Strange behavior of Regular Expression matching numbers in Python 3

Question

Can anybody find a mistake in the following regex:

regex = ([.0-9]+[/–_\":・’ー‘‐`─'.,-\0-9]*)

My intention is to match "numerical" strings of any kind but if the number is followed by e.g. a letter I just want to get the number.

When I use it with the following sentences:
s1 = Bla bla 805P bla 1080P; bla bla
s2 = Bla bla 5600p bla 5400p
It finds 805P and 1080P in s1 and 5600 and 5400 in s2.
You can check it using http://regexpal.com

I also used this regex in Regex Buddy and it gives me a description that says nothing about letters.
Does anybody have any idea why am I catching P and P; in s1 if there is no letters included in the second group of characters?

It should be a double backslash, i.e. escaped backslash. I think it displays it as a single backslash for whatever reason. When I edit my post there are two backslashes before 0. — user3055017
– user3055017, Commented Aug 6, 2014 at 8:41

Amadan · Accepted Answer · 2014-08-06 08:07:31Z

2

A part of your regexp says [... ,-\\ ...], which will include any characters between comma and backslash, which includes the following characters:

,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\

This, incidentally, includes P.

To match a literal minus sign in a character class, it needs to be first, last or escaped. E.g.

[- ... ,\\ ... ]
[ ... ,\\ ... -]
[ ... ,\-\\ ... ]

would all be valid ways to write what you intended.

answered Aug 6, 2014 at 8:07

Amadan

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jonrsharpe · Accepted Answer · 2014-08-06 08:06:43Z

2

You're escaping the 0, not the -, with -\0. Also, that is very complex; all you need is

pattern = "([.0-9]+[^A-Za-z]*)"

I.e. "One or more numbers or periods followed by as many non-letter characters as possible". You can add more characters to ignore to that second block if required. See demo

answered Aug 6, 2014 at 8:06

jonrsharpe

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