Python 提取特定字符后的文本
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Extract text after specific character
提问by user2990084
I need to extract the word after the @
我需要提取后面的单词 @
How can I do that? What I am trying:
我怎样才能做到这一点？我在尝试什么：
text="Hello there @bob !"
user=text[text.find("@")+1:]
print user
output:
输出：
bob !
But the correct output should be:
但正确的输出应该是：
bob
采纳答案by Shashank
A regex solution for fun:
一个有趣的正则表达式解决方案：
>>> import re
>>> re.findall(r'@(\w+)', '@Hello there @bob @!')
['Hello', 'bob']
>>> re.findall(r'@(\w+)', 'Hello there bob !')
[]
>>> (re.findall(r'@(\w+)', 'Hello there @bob !') or None,)[0]
'bob'
>>> print (re.findall(r'@(\w+)', 'Hello there bob !') or None,)[0]
None
The regex above will pick up patterns of one or more alphanumeric characters following an '@' character until a non-alphanumeric character is found.
上面的正则表达式将选择一个或多个字母数字字符的模式，直到找到一个非字母数字字符。
Here's a regex solution to match one or more non-whitespace characters if you want to capture a broader range of substrings:
如果您想捕获更广泛的子字符串，这是匹配一个或多个非空白字符的正则表达式解决方案：
>>> re.findall(r'@(\S+?)', '@Hello there @bob @!')
['Hello', 'bob', '!']
Note that when the above regex encounters a string like @xyz@abcit will capture xyz@abcin one result instead of xyzand abcseparately. To fix that, you can use the negated \scharacter class while also negating @characters:
需要注意的是，当上述正则表达式遇到像绳子@xyz@abc将捕获xyz@abc的一个结果，而不是xyz和abc独立。要解决这个问题，您可以使用否定\s字符类，同时也否定@字符：
>>> re.findall(r'@([^\s@]+)', '@xyz@abc some other stuff')
['xyz', 'abc']
And here's a regex solution to match one or more alphabet characters only in case you don't want any numbers or anything else:
这是一个正则表达式解决方案，仅在您不需要任何数字或其他任何内容时才匹配一个或多个字母字符：
>>> re.findall(r'@([A-Za-z]+)', '@Hello there @bobv2.0 @!')
['Hello', 'bobv']
回答by ODiogoSilva
So you want the word starting after @ up to a whitespace?
所以你想要从@ 开始的单词到一个空格？
user=text[text.find("@")+1:].split()[0]
print(user)
bob
EDIT: as @bgstech note, in cases where the string does not have a "@", make a check before:
编辑：作为@bgstech 注意，在字符串没有“@”的情况下，请先检查：
if "@" in text:
user=text[text.find("@")+1:].split()[0]
else:
user="something_else_appropriate"
