检查Java中是否可以将String解析为Double的最快方法 [英] Fastest way to check if a String can be parsed to Double in Java

问题描述

我知道有百万种做法，但最快的是什么？这应该包括科学记数法。

注意：我不想将值转换为Double，我只想知道是否可能。即 private boolean isDouble（String value）。

解决方案

它使用与Double类使用相同的正则表达式。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

以下是代码部分：

避免在无效字符串上调用此方法并抛出NumberFormatException，下面的正则表达式可用于屏蔽输入字符串：

  final String Digits =（\\p {Digit} +）; 
 final String HexDigits =（\\p {XDigit} +）; 
 
 //一个指数是'e'或'E'，后跟一个可选的
 //有符号的十进制整数。 
 final String Exp =[eE] [+  - ]？+数字; 
 final String fpRegex = 
（[\\\x00-\\x20] *+ //可选的前导空格
[+  - ]？（+ //可选符号字符
NaN |+ //NaNstring 
Infinity |+ //Infinitystring 
 
 //十进制浮点数代表一个有限的正值
 //没有前导符号的字符串最多有五个基本部分：
 // Digits。Digits ExponentPart FloatTypeSuffix 
 // 
 //由于此方法允许整数唯一的字符串作为输入
 //除了浮点文字的字符串之外，
 //下面的两个子模式是简化语法
 //从Java语言规范，第二
 //版，第3.10.2节。
 
 // Digits ._opt Digits_opt ExponentPart_opt FloatTypeSuffix_opt 
（（+ Digits +（\\。）？（+ Digits +？）（+ Exp +）？）|+ 
 
 //。数字ExponentPart_opt FloatTypeSuffix_opt 
（\\。（+ Digits +）（+ Exp +）？）|+ 
 
 //十六进制字符串
（ + 
 // 0 [xX] HexDigits ._opt BinaryExponent FloatTypeSuffix_opt 
（0 [xX]+ HexDigits +（\\。）？）|+ 
 
 // 0 [xX] HexDigits_opt。HexDigits BinaryExponent FloatTypeSuffix_opt 
（0 [xX]+ HexDigits +？（\\。）+ HexDigits +）+ 
 
 $ b[+] $$$$$）+ 
 x20] *）; //可选尾随的空格
 
 if（Pattern.matches（fpRegex，myString））
 Double.valueOf（myString）; //不会抛出NumberFormatException 
 else {
 //执行适当的替代动作
} 
  

I know there's a million ways of doing this but what is the fastest? This should include scientific notation.

NOTE: I'm not interested in converting the value to Double, i'm only interested in knowing if it's possible. i.e. private boolean isDouble(String value).

解决方案

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
        }

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