正则表达式,如何在12/24小时时间戳中去除所有非字母数字字符但保留冒号?
我有一个这样的字符串:
Today, 3:30pm - Group Meeting to discuss "big idea"
我该如何构建一个正则表达式,让它解析后返回:
Today 3:30pm Group Meeting to discuss big idea
我希望它能去掉所有不是字母或数字的字符,除了那些出现在12小时或24小时时间格式中的字符。
6 个回答
1
我猜你是想保留空格的,这段代码是用Python写的,但它是PCRE格式,所以应该可以在其他地方使用。
import re
x = u'Today, 3:30pm - Group Meeting to discuss "big idea"'
re.sub(r'[^a-zA-Z0-9: ]', '', x)
输出结果是:'今天 3:30pm 组会讨论大点子'
如果你想要一个稍微干净一点的结果(没有重复的空格)
import re
x = u'Today, 3:30pm - Group Meeting to discuss "big idea"'
tmp = re.sub(r'[^a-zA-Z0-9: ]', '', x)
re.sub(r'[ ]+', ' ', tmp)
输出结果是:'今天 3:30pm 组会讨论大点子'
2
这是关于Python编程的内容。
import string
punct=string.punctuation
s='Today, 3:30pm - Group Meeting:am to discuss "big idea" by our madam'
for item in s.split():
try:
t=time.strptime(item,"%H:%M%p")
except:
item=''.join([ i for i in item if i not in punct])
else:
item=item
print item,
这是程序的输出结果。
$ ./python.py
Today 3:30pm Group Meetingam to discuss big idea by our madam
# change to s='Today, 15:30pm - Group 1,2,3 Meeting to di4sc::uss3: 2:3:4 "big idea" on 03:33pm or 16:47 is also good'
$ ./python.py
Today 15:30pm Group 123 Meeting to di4scuss3 234 big idea on 03:33pm or 1647 is also good
注意:这个方法应该改进一下,只在需要的时候检查时间是否有效(通过设置条件),不过我现在就先这样写了。
8
# this: D:DD, DD:DDam/pm 12/24 hr
re = r':(?=..(?<!\d:\d\d))|[^a-zA-Z0-9 ](?<!:)'
冒号前面必须至少有一个数字,后面至少要有两个数字,这样才能表示时间。其他的冒号都会被当作普通文本中的冒号。
它是怎么工作的
: // match a colon
(?=.. // match but not capture two chars
(?<! // start a negative look-behind group (if it matches, the whole fails)
\d:\d\d // time stamp
) // end neg. look behind
) // end non-capture two chars
| // or
[^a-zA-Z0-9 ] // match anything not digits or letters
(?<!:) // that isn't a colon
然后当这个规则应用到这段搞笑的文字上:
Today, 3:30pm - Group 1,2,3 Meeting to di4sc::uss3: 2:3:4 "big idea" on 03:33pm or 16:47 is also good
...就会把它变成:
Today, 3:30pm Group 123 Meeting to di4scuss3 234 big idea on 03:33pm or 16:47 is also good