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[英]Regular expression to extract one word before a given word from a word list in Python
[英]Regular Expression return the word before a string - python
我正在編寫一個python腳本,想要匹配一個大文件中的所有Group對象名稱,原始數據的示例如下:
IT_PC (Group) -Host: 192.168.103.144 -Host: 192.168.103.145 -Network: 192.168.103.0 255.255.255.0 HR_PC (Group) -Host: 192.168.65.145 -Host: 192.168.62.146 -Host: 192.168.62.154
Finance_PC (Group) -Finance_PC_192.168.41.125
Testing_PC (Group) -Host: 192.168.129.1 -Host: 192.168.129.97 -Host: 192.168.59.81 -Host: 192.168.59.82
我所需的輸出應如下所示:
IT_PC (Group)
HR_PC (Group)
Finance_PC (Group)
Testing_PC (Group)
我正在嘗試使用以下正則表達式來匹配我所需的結果,但它僅返回第一個['IT_PC(Group)']。 對我有什么建議嗎,謝謝。
source = "IT_PC (Group) -Host: 192.168.103.144 -Host: 192.168.103.145 -Network: 192.168.103.0 255.255.255.0 HR_PC (Group) -Host: 192.168.65.145 -Host: 192.168.62.146 -Host: 192.168.62.154 Finance_PC (Group) -Finance_PC_192.168.41.125 Testing_PC (Group) -Host: 192.168.129.1 -Host: 192.168.129.97 -Host: 192.168.59.81 -Host: 192.168.59.82"
data = ".*? (?= \(group\))"
a = re.findall(data, source)
print a
我將使用\\w+(?= \\(Group\\))
表達式,該表達式將匹配一個或多個字母數字字符( A-Za-z0-9_
),后跟一個空格和一個(Group)
:
>>> re.findall(r"\w+(?= \(Group\))", source)
['IT_PC', 'HR_PC', 'Finance_PC', 'Testing_PC']
或者,您甚至可以更詳細地說明組對象名稱格式,並在下划線后要求一個或多個大寫字母:
>>> re.findall(r"\w+_[A-Z]+(?= \(Group\))", source)
['IT_PC', 'HR_PC', 'Finance_PC', 'Testing_PC']
或者,如果您還需要Group
:
>>> re.findall(r"\w+ \(Group\)", source)
['IT_PC (Group)', 'HR_PC (Group)', 'Finance_PC (Group)', 'Testing_PC (Group)']
>>> re.findall(r"\w+_[A-Z]+ \(Group\)", source)
['IT_PC (Group)', 'HR_PC (Group)', 'Finance_PC (Group)', 'Testing_PC (Group)']
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