如何确定一个字符在Java中是否是字母？

unicode java

2022-09-01 15:21:21

如何检查单字符字符串是否为字母 - 包括任何带重音符号的字母？

我最近必须解决这个问题，所以我会在最近的VB6问题提醒我之后自己回答。

答案 1

Character.isLetter（）比 string.matches（）快得多，因为 string.matches（）每次都会编译一个新的模式。即使缓存该模式，我认为isLetter（）仍然会击败它。

编辑：只是再次遇到这个，并认为我会尝试提出一些实际数字。这是我在基准测试中的尝试，检查所有三种方法（有和没有缓存，和）。我还确保检查了有效和无效的字符，以免扭曲事物。matches()PatternCharacter.isLetter()

import java.util.regex.*;

class TestLetter {
    private static final Pattern ONE_CHAR_PATTERN = Pattern.compile("\\p{L}");
    private static final int NUM_TESTS = 10000000;

    public static void main(String[] args) {
        long start = System.nanoTime();
        int counter = 0;
        for (int i = 0; i < NUM_TESTS; i++) {
            if (testMatches(Character.toString((char) (i % 128))))
                counter++;
        }
        System.out.println(NUM_TESTS + " tests of Pattern.matches() took " +
                (System.nanoTime()-start) + " ns.");
        System.out.println("There were " + counter + "/" + NUM_TESTS +
                " valid characters");
        /*********************************/
        start = System.nanoTime();
        counter = 0;
        for (int i = 0; i < NUM_TESTS; i++) {
            if (testCharacter(Character.toString((char) (i % 128))))
                counter++;
        }
        System.out.println(NUM_TESTS + " tests of isLetter() took " +
                (System.nanoTime()-start) + " ns.");
        System.out.println("There were " + counter + "/" + NUM_TESTS +
                " valid characters");
        /*********************************/
        start = System.nanoTime();
        counter = 0;
        for (int i = 0; i < NUM_TESTS; i++) {
            if (testMatchesNoCache(Character.toString((char) (i % 128))))
                counter++;
        }
        System.out.println(NUM_TESTS + " tests of String.matches() took " +
                (System.nanoTime()-start) + " ns.");
        System.out.println("There were " + counter + "/" + NUM_TESTS +
                " valid characters");
    }

    private static boolean testMatches(final String c) {
        return ONE_CHAR_PATTERN.matcher(c).matches();
    }
    private static boolean testMatchesNoCache(final String c) {
        return c.matches("\\p{L}");
    }
    private static boolean testCharacter(final String c) {
        return Character.isLetter(c.charAt(0));
    }
}

我的输出：

10000000 tests of Pattern.matches() took 4325146672 ns.
There were 4062500/10000000 valid characters
10000000 tests of isLetter() took 546031201 ns.
There were 4062500/10000000 valid characters
10000000 tests of String.matches() took 11900205444 ns.
There were 4062500/10000000 valid characters

因此，即使使用缓存，这几乎要好8倍。（未缓存比缓存差近3倍。Pattern

答案 2

只需检查字母是否在A-Z中，因为它不包括带有重音符号的字母或其他字母表中的字母。

我发现您可以将正则表达式类用于“Unicode字母”或其区分大小写的变体之一：

string.matches("\\p{L}"); // Unicode letter
string.matches("\\p{Lu}"); // Unicode upper-case letter

您也可以使用 Character 类执行此操作：

Character.isLetter(character);

但是如果您需要检查多个字母，那就不太方便了。