检查String是否可以在Java中解析为Double的最快方法

Question

我知道有一百万种方法可以做到这一点,但最快的是什么？这应该包括科学记数法.

注意:我对将值转换为Double不感兴趣,我只想知道它是否可能.即private boolean isDouble(String value).

Answer 1

您可以使用Double类使用的相同正则表达式来检查它.这里有很好的记录:

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

这是代码部分:

为了避免在无效字符串上调用此方法并抛出NumberFormatException,可以使用下面的正则表达式来筛选输入字符串:

  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
        }

Answer 2

有一个方便的NumberUtils#isNumber在Apache的百科全书郎.这有点牵强:

有效数字包括用0x限定符标记的十六进制,科学记数法和用类型限定符标记的数字(例如123L).

但我想它可能比正则表达式或抛出和捕获异常更快.