Python中的非贪婪正则表达式

鉴于案文：

'Adf Adf asdf asdf asfdfhttps://.com/abcabcabc kdfja ladsjfladsjf LADSJF ladsjfl adsfadf adf asdf asdf asfdfhttps://.com/abcabcabc\n kdfja ladsjfladsjf LADSJF ladsjfl adsfhttps://.com/djflkajdsfl\n\n djldjfld djfladjf ldfdjlkfj ldfj'

如何匹配表单中的任何urlhttps://.com/subdir[直到碰到空格或新行、逗号或句号]

尝试：

re.findall('http.*',s) 
['https://<somepage>.com/abcabcabc kdfja ladsjfladsjf ladksjf ladsjfl adsfadf adf asdf asdf asfdf https://<somepage>.com/abcabcabc', 'https://<somepage>.com/djflkajdsfl']

re.findall('http.* ',s) 
['https://<somepage>.com/abcabcabc kdfja ladsjfladsjf ladksjf ladsjfl adsfadf adf asdf asdf asfdf ']

re.findall('http.* ?',s) 
['https://<somepage>.com/abcabcabc kdfja ladsjfladsjf ladksjf ladsjfl adsfadf adf asdf asdf asfdf https://<somepage>.com/abcabcabc', 'https://<somepage>.com/djflkajdsfl']

re.findall('http.* {1}?',s) 
['https://<somepage>.com/abcabcabc kdfja ladsjfladsjf ladksjf ladsjfl adsfadf adf asdf asdf asfdf ']

re.findall('http.* +?',s) 
['https://<somepage>.com/abcabcabc kdfja ladsjfladsjf ladksjf ladsjfl adsfadf adf asdf asdf asfdf ']

re.findall('http.*[^ \n]',s) 
['https://<somepage>.com/abcabcabc kdfja ladsjfladsjf ladksjf ladsjfl adsfadf adf asdf asdf asfdf
https://<somepage>.com/abcabcabc', 'https://<somepage>.com/djflkajdsfl']

re.findall('http.*[^ \\n]',s) ['https://<somepage>.com/abcabcabc kdfja ladsjfladsjf ladksjf ladsjfl adsfadf adf asdf asdf asfdf
https://<somepage>.com/abcabcabc', 'https://<somepage>.com/djflkajdsfl']

re.findall('http.*[^ \\\n]',s) ['https://<somepage>.com/abcabcabc kdfja ladsjfladsjf ladksjf ladsjfl adsfadf adf asdf asdf asfdf
https://<somepage>.com/abcabcabc', 'https://<somepage>.com/djflkajdsfl']

re.findall('http.* *?',s) ['https://imgur.com/abcabcabc kdfja ladsjfladsjf ladksjf ladsjfl adsfadf adf asdf asdf asfdf https://imgur.com/abcabcabc', 'https://somepage.com/djflkajdsfl']

re.findall（'http.'，s）
['https://.com/abcabcabc kdfja ladsjfladsjf LADSJF ladsjfl adsfadf adf asdf asdf asfdfhttps://.com/abcabcabc', 'https://.com/djflkajdsfl']
关于findall（'http.*'，s）
['https://.com/abcabcabc kdfja ladsjfladsjf LADSJF ladsjfl adsfadf adf asdf asdf asfdf']
关于findall（'http.*？'，s）
['https://.com/abcabcabc kdfja ladsjfladsjf LADSJF ladsjfl adsfadf adf asdf asdf asfdfhttps://.com/abcabcabc', 'https://.com/djflkajdsfl']
关于findall（'http.*{1}？'，s）
['https://.com/abcabcabc kdfja ladsjfladsjf LADSJF ladsjfl adsfadf adf asdf asdf asfdf']
关于findall（'http.*+？'，s）
['https://.com/abcabcabc kdfja ladsjfladsjf LADSJF ladsjfl adsfadf adf asdf asdf asfdf']
关于findall（'http.*[^\n]'，s）
['https://.com/abcabcabc kdfja ladsjfladsjf LADSJF ladsjfl adsfadf adf asdf asdf asfdf
https://.com/abcabcabc', 'https://.com/djflkajdsfl']
关于findall（'http.*[^\\n]'，s）['https://.com/abcabcabc kdfja ladsjfladsjf LADSJF ladsjfl adsfadf adf asdf asdf asfdf
https://.com/abcabcabc', 'https://.com/djflkajdsfl']
关于findall（'http.*[^\\\n]'，s）['https://.com/abcabcabc kdfja ladsjfladsjf LADSJF ladsjfl adsfadf adf asdf asdf asfdf
https://.com/abcabcabc', 'https://.com/djflkajdsfl']
关于findall（'http.*？'，s）['https://imgur.com/abcabcabc kdfja ladsjfladsjf LADSJF ladsjfl adsfadf adf asdf asdf asfdfhttps://imgur.com/abcabcabc', 'https://somepage.com/djflkajdsfl']

尝试以下操作：

re.findall('http[^ \n,]*',s)

re.findall('http[^ \\n,]*/[^ \\n,\.]*',s)

您可以查看此操作

因为您使用的是

，所以懒惰（

*？

）和贪婪（

*

）都不适合您。懒惰将只移动一个角色然后停止，而贪婪将继续直到结束

相反，您希望指定不需要的字符。（

[^\n，]

）并对其进行搜索。由于您希望在这些字符的第一个实例处停止，因此需要使用贪婪搜索来完成此操作

由于

字符在URL中是合法的，因此很难基于此限制字符串。由于您始终希望包含子目录，因此可以通过以下操作完成此操作：

re.findall('http[^ \n,]*',s)

re.findall('http[^ \\n,]*/[^ \\n,\.]*',s)

---

您可以实际查看它。

第一个示例中的问题不是regexp匹配的空格太多；它在空格前匹配了太多的字母。因此，不要将“非贪婪的”

？

修饰符放在空格之后，将其放在

*

之后，因为这是当前匹配过多的内容

py3.7 >>> re.findall('http.*? ', s)
['https://.com/abcabcabc ']

---

另一方面，

[^\n]

不是任何类型的修饰符–它本身就是一个完全匹配的表达式。因此，将其放在现有表达式之后不会使其匹配度降低；现在有两个匹配表达式，它们一起匹配更多

您必须使用它来代替匹配过多的表达式，即代替

：

py3.7 >>> re.findall('http[^ \n]*', s)
['https://.com/abcabcabc', 'https://.com/abcabcabc', 'https://.com/djflkajdsfl']