有没有更好的选择来检查子字符串是否为数字？

Question

我正在尝试解析一个长字符串。在字符串中，有数字和其他种类的符号（文本）。我需要分别存储单词和数字。首先，我将字符串用空格分隔。

String fullString = "Some 3 random 5.6 text 12 with -4.5 numbers -1 in 0 beetween";
String [] words = fullString.split(" ");

我已经创建了一个方法来检查 Double.parseDouble() 方法是否在 words 变量中存储的每个子字符串上出现和异常。

private static boolean isNumeric(String str) {
    if (str == null) {
       return false;
    }
    try {
       double parseDouble = Double.parseDouble(str);
    } catch (NumberFormatException numberFormatException) {
       return false;
    }
    return true;
}

在我的代码中，当我得到 true 时，我将不得不运行两次 Double.parseDouble(str) 方法。

for(String word: words){
    if(isNumeric(word)){
        numbers.push = Double.parseDouble(word);
    }
}

我想我正在寻找以下两个选项中的任何一个：

1. 一种更简单的方法来检查存储在words 中的子字符串是否为数字。
2. 另一种只解析一次双精度的方法。

Answer 1

在您的情况下，如果使用 Double.parseDouble() 是一个足够好的指标，如果字符串是否为数字，并且您只想避免运行相同的方法两次，一次用于验证，一次用于实际解析，而不是使用 MgntUtils Open TextUtils 类中具有以下方法的源库：

public static double parseStringToDouble(java.lang.CharSequence num,
                                         double defaultValue)

你可以这样使用它：

Double num = TextUtils.parseStringToDouble(parsedStr, Double.MIN_VALUE);
if(Double.MIN_VALUE.equals(num)) {
  treatStringAsNotNumeric(parsedStr);
} else {
  treatStringAsNumeric(num /* or may be parsedStr*/);
}

该库以Maven artifact 和Github 的形式提供（包括源代码和Javadoc）。 Javadoc 也可以在线获取here

免责声明：库是我写的

Answer 2

关于字符串的问题在于，可以在字符串中以多种不同格式提供数值，其中一些格式永远不会被数字解析器（如Integer.parseInt() 或Double.parseDouble()）传递。下面的 isNumeric() 方法可以处理相当多的最常见的方法，至少是在北美使用的方法。需要对String#matches() 方法中以下isNumeric() 方法代码中使用的Regular Expression 进行一些修改，以处理世界其他地区使用的其他特定数字符号格式。

isNumeric() 方法：

/**
 * Returns true if the string is a numerical type value.
 * 
 * The Regular Expression (regex) used here covers just about any provided 
 * numerical values like:<pre>
 * 
 *    "05", "0",   "18", "-12", "-33.0", "0.33", "-22.8", ".675E8", 
 *    "-.322", "44.343e6", "+453", "+94.234, 1,234.87, 32,043,001"
 * </pre>
 * 
 * @param numericalString (String)
 * 
 * @return (boolean)
 */
public static boolean isNumeric(String numericalString) {
    /* Utilization of the String#matches() method.
       The String#replace() method is used to remove commas in numbers.
       These commas should also be removed before parsing to a specific 
       numerical data type (wherever that might be in your code).                         */
    return numericalString.replace(",", "").matches("[-+]?(\\d+)?(\\.(\\d+)?((?i)E\\d+)?)?");
}

如何使用 isNumeric() 方法：

// The string to work against.
String fullString = "Some 3 random 5.6 text 12 with -4.5 numbers -1 "
                  + "in 0 between. These are added 05 0 18 -12 -33.0 0.33 "
                  + "-22.8 .675E8 -.322 44.343e6 +453 +94.234 1,234.87 "
                  + "32,043,001 as well!";
    
/* Split the contents of fullString based on one or more 
   whitespaces as delimiter. This takes care of any strings
   that may have double-spacing anywhere within it.      */
String [] words = fullString.split("\\s+");
    
/* Because we may never know how large any given string may 
   be, We use the List Interface or ArrayList object to 
   create a collection of both words and double numerical 
   values. Either of these object can grow dynamically (as
   can may other objects). Arrays can not grow dynamically.
   You can convert these Lists to Arrays later if you really 
   like.                                      */
List<String> wordsList = new ArrayList<>();
List<Double> numbersList = new ArrayList<>();
    
/* Iterate through the 'words' array and add alpha character
   words to the 'wordsList' list and numerical values (if any)
   to the 'numbersList' list. In this example all string number
   values will be converted to Double data type (also handles 
   Exponents).         */
for (String wordStrg : words) {
    if (isNumeric(wordStrg)) {
        // The String#replace() method id used to remove commas in numbers.
        numbersList.add(Double.parseDouble(wordStrg.replace(",", "")));
    }
    else {
        wordsList.add(wordStrg);
    }
}
    
// Show what is in the Lists:
System.out.println("The Words List:");
System.out.println(wordsList);
System.out.println();
System.out.println("The Numbers List (as Double):");
System.out.println(numbersList);
    
// ==============================================================
// If for some reason you want to convert the Lists 
// to arrays then you can do it this way:
String[] onlyWords = wordsList.toArray(new String[wordsList.size()]);
Double[] onlyNumbers = numbersList.toArray(new Double[numbersList.size()]);
    
// Display The Arrays:
System.out.println();
System.out.println("The Words Array:");
System.out.println(Arrays.toString(onlyWords));
System.out.println();
System.out.println("The Numbers Array (as Double):");
System.out.println(Arrays.toString(onlyNumbers));

Answer 3

假设你有 commons-lang3 你可以做；

扩展 -> 也匹配像 0xFF 0e0 0L 这样的数字

Arrays.stream(words).filter(NumberUtils::isCreatable).toArray()

简单版本 -> 匹配 0 -0 0.0 .0

Arrays.stream(words).filter(Pattern.compile("^-?(\\d+|(\\.\\d+)|(\\d+\\.\\d+))$").asPredicate()).toArray()