[英]get a specific substring after and before a character in a string python
我想從解密字符串中分出時間。 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
包含我要查找的10個描述的首字母:
然后在它之前和之后取兩個數字,因此我可以分配時間
time = ''.join(filter(lambda x: x, desc[0:10].replace(" " , "")))
print "Time:" , time , time.rsplit(':', 1)[0]
time.rsplit(':', 1)[0]
返回之前的所有數字:
time.rsplit(':', 1)[0]
返回以下所有數字:
如何界定之后且只有兩個數字拆分:
?這是一個好辦法嗎? 我嘗試過使用regex是更好的主意,但這有點復雜嗎?
您可以嘗試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()
輸出將是:
08:59
19:59
19:55
08:59
然后,從此輸出中,您可以輕松地拆分小時和分鍾
如何定義在之后和之前僅拆分兩個數字:
使用正則表達式; 以下正則表達式模式正好匹配2位數字,后跟一個冒號,再恰好兩位數(這實際上是您要的):
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)
產量
hour = 19, min = 55
您可以re.findall()
:
re.findall(r'([\d.]+):([\d.]+)',desc)
如果字符串彼此分開,則可以使用如下的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....'
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