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    <p>Your regex matches everything from the first s to the last s, so if you replace the match with "no", you get "thinotring". </p> <p>The parentheses don't limit the match, they capture the text matched by whatever is inside them in a special variable called backreference. In your example, backreference number 1 would contain <code>is a</code>. You can refer to a backreference later in the same regex using backslashes and the number of the backreference: <code>\1</code>.</p> <p>What you probably want is lookaround:</p> <pre><code>re.sub(r"(?&lt;=s ).*?(?= s)", "no", "this is a string") </code></pre> <p><code>(?&lt;=s )</code> means: Assert that it is possible to match <code>s</code> before the current position in the string, but don't make it part of the match.</p> <p>Same for <code>(?= s)</code>, but it asserts that the string will continue with <code>s</code> after the current position.</p> <p>Be advised that lookbehind in Python is limited to strings of fixed length. So if that is a problem, you can sort of work around this using...<em>backreferences</em>!</p> <pre><code>re.sub(r"(s ).*?( s)", r"\1no\2", "this is a string") </code></pre> <p>OK, this is a contrived example, but it shows what you can do. From your edit, it's becoming apparent that you're trying to parse HTML with regex. Now that is not such a good idea. Search SO for "regex html" and you'll see why.</p> <p>If you still want to do it:</p> <pre><code>re.sub(r"(&lt;a.*?href=['"])((?!http).*?['"].*?&gt;)", r'\1http://\2', string) </code></pre> <p>might work. But this is extremely brittle.</p>
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    1. USTim Pietzcker
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    1. COSadly that won't work, see my new example in the edited question.
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      1. USKevin Renskers
    2. COI came to the same solution as you and eumiro did. Thanks all!
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