How to Check That a String Is Parseable to a Double

How to check that a string is parseable to a double?

The common approach would be to check it with a regular expression like it's also suggested inside the Double.valueOf(String) documentation.

The regexp provided there (or included below) should cover all valid floating point cases, so you don't need to fiddle with it, since you will eventually miss out on some of the finer points.

If you don't want to do that, try catch is still an option.

The regexp suggested by the JavaDoc is included below:

final String Digits     = "(\\p{Digit}+)";
final String HexDigits  = "(\\p{XDigit}+)";
// an exponent is 'e' or 'E' followed by an optionally 
// signed decimal integer.
final String Exp        = "[eE][+-]?"+Digits;
final String fpRegex    =
    ("[\\x00-\\x20]*"+ // Optional leading "whitespace"
    "[+-]?(" +         // Optional sign character
    "NaN|" +           // "NaN" string
    "Infinity|" +      // "Infinity" string

    // A decimal floating-point string representing a finite positive
    // number without a leading sign has at most five basic pieces:
    // Digits . Digits ExponentPart FloatTypeSuffix
    // 
    // Since this method allows integer-only strings as input
    // in addition to strings of floating-point literals, the
    // two sub-patterns below are simplifications of the grammar
    // productions from the Java Language Specification, 2nd 
    // edition, section 3.10.2.

    // Digits ._opt Digits_opt ExponentPart_opt FloatTypeSuffix_opt
    "((("+Digits+"(\\.)?("+Digits+"?)("+Exp+")?)|"+

    // . Digits ExponentPart_opt FloatTypeSuffix_opt
    "(\\.("+Digits+")("+Exp+")?)|"+

    // Hexadecimal strings
    "((" +
    // 0[xX] HexDigits ._opt BinaryExponent FloatTypeSuffix_opt
    "(0[xX]" + HexDigits + "(\\.)?)|" +

    // 0[xX] HexDigits_opt . HexDigits BinaryExponent FloatTypeSuffix_opt
    "(0[xX]" + HexDigits + "?(\\.)" + HexDigits + ")" +

    ")[pP][+-]?" + Digits + "))" +
    "[fFdD]?))" +
    "[\\x00-\\x20]*");// Optional trailing "whitespace"

if (Pattern.matches(fpRegex, myString)){
    Double.valueOf(myString); // Will not throw NumberFormatException
} else {
    // Perform suitable alternative action
}

Fastest way to check if a String can be parsed to Double in Java

You can check it using the same regular expression the Double class uses. It's well documented here:

http://docs.oracle.com/javase/6/docs/api/java/lang/Double.html#valueOf%28java.lang.String%29

Here is the code part:

To avoid calling this method on an invalid string and having a NumberFormatException be thrown, the regular expression below can be used to screen the input string:

  final String Digits     = "(\\p{Digit}+)";
  final String HexDigits  = "(\\p{XDigit}+)";

        // an exponent is 'e' or 'E' followed by an optionally 
        // signed decimal integer.
        final String Exp        = "[eE][+-]?"+Digits;
        final String fpRegex    =
            ("[\\x00-\\x20]*"+  // Optional leading "whitespace"
             "[+-]?(" + // Optional sign character
             "NaN|" +           // "NaN" string
             "Infinity|" +      // "Infinity" string

             // A decimal floating-point string representing a finite positive
             // number without a leading sign has at most five basic pieces:
             // Digits . Digits ExponentPart FloatTypeSuffix
             // 
             // Since this method allows integer-only strings as input
             // in addition to strings of floating-point literals, the
             // two sub-patterns below are simplifications of the grammar
             // productions from the Java Language Specification, 2nd 
             // edition, section 3.10.2.

