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    <p>A single character is a valid regular expression. A single character that is not "magic" matches itself. If you can identify a single character that will never, ever appear in your text, you could make a pattern from that.</p> <p>How about ASCII NUL, character 0?</p> <p>I stuck in one more string in your test program, the string: <code>'\0'</code></p> <p>It was about as fast as your best pattern: <code>b(?&lt;!b)</code></p> <p>Okay, you already have a character after the end of the string. How about a character <em>before</em> the start of the string? That's impossible: <code>'x^'</code></p> <p>Aha! That's faster than checking for a character after end of string. But it's about as fast as your best pattern.</p> <p>I suggest replacing the <code>b</code> with an ASCII NUL and calling it good. When I tried that pattern: <code>\0(?&lt;!\0)</code></p> <p>It wins by a tiny fraction. But really, on my computer, all the ones discussed above are so close together that there isn't much to distinguish them.</p> <p>Results:</p> <pre><code>Pattern Time \0(?&lt;!\0) 0.098 \0 0.099 x^ 0.099 b(?&lt;!b) 0.099 ^(?&lt;=x) 1.416 $b 1.446 $a 1.447 \Za 1.462 \Zb 1.465 [^\s\S] 2.280 a(?&lt;!a) 2.843 </code></pre> <p>That was <em>fun</em>. Thanks for posting the question.</p> <p>EDIT: Ah <em>hah</em>! I rewrote the program to test with real input data, and got a different result.</p> <p>I downloaded "The Complete Works of William Shakespeare" from Project Gutenberg as a text file. (Weird, it gave an error on <code>wget</code> but let my browser get it... some sort of measure to protect against automated copying?) URL: <a href="http://www.gutenberg.org/cache/epub/100/pg100.txt" rel="nofollow">http://www.gutenberg.org/cache/epub/100/pg100.txt</a></p> <p>Here are the results, followed by the modified program as I ran it.</p> <pre><code>Pattern Time \0(?&lt;!\0) 0.110 \0 0.118 x^ 0.119 b(?&lt;!b) 0.143 a(?&lt;!a) 0.275 ^(?&lt;=x) 1.577 $b 1.605 $a 1.611 \Za 1.634 \Zb 1.634 [^\s\S] 2.441 </code></pre> <p>So yeah I'm definitely going with that first one.</p> <pre><code>#!/usr/bin/env python import re import time tests = [ r'x^', r'\0', r'[^\s\S]', r'^(?&lt;=x)', r'a(?&lt;!a)', r'b(?&lt;!b)', r'\0(?&lt;!\0)', r'\Za', r'\Zb', r'$a', r'$b' ] timing = [] #text = 'a' * 50000000 text = open("/tmp/pg100.txt").read() text = text * 10 for t in tests: pat = re.compile(t) start = time.time() pat.search(text) dur = time.time() - start timing.append((t, dur)) timing.sort(key=lambda x: x[1]) print('%-30s %s' % ('Pattern', 'Time')) for t, dur in timing: print('%-30s %0.3f' % (t, dur)) </code></pre>
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    1. COIt is possible to throw in example to make the `\0` matches, and make `\0(?<!\0)`, `x^`, `b(?<!b)` slows down to other.
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    2. COsingle character regexes _can_ match some string, so they aren't really correct answers. As for `\0(?<!\0)`, that's what I've been using and it's as good as it gets so far. Seeing your `x^`, I thought `a\A` on my original input would be the fastest but it's not. I'm confused as to why the engine keeps looking after the first character. `\A` only matches at the start.
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    3. CO@Mansour: Good so far, since you are assuming that `\0` won't appear in text. If you say that you assume the text won't contain `\0`, then it is fair. But no assumption = anything goes. So I can specify `\0` * big number and it will comes in last.
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