正则表达式:匹配开始或空格

使用“或” | “运营商:

>>> re.sub(r'(^|\W)GBP([\W\d])', u'\g<1>£\g<2>', text)
u'\xa3 5 Off when you spend \xa375.00'

\b 是单词边界,可以是空白、行的开头或非字母数字符号( \bGBP\b ).

word boundary (字符串的开头已经是),GBP之后是一个数值或字边界:

re.sub(u'\bGBP(?=\b|\d)', u'£', text)

lookahead . 是否足够包容?

左侧空白边界(字符串中的一个位置,它要么是字符串开头,要么在空白字符后面)可以用

(?<!\S)   # A negative lookbehind requiring no non-whitespace char immediately to the left of the current position
(?<=\s|^) # A positive lookbehind requiring a whitespace or start of string immediately to the left of the current position
(?:\s|^)  # A non-capturing group matching either a whitespace or start of string 
(\s|^)    # A capturing group matching either a whitespace or start of string

见 regex demo . Python 3 demo :

import re
rx = r'(?<!\S)GBP([\W\d])'
text = 'GBP 5 Off when you spend GBP75.00'
print( re.sub(rx, r'£\1', text) )
# => £ 5 Off when you spend £75.00

注意:您可以使用 \1 \g<1> 在替换模式中,因为当后面没有数字时,不需要明确的反向引用。

(?!\S)   # A negative lookahead requiring no non-whitespace char immediately to the right of the current position
(?=\s|$) # A positive lookahead requiring a whitespace or end of string immediately to the right of the current position
(?:\s|$)  # A non-capturing group matching either a whitespace or end of string 
(\s|$)    # A capturing group matching either a whitespace or end of string

我想你在找我 '(^|\W)GBP([\W\d])'

是的,为什么不呢?

re.sub(u'^\W*GBP...

匹配字符串的开头,0个或更多空格,然后是GBP。。。

:哦,我想你需要替换,用 | :

re.sub(u'(^|\W)GBP...

它在Perl中工作:

$text = 'GBP 5 off when you spend GBP75';
$text =~ s/(\W|^)GBP([\W\d])/$1\$$2/g;
printf "$text\n";

$ 5 off when you spend $75

请注意,我规定匹配应该是全局的,以获得所有的匹配。