为什么我不能用这个python正则表达式摆脱L？

Question

I'm trying to get rid of the Ls at the ends of integers with a regular expression in python:

import re
s = '3535L sadf ddsf df 23L 2323L'
s = re.sub(r'\w(\d+)L\w', '\1', s)

However, this regex doesn't even change the string. I've also tried s = re.sub(r'\\w\\d+(L)\\w', '', s) since I thought that maybe the L could be captured and deleted, but that didn't work either.

Answer 1

\\w = [a-zA-Z0-9_]

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In other words, \\w does not include whitespace characters. Each L is at the end of the word and therefore doesn't have any "word characters" following it. Perhaps you were looking for word boundaries ?

re.sub(r'\b(\d+)L\b', '\1', s)

Demo

Answer 2

I'm not sure what you're trying to do with those \\w s in the first place, but to match a string of digits followed by an L , just use \\d+L , and to remove the L you just need to put the \\d+ part in a capture group so you can sub it for the whole thing:

>>> s = '3535L sadf ddsf df 23L 2323L'
>>> re.sub(r'(\d+)L', r'\1', s)
'3535 sadf ddsf df 23 2323'

Here's the regex in action:

(\d+)L

Debuggex Demo

Of course this will also convert, eg, 123LBQ into 123BQ , but I don't see anything in your examples or in your description of the problem that indicates that this is possible, or which possible result you want for that, so…

Answer 3

You can use look behind assertion

>>> s = '3535L sadf ddsf df 23L 2323L'
>>> s = re.sub(r'\w(?<=\d)L\b', '', s)
>>> s
'353 sadf ddsf df 2 232'

(?<=\\d)L asserts that the L is presceded by a digit, in which case replace it with null ''

Answer 4

Try this: re.sub(r'(?<=\\d)L', '\\1', s)

This uses a lookbehind to find a digit followed by an "L".

Answer 5

Why not use a - IMO more readable - generator expression ?

>>> s = '3535L sadf ddsf df 23L 2323L'
>>> ' '.join(x.rstrip('L') if x[-1:] =='L' and  x[:-1].isdigit() else x for x in s.split())
'3535 sadf ddsf df 23 2323'