[英]multiline regex pattern match
I have the following multiline(?) string that I get from the output of a process. 我有以下从过程输出中获取的multiline(?)字符串。
04/18@14:22 - RESPONSE from 192.68.10.1 :
04/18 @ 14:22-从192.68.10.1起响应:
04/18@14:22 - RESPONSE from 192.68.10.1 :04/18 @ 14:22-从192.68.10.1起响应:
TSB1 File Name: OCAP_TSB_76 04/18@14:22 - RESPONSE from 192.68.10.1 : TSB1 Duration: 1752 seconds 04/18@14:22 - RESPONSE from 192.68.10.1 : TSB1 Bit Rate: 3669 kbps 04/18@14:22 - RESPONSE from 192.68.10.1 :TSB1文件名:OCAP_TSB_76 04/18 @ 14:22-从192.68.10.1起响应:TSB1持续时间:1752秒04/18 @ 14:22-从192.68.10.1起响应:TSB1比特率:3669 kbps 04/18 @ 14: 22-从192.68.10.1起的响应:
04/18@14:22 - RESPONSE from 192.68.10.1 :04/18 @ 14:22-从192.68.10.1起响应:
TSB2 File Name: OCAP_TSB_80 04/18@14:22 - RESPONSE from 192.68.10.1 : TSB2 Duration: 56 seconds 04/18@14:22 - RESPONSE from 192.68.10.1 : TSB2 Bit Rate: 3675 kbps 04/18@14:22 - RESPONSE from 192.68.10.1 :TSB2文件名:OCAP_TSB_80 04/18 @ 14:22-从192.68.10.1起响应:TSB2持续时间:56秒04/18 @ 14:22-从192.68.10.1起响应:TSB2比特率:3675 kbps 04/18 @ 14: 22-从192.68.10.1起的响应:
I am trying to extract just the values in 'seconds' and 'kbps'. 我正在尝试仅提取“秒”和“ kbps”中的值。
This is what I have so far. 到目前为止,这就是我所拥有的。
>>> cpat = re.compile(r"\.*RESPONSE from[^:]+:\s*TSB[\d] Duration:\s*(\d+) seconds\.*?RESPONSE from[^:]+:\s*TSB[\d] Bit Rate:\s*(\d+) kbps", re.DOTALL)
>>> m = re.findall(cpat,txt)
>>> m
[]
I find matches if I break the regex into separate parts. 如果将正则表达式分解为单独的部分,我会找到匹配项。 But, I am looking to find matches like below
但是,我正在寻找符合以下条件的比赛
m [(1752,3669),(52,3675)]
m [(1752,3669),(52,3675)]
Thanks a lot! 非常感谢!
re.compile(r"\.*RESPONSE from[^:]+:\s*TSB[\d] Duration:\s*(\d+) seconds\.*?RESPONSE from[^:]+:\s*TSB[\d] Bit Rate:\s*(\d+) kbps", re.DOTALL)
^
I think that this dot was not meant to be escaped (because otherwise, it will be matching literal dots instead of any character. Try with: 我认为该点不是要转义的(否则,它将匹配文字点而不是任何字符。请尝试:
re.compile(r"\.*RESPONSE from[^:]+:\s*TSB[\d] Duration:\s*(\d+) seconds.*?RESPONSE from[^:]+:\s*TSB[\d] Bit Rate:\s*(\d+) kbps", re.DOTALL)
Also, there are some unnecessary parts in your regex that you can remove and still ensure the matches you're looking for. 另外,您的正则表达式中有一些不必要的部分可以删除,但仍然可以确保找到所需的匹配项。 I removed them in the below regex:
我在下面的正则表达式中删除了它们:
re.compile(r"RESPONSE from[^:]+:\s*TSB\d Duration:\s*(\d+) seconds.*?RESPONSE from[^:]+:\s*TSB\d Bit Rate:\s*(\d+) kbps", re.DOTALL)
Namely: 即:
.*
at the start of the regex with re.findall
. re.findall
在正则表达式的开头不需要.*
。 \\d
within square brackets if it is alone. \\d
无需将\\d
放在方括号中。 This code gives what you want: 这段代码给出了您想要的:
import re 汇入
data = '''
04/18@14:22 - RESPONSE from 192.68.10.1 :
04/18@14:22 - RESPONSE from 192.68.10.1 :
TSB1 File Name: OCAP_TSB_76 04/18@14:22 - RESPONSE from 192.68.10.1 : TSB1 Duration: 1752 seconds 04/18@14:22 - RESPONSE from 192.68.10.1 : TSB1 Bit Rate: 3669 kbps 04/18@14:22 - RESPONSE from 192.68.10.1 :
04/18@14:22 - RESPONSE from 192.68.10.1 :
TSB2 File Name: OCAP_TSB_80 04/18@14:22 - RESPONSE from 192.68.10.1 : TSB2 Duration: 56 seconds 04/18@14:22 - RESPONSE from 192.68.10.1 : TSB2 Bit Rate: 3675 kbps 04/18@14:22 - RESPONSE from 192.68.10.1 :
'''
output = []
block_pattern = re.compile(r'(\d+\/\d+@\d+:\d+ - RESPONSE.*?)(.*)')
seconds_speed_pattern = re.compile(r'TSB.*Duration:(.*)seconds.*TSB.*Bit Rate:(.*)kbps')
blocks = re.findall(block_pattern, data)
for block in blocks:
ss_data = re.findall(seconds_speed_pattern, block[1])
if ss_data:
output.append(ss_data[0])
print output
This prints 此打印
[(' 1752 ', ' 3669 '), (' 56 ', ' 3675 ')]
In order to convert those values from str
to int
s just do: 为了将这些值从
str
转换为int
只需执行以下操作:
output = [(int(a.strip()), int(b.strip())) for a, b in output]
This gives: 这给出:
[(1752, 3669), (56, 3675)]
result = re.findall(r"(?sim)Duration: (\d+).*?Rate: (\d+)", subject)
Options: dot matches newline; case insensitive; ^ and $ match at line breaks
Match the characters “Duration: ” literally «Duration: »
Match the regular expression below and capture its match into backreference number 1 «(\d+)»
Match a single digit 0..9 «\d+»
Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
Match any single character «.*?»
Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
Match the characters “Rate: ” literally «Rate: »
Match the regular expression below and capture its match into backreference number 2 «(\d+)»
Match a single digit 0..9 «\d+»
Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
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