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    <pre><code>re.search(r'\n', r'this\nis\nit') </code></pre> <p>As you said, "there will never be a regular expression in the second string." So we need to look at these strings differently: the first string is a regex, the second just a string. Usually your second string will not be raw, so any backslashes are Python-escapes, not regex-escapes.</p> <p>So the first string consists of a literal "\" and an "n". This is interpreted by the regex parser as a newline (<a href="http://docs.python.org/library/re.html" rel="nofollow">docs</a>: "Most of the standard escapes supported by Python string literals are also accepted by the regular expression parser"). So your regex will be searching for a newline character.</p> <p>Your second string consists of the string "this" followed by a literal "\" and an "n". So this string does not contain an actual newline character. Your regex will not match.</p> <p>As for your second regex:</p> <pre><code>re.search(r'\\n', r'this\nis\nit') </code></pre> <p>This version matches because your regex contains three characters: a literal "\", another literal "\" and an "n". The regex parser interprets the two slashes as a single "\" character, followed by an "n". So your regex will be searching for a "\" followed by an "n", which is found within the string. But that isn't very helpful, since it has nothing to do with newlines.</p> <p>Most likely what you want is to drop the <code>r</code> from the second string, thus treating it as a normal Python string.</p> <pre><code>re.search(r'\n', 'this\nis\nit') </code></pre> <p>In this case, your regex (as before) is searching for a newline character. And, it finds it, because the second string contains the word "this" followed by a newline.</p>
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    1. COThanks for the clarity! I did understand what was happening, but I don't understand the particular cases. To be more specific: I want to ensure that there are no "weird characters" being captured in my code. Therefore, I want to make sure there are no newlines, tabs, etc. in the string to be checked (the 2nd string). So in this case, let's say the string I'm testing is "temp", how would I test the string? E.g., could I do: `if re.search(r'\n',temp)` to test for a newline in temp? What if temp has a "\n" in it, but as a literal "\" + "n"... I don't understany how Python differentiates.
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      1. USMike Williamson
    2. COWell if `temp` is a string variable, then that will be fine. The escaping is only relevant for *string literals* -- that is, actual quoted representations of strings in the source code. If you have a variable with a string in it, then there's no escaping. There's just newline characters, backslash characters and 'n' characters. If `temp` has a newline in it (ASCII 10), then your regex will find it. If `temp` has a backslash (ASCII 92) then an 'n' (ASCII 110), then your regex will not find a newline. Use `map(ord, temp)` to see the ASCII values in `temp`, which should make it clear.
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    3. CO@mguica Thanks so much! Your comment above was super-detailed & clear, which is what I'd needed. :)
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