parsed string to contain an exponent
that begins with the "E" or "e" character and that is followed by an
optional positive or negative sign and an integer. In other words, it
successfully parses strings in the form nnnExx, nnnE+xx, and nnnE-xx.
It does not allow a decimal separator or sign in the significand or
mantissa; to allow these elements in the string to be parsed, use the
AllowDecimalPoint and AllowLeadingSign flags, or use a composite style
that includes these individual flags.
Since decimal separator ("." in your string) can vary from culture to culture
it's safier to use InvariantCulture. Do not forget to allow this decimal
separator (NumberStyles.Float)
decimal h = Decimal.Parse(
"2.09550901805872E-05",
NumberStyles.Float | NumberStyles.AllowExponent,
CultureInfo.InvariantCulture);
Perharps, more convenient code is when we use NumberStyles.Any:
decimal h = Decimal.Parse(
"2.09550901805872E-05",
NumberStyles.Any,
CultureInfo.InvariantCulture);
use System.Globalization.NumberStyles.Any
decimal h2 = Decimal.Parse("2.09550901805872E-05", System.Globalization.NumberStyles.Any);
decimal h = Convert.ToDecimal("2.09550901805872E-05");
decimal h2 = decimal.Parse("2.09550901805872E-05", System.Globalization.NumberStyles.Any)
This thread was very helpful to me. For the benefit of others, here is complete code:
var scientificNotationText = someSourceFile;
// FileTimes are based solely on nanoseconds.
long fileTime = 0;
long.TryParse(scientificNotationText, NumberStyles.Any, CultureInfo.InvariantCulture,
out fileTime);
var dateModified = DateTime.FromFileTime(fileTime);
Decimal h2 = 0;
Decimal.TryParse("2.005E01", out h2);
This is probably dumb but it's giving me a hard time. I need to convert/format a double to string with a mandatory decimal point.
1 => 1.0
0.2423423 => 0.2423423
0.1 => 0.1
1234 => 1234.0
Basically, I want to output all decimals but also make sure the rounded values have the redundant .0 too. I am sure there is a simple way to achieve this.
Use double.ToString("N1"):
double d1 = 1d;
double d2 = 0.2423423d;
double d3 = 0.1d;
double d4 = 1234d;
Console.WriteLine(d1.ToString("N1"));
Console.WriteLine(d2.ToString("N1"));
Console.WriteLine(d3.ToString("N1"));
Console.WriteLine(d4.ToString("N1"));
Demo
Standard Numeric Format Strings
The Numeric ("N") Format Specifier
Update
(1.234).ToString("N1") produces 1.2 and in addition to removing additional decimal digits, it also adds a thousands separator
Well, perhaps you need to implement a custom NumberFormatInfo object which you can derive from the current CultureInfo and use in double.ToString:
var culture = CultureInfo.CurrentCulture;
var customNfi = (NumberFormatInfo)culture.NumberFormat.Clone();
customNfi.NumberDecimalDigits = 1;
customNfi.NumberGroupSeparator = "";
Console.WriteLine(d1.ToString(customNfi));
Note that you need to clone it since it's readonly by default.
Demo
There is not a built in method to append a mandatory .0 to the end of whole numbers with the .ToString() method, as the existing formats will truncate or round based on the number of decimal places you specify.
My suggestion is to just roll your own implementation with an extension method
public static String ToDecmialString(this double source)
{
if ((source % 1) == 0)
return source.ToString("f1");
else
return source.ToString();
}
And the usage:
double d1 = 1;
double d2 = 0.2423423;
double d3 = 0.1;
double d4 = 1234;
Console.WriteLine(d1.ToDecimalString());
Console.WriteLine(d2.ToDecimalString());
Console.WriteLine(d3.ToDecimalString());
Console.WriteLine(d4.ToDecimalString());
Results in this output:
1.0
0.2423423
0.1
1234.0
You could do something like this: if the number doesn't have decimal points you can format its output to enforce one decimal 0 and if it has decimal places, just use ToString();
double a1 = 1;
double a2 = 0.2423423;
string result = string.Empty;
if(a1 - Math.Floor(a1) >0.0)
result = a1.ToString();
else
result = a1.ToString("F1");
if (a2 - Math.Floor(a2) > 0.0)
result = a2.ToString();
else
result = a2.ToString("F1");
When you use "F" as formatting, the output won't contain thousands separator and the number that follows it specifies the number of decimal places.
Use ToString("0.0###########################").
It does work. I found it in duplicate of your question decimal ToString formatting which gives at least 1 digit, no upper limit
Double provides a method ToString() where you can pass an IFormatProvider-object stating how you want your double to be converted.
Additionally, it should display trailing 0 at all costs.
value = 16034.125E21;
// Display value using the invariant culture.
Console.WriteLine(value.ToString(CultureInfo.InvariantCulture));
// Display value using the en-GB culture.
Console.WriteLine(value.ToString(CultureInfo.CreateSpecificCulture("en-GB")));
// Display value using the de-DE culture.
Console.WriteLine(value.ToString(CultureInfo.CreateSpecificCulture("de-DE")));
// This example displays the following output to the console:
// -16325.62015
// -16325.62015
// -16325,62015
// 1.6034125E+25
// 1.6034125E+25
// 1,6034125E+25
Here is the documentation from MSDN.
You can cast to string and then appen ".0" if there was no decimal point given
string sValue=doubleValue.ToString();
if(!sValue.Contains('.'))
sValue+=".0";
EDIT:
As mentioned in the comments '.' may not be the decimal seperator in the current culture. Refer to this article to retrieve the actual seperator if you want make your code save for this case.
I have a decimal number like this:
62.000,0000000
I need to cast that decimal into int; it will always have zero decimal numbers; so I wont loose any precision. What I want is this:
62.000
Stored on an int variable in c#.
I try many ways but it always give me an error, that the string isn't in the correct format, this is one of the ways i tryied:
int.Parse("62.000,0000000");
Hope you can help me, thanks.
You need to parse it in the right culture. For example, the German culture uses "." as a group separator and "," as the decimal separator - so this code works:
CultureInfo german = new CultureInfo("de-DE");
decimal d = decimal.Parse("62.000,0000000", german);
int i = (int) d;
Console.WriteLine(i); // Prints 62000
Is that what you're after? Note that this will then fail if you present it with a number formatted in a different culture...
EDIT: As Reed mentions in his comment, if your real culture is Spain, then exchange "de-DE" for "es-ES" for good measure.
You need to specify your culture correctly:
var ci = CultureInfo.CreateSpecificCulture("es-ES");
int value = (int) decimal.Parse("60.000,000000", ci);
If you do this, it will correctly convert the number.
Why don't you just cast it to int instead of parsing?