通过re.split
。
>>> s = 'Arrived at location NEWARK NJ US'
>>> re.split(r'(?<=[a-z])\s+(?=[^a-z]*$)', s)
['Arrived at location', 'NEWARK NJ US']
(?<=[a-z])
为小写字母后,看起来。
\s+
,那么符合下列一个或多个空格字符,
- 只有
- 它后面
(?=..)
[^a-z]*
任何字符,但一个小写字母,零次或多次的不是。
$
紧接着一行结束。所以它与紧接在最后一个小写字母后面的空格相匹配。
OR
>>> s = "Arrived at location NEWARK NJ US"
>>> part1, part2 = re.findall(r'(.*[a-z])\s+(.*)', s)[0]
>>> part1
'Arrived at location'
>>> part2
'NEWARK NJ US'
OR
>>> s = "Arrived at location NEWARK NJ US"
>>> part1, part2 = re.search(r'(.*[a-z])\s+(.*)', s).groups()
>>> part1
'Arrived at location'
>>> part2
'NEWARK NJ US'
OR
re.split将返回定界符如果分隔符是一个捕获GR内本牛津大学出版社。
>>> part1, part2 = [i for i in re.split(r'(.*[a-z])\s+', s) if not i == '']
>>> part1
'Arrived at location'
>>> part2
'NEWARK NJ US'
可读在旁观者:-) – foosion 2015-03-13 11:32:53
@foosion的眼睛:那是正则表达式为您服务。 – 2015-03-13 11:34:20
@MartijnPieters也许我应该在Q中说“简单的正则表达式”,但你可能会给出相同的评论。 :-) – foosion 2015-03-13 11:36:32