Question

使用re.findall，我想提取分配给每个PCR的值。

>>> z
'PCR-09: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 \r\nPCR-10: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 \r\nPCR-11: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 \r\nPCR-12: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 \r\nPCR-13: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 \r\nPCR-14: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 \r\nPCR-15: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 \r\nPCR-16: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 \r\n

>>> print z
PCR-09: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 
PCR-10: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 
PCR-11: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 
PCR-12: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 
PCR-13: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 
PCR-14: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 
PCR-15: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 
PCR-16: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00

最初，我试过这个，但是有人可以指出正则表达式使用了什么问题吗？

>>> re.search('PCR-09:(.*?)', z).groups()
('',)

非贪婪的expr (.*?)是否应该匹配所有字符，直到找到换行符？

通过稍加修改的正则表达式，我得到了所需的结果：

>>> re.search('PCR-09:(.*?)\s\r\n', z).groups()
(' 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00',)

在同一行上，这不起作用：

>>> re.findall(r'(PCR-\d+):(.*?)', z)
[('PCR-09', ''), ('PCR-10', ''), ('PCR-11', ''), ('PCR-12', ''), ('PCR-13', ''), ('PCR-14', ''), ('PCR-15', ''), ('PCR-16', ''),

但这样做：

>>> re.findall(r'(PCR-\d+):(.*?)\s\r\n', z,re.DOTALL)
[('PCR-09', ' 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00'), ('PCR-10', ' 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00'), ('PCR-11', ' 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00'), ('PCR-12', ' 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00'), ('PCR-13', ' 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00'), ('PCR-14', ' 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00'), ('PCR-15', ' 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00'), ('PCR-16', ' 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00'),

希望有人能解释我的方法有什么问题。

由于

Answer 1

r'PCR-09:(.*?)'与您的预期不符的原因是非贪婪的正则表达式一旦有效就会停止。

因此(.*?)可以匹配''，因此正则表达式会立即停止。

相比之下，r'(PCR-\d+):(.*?)\s\r\n'非贪婪，但因为它需要找到`\ s \ r \ n'，它会强制扩展工作。

我建议使用贪婪的正则表达式，其中只包含您希望找到的字符：r'(PCR-\d+):([0-9 ]*)'。

Answer 2

模式PCR-09:(.*?)告诉Python在PCR-09:之后非贪婪地匹配零个或多个字符。所以，它恰好这样做并匹配零个字符。

你需要让你的正则表达式贪心才能匹配换行符之前的所有内容：

>>> re.search('PCR-09:(.*)', z).groups()
(' 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 \r',)
>>>

请注意，您的PCR-09:(.*?)\s\r\n模式有效，因为它告诉Python在PCR-09:和之后获得零个或多个字符 \s\r\n。换句话说，获取它们之间的所有内容。

Answer 3

尝试使用：split

[ x.split(':') for x in z.split('\r\n')]

输出：

[['PCR-09', ' 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 '], ['PCR-10', ' 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 '], ['PCR-11', ' 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 '], ['PCR-12', ' 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 '], ['PCR-13', ' 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 '], ['PCR-14', ' 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 '], ['PCR-15', ' 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 '], ['PCR-16', ' 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 '], ['']]

使用正则表达式

re.findall('(PCR-\d+)(.*)',z)

用非贪婪的正则表达式捕获部分文本

3 个答案: