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在字符串python中的字符之前和之后获取特定的子字符串

[英]get a specific substring after and before a character in a string python

I want to split time from decription strings. 我想从解密字符串中分出时间。 desc can have this kind of strings: desc可以具有以下字符串:

desc = '08:59 Hudh aes ....' 
desc = '19:59Aksi jeh....' 
desc = 'just letters without time'
desc = '21.19:55Aksi hsue....'
desc = '256.08:59Aksi mane....'

time contains 10 first letters of description I want to find : and then to take two numbers before it and after it, so i can split the time time包含我要查找的10个描述的首字母:然后在它之前和之后取两个数字,因此我可以分配时间

time = ''.join(filter(lambda x: x, desc[0:10].replace(" " , "")))
print  "Time:" , time , time.rsplit(':', 1)[0]

time.rsplit(':', 1)[0] returns all numbers before : time.rsplit(':', 1)[0]返回之前的所有数字:

time.rsplit(':', 1)[0] returns all numbers after : time.rsplit(':', 1)[0]返回以下所有数字:

How to define to split only two numbers after and before : ?Is this a good way? 如何界定之后且只有两个数字拆分: ?这是一个好办法吗? Is better idea to use regex, i tried but it's a little complicated? 我尝试过使用regex是更好的主意,但这有点复杂吗?

You can try re.finditer() 您可以尝试re.finditer()

 test = """desc = '08:59 Hudh aes ....' 
 desc = '19:59Aksi jeh....' 
 desc = 'just letters without time'
 desc = '21.19:55Aksi hsue....'
 desc = '256.08:59Aksi mane....'
 """
 pattern = r"\d{2}:\d{2}"
 for m in re.finditer(pattern,test):
    print m.group()

Output will be: 输出将是:

08:59 08:59
19:59 19:59
19:55 19:55
08:59 08:59

Then from this output you can split the hours and minutes easily 然后,从此输出中,您可以轻松地拆分小时和分钟

How to define to split only two numbers after and before : 如何定义在之后和之前仅拆分两个数字:

With regex; 使用正则表达式; the following regex pattern matches exactly 2 digits, followed by a colon, followed by exactly two digits (which is literally what you ask for): 以下正则表达式模式正好匹配2位数字,后跟一个冒号,再恰好两位数(这实际上是您要的):

import re

desc = '21.19:55Aksi hsue....'
m = re.search(r'(\d{2}):(\d{2})', desc)
if m:
    hour, min = m.groups()
else:
    # no time in desc string
    pass
print 'hour = {}, min = {}'.format(hour, min)

Output 产量

hour = 19, min = 55

您可以re.findall()

re.findall(r'([\d.]+):([\d.]+)',desc)

If the strings are separated each other you can use a re.search method like this: 如果字符串彼此分开,则可以使用如下的re.search方法:

import re
[hour, min, desc] = re.search(r'(?:(?:\d+\.)?(\d{2}):(\d{2}))?[ \t]*(.*)',desc).groups()

# --
# desc  = '08:59 Hudh aes ....'
# hour -> '08'
# min  -> '59'
# desc -> 'Hudh aes ....', the space before the description are stripped too
# --
# desc  = '19:59Aksi jeh....'
# hour -> '19'
# min  -> '59'
# desc -> 'Aksi jeh....'
# --
# desc  = 'just letters without time'
# hour -> None
# min  -> None
# desc -> 'just letters without time'
# --
# desc = '256.08:59Aksi mane....'
# hour -> '08'
# min  -> '59'
# desc -> 'Aksi mane....'

Online regex demo 在线正则表达式演示

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