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    <p>Character.isLetter() is much faster than string.matches(), because string.matches() compiles a new Pattern every time. Even caching the pattern, I think isLetter() would still beat it.</p> <hr> <p><strong>EDIT:</strong> Just ran across this again and thought I'd try to come up with some actual numbers. Here's my attempt at a benchmark, checking all three methods (<code>matches()</code> with and without caching the <code>Pattern</code>, and <code>Character.isLetter()</code>). I also made sure that there were both valid and invalid characters checked, so as not to skew things.</p> <pre><code>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 &lt; 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 &lt; 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 &lt; 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)); } } </code></pre> <p>And my output:</p> <pre>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</pre> <p>So that's almost 8x better, even with a cached <code>Pattern</code>. (And uncached is nearly 3x worse than cached.)</p>
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    1. POHow to determine whether a character is a letter in Java?
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    1. USMichael Myers
    1. USMichael Myers
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      1. VTUpMod
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    1. COYou should use `c.codePointAt(0)` rather than `c.charAt(0)` in `testCharacter()`; otherwise it'll fail for characters outside the BMP.
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      1. USMechanical snail
 

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