\p{N}[\p{N}.]*
It's used in a word counting algorithm but not sure if that's what it's doing, and whether or not its accurate.
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First off the actual regex is just the part between the two forward slashes. As for the regex, i have not seen the \p (special/escaped character p) or {N} (cardinality) before either. – WestCoastProjects Mar 08 '13 at 02:11
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Can you not just test it on small amounts of data?... – Simon Whitehead Mar 08 '13 at 02:12
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Include the language/environment. Not all regular expressions are the same. – Mar 08 '13 at 04:37
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Looks like it matches unicode numbers kind of like \d[\d.]*. It does not look like a very good expression since it will match things like 1234 and 1..2..3.. but that depends on your data. I mean you may not have things like this in it.
\p{N} or \p{Number}: any kind of numeric character in any script.
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Like, if I matched it against "How now, brown cow", it couldn't be used to tell me that there are 4 words in the string? – Mar 08 '13 at 02:23
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Thanks again @jdb - almost there. Can you please see my original (edited) post? Another user mentioned that I might have posted some Ruby-specific escape sequences in my original post, so I deleted them out. Can you look at it and confirm that the original regex I posted doesn't change your opinion of what this regex does? Thanks again! – Mar 08 '13 at 02:33
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I have run `\p{N}[\p{N}.]*` in Java and it does exactly what I said. `"How now, brown cow".replaceAll("\\p{N}[\\p{N}.]*", "x");` prints `How now, brown cow`. And "123".replaceAll("\\p{N}[\\p{N}.]*", "x");` prints `x` – jdb Mar 08 '13 at 02:37