python string splitting

Question

I have an input string like this: a1b2c30d40 and I want to tokenize the string to: a, 1, b, 2, c, 30, d, 40.

I know I can read each character one by one and keep track of the previous character to determine if I should tokenize it or not (2 digits in a row means don't tokenize it) but is there a more pythonic way of doing this?

Cat Plus Plus · Accepted Answer · 2011-01-30 17:01:04Z

13

>>> re.split(r'(\d+)', 'a1b2c30d40')
['a', '1', 'b', '2', 'c', '30', 'd', '40', '']

On the pattern: as the comment says, \d means "match one digit", + is a modifier that means "match one or more", so \d+ means "match as much digits as possible". This is put into a group (), so the entire pattern in context of re.split means "split this string using as much digits as possible as the separator, additionally capturing matched separators into the result". If you'd omit the group, you'd get ['a', 'b', 'c', 'd', ''].

edited Jan 30, 2011 at 17:01

answered Jan 30, 2011 at 16:04

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6 Comments

Hery Over a year ago

Umm I don't understand regex very well. Do you mind putting some explanation of the (\d+) pattern?

user395760 Over a year ago

It splits on numbers/consecutive digits (\d is 0-9, + is one or more).

6502 Over a year ago

+1: but I think it would be better to use a raw string for the regexp

Cat Plus Plus Over a year ago

@6502: it matters only if there are any Python-interpreted escapes (e.g. \n or \t) in the pattern (to avoid escaping the `\` in them).

6502 Over a year ago

@Piotr: I know... but I think your answer would be more instructive with a raw string for two reasons: 1) in my experience the fact that Python leaves the backslash in the string if the escaped code is not valid is often surprising: open("c:\zer\woob.dat") works and open("c:\temp\buf.dat") doesn't 2) there are much more special chars than \t and \n and specifically '\b' is one of them and r'\b' is instead special for a regexp. By using that fancy r'(\d+)' may be the reader will take the time to read about the escaping problem...

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