             // Digits ._opt Digits_opt ExponentPart_opt FloatTypeSuffix_opt
             "((("+Digits+"(\\.)?("+Digits+"?)("+Exp+")?)|"+

             // . Digits ExponentPart_opt FloatTypeSuffix_opt
             "(\\.("+Digits+")("+Exp+")?)|"+

       // Hexadecimal strings
       "((" +
        // 0[xX] HexDigits ._opt BinaryExponent FloatTypeSuffix_opt
        "(0[xX]" + HexDigits + "(\\.)?)|" +

        // 0[xX] HexDigits_opt . HexDigits BinaryExponent FloatTypeSuffix_opt
        "(0[xX]" + HexDigits + "?(\\.)" + HexDigits + ")" +

        ")[pP][+-]?" + Digits + "))" +
             "[fFdD]?))" +
             "[\\x00-\\x20]*");// Optional trailing "whitespace"

  if (Pattern.matches(fpRegex, myString))
            Double.valueOf(myString); // Will not throw NumberFormatException
        else {
            // Perform suitable alternative action
        }

How to find if a Scala String is parseable as a Double or not?

For Scala 2.13+ see Xavier's answer below. Apparently there's a toDoubleOption method now.

For older versions:

def parseDouble(s: String) = try { Some(s.toDouble) } catch { case _ => None }

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Fancy version (edit: don't do this except for amusement value; I was a callow youth years ago when I used to write such monstrosities):

case class ParseOp[T](op: String => T)
implicit val popDouble = ParseOp[Double](_.toDouble)
implicit val popInt = ParseOp[Int](_.toInt)
// etc.
def parse[T: ParseOp](s: String) = try { Some(implicitly[ParseOp[T]].op(s)) } 
                                   catch {case _ => None}

scala> parse[Double]("1.23")
res13: Option[Double] = Some(1.23)

scala> parse[Int]("1.23")
res14: Option[Int] = None

scala> parse[Int]("1")
res15: Option[Int] = Some(1)

Check if a string value is equal to a double?

Java specifies a regular expression that can be used to validate strings being passed to Double.parseDouble or Double.valueOf in the Javadoc of Double.valueOf:

final String Digits     = "(\\p{Digit}+)";
final String HexDigits  = "(\\p{XDigit}+)";
// an exponent is 'e' or 'E' followed by an optionally
// signed decimal integer.
final String Exp        = "[eE][+-]?"+Digits;
final String fpRegex    =
  ("[\\x00-\\x20]*"+  // Optional leading "whitespace"
   "[+-]?(" + // Optional sign character
   "NaN|" +           // "NaN" string
   "Infinity|" +      // "Infinity" string

   // A decimal floating-point string representing a finite positive
   // number without a leading sign has at most five basic pieces:
   // Digits . Digits ExponentPart FloatTypeSuffix
   //
   // Since this method allows integer-only strings as input
   // in addition to strings of floating-point literals, the
   // two sub-patterns below are simplifications of the grammar
   // productions from section 3.10.2 of
   // The Java™ Language Specification.

   // Digits ._opt Digits_opt ExponentPart_opt FloatTypeSuffix_opt
   "((("+Digits+"(\\.)?("+Digits+"?)("+Exp+")?)|"+

   // . Digits ExponentPart_opt FloatTypeSuffix_opt
   "(\\.("+Digits+")("+Exp+")?)|"+

   // Hexadecimal strings
   "((" +
    // 0[xX] HexDigits ._opt BinaryExponent FloatTypeSuffix_opt
    "(0[xX]" + HexDigits + "(\\.)?)|" +

    // 0[xX] HexDigits_opt . HexDigits BinaryExponent FloatTypeSuffix_opt
    "(0[xX]" + HexDigits + "?(\\.)" + HexDigits + ")" +

    ")[pP][+-]?" + Digits + "))" +
   "[fFdD]?))" +
   "[\\x00-\\x20]*");// Optional trailing "whitespace"

 if (Pattern.matches(fpRegex, myString))
   Double.valueOf(myString); // Will not throw NumberFormatException
 else {
   // Perform suitable alternative action
 }

How to check if a string is parsable to float

As mentioned before, use Float#parseFloat(String) but don't forget to replace the commas:

try {
    float val = Float.parseFloat(string.replace(',', '.'));
} catch (NumberFormatException e) {
    // Executed iff the string couldn't be parsed.
}

Inspired by this great answer, you can also check out the corresponding Float#valueOf(String) if you prefer a nasty regex.

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