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    <p>You could capture the number and check the length before adding 0, but you might be able to use this instead:</p> <pre><code>import re txt = 'file8, file9, file10' pat = r"(?&lt;!\d)(\d)(?=,|$)" regexp = re.compile(pat) print(regexp.sub(r"0\1", txt)) </code></pre> <p><a href="http://regex101.com/r/bA3xC6" rel="nofollow">regex101 demo</a></p> <p><code>(?&lt;! ... )</code> is called a negative lookbehind. This prevents (negative) a match if the pattern after it has the pattern in the negative lookbehind matches. For example, <code>(?&lt;!a)b</code> will match all <code>b</code> in a string, except if it has an <code>a</code> before it, meaning <code>bb</code>, <code>cb</code> matches, but <code>ab</code> doesn't match. <code>(?&lt;!\d)(\d)</code> thus matches a digit, unless it has another digit before it.</p> <p><code>(\d)</code> is a single digit, enclosed in a capture group, denoted by simple parentheses. The captured group gets stored in the first capture group.</p> <p><code>(?= ... )</code> is a positive lookahead. This matches <em>only</em> if the pattern inside the positive lookahead matches after the pattern before this positive lookahead. In other words, <code>a(?=b)</code> will match all <code>a</code> in a string only if there's a <code>b</code> after it. <code>ab</code> matches, but <code>ac</code> or <code>aa</code> don't.</p> <p><code>(?=,|$)</code> is a positive lookahead containing <code>,|$</code> meaning either a comma, or the end of the string.</p> <p><code>(?&lt;!\d)(\d)(?=,|$)</code> thus matches any digit, as long as there's no digit before it and there's a comma after it, or if that digit is at the end of the string.</p>
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    1. COdammit, i wrote the exact same code, but you beat me to this :P
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    2. CO@GamesBrainiac Oops, sorry ^^;
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    3. COOkay, I have to ask, can you explain the logic behind all the symbols in the pat? I know \d is digit and | alternative, but the rest - especially in that order confuses me? I don't think I would have ever come on that on my own
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