从末端找到字符串中的第一匹匹配

发布于 2025-02-13 04:12:10 字数 641 浏览 1 评论 0原文

我有一个字符串

Manager of Medical Threat Devlop at Micro

，我想找到，的的任何单词。在这里，我想获得['micro']（在 word的最后一个之后，在字符串的末尾）。

当前代码

如果我应用r'（？：for | ot | of）\ s+（。*）'我会得到不正确的[ micro']。

更多示例：

canno - ＆gt; CANNO
Medicalof threant devlop的经理 - ＆gt; CANNO
Medical For Thraist DevLop的经理 - ＆gt; Canno
threat devlop在Canno Matt - ＆gt; Canno Matt

原文

I have a string

Manager of Medical Threat Devlop at Micro

I want to find any words that go after at, for, of. Here, I want to get the ['Micro'] (that is at the end of string, after the last at word).

Current code

If I apply r'(?:for|at|of)\s+(.*)' I will get incorrect ['Medical Threat Devlop at Micro'].

More examples:

Manager of Medical Threat Devlop at Canno -> Canno
Manager of Medicalof Threat Devlop of Canno -> Canno
Manager of Medicalfor Threat Devlop for Canno -> Canno
Threat Devlop at Canno Matt -> Canno Matt

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一念一轮回 2025-02-20 04:12:10

您可以使用

re.findall(r'.*\b(?:for|at|of)\s+(.*)', text)

regex demo 。 详细信息：

。* - 除线断裂字符以外的任何零或更多字符，尽可能多的
\ b - 一个Word Boundare
（？：for | at | of） - for ， at 或的
\ s+ -一个或多个空格
（。*） - 第1组：除折扣以外的任何零或更多字符，尽可能多。

另一个将获得相同结果的正则是

re.findall(r'\b(?:for|at|of)\s+((?:(?!\b(?:for|at|of)\b).)*)
详细信息：

  \ b   -  word boundard boundary
 （？for | AT | of）  -  对于， 或 
  \ s+  - 一个或多个whitepsepaces
 （（？：（？） ，或的    的总体 yat a 尽可能多的事件， 整个单词char sequence
  $ 代码>  - 字符串的结尾。

注意，您还可以使用 re.Search ，因为您期望一个匹配：
match = re.search(r'.*\b(?:for|at|of)\s+(.*)', text)
if match:
    print(match.group(1))

, text)

详细信息：

\ b - word boundard boundary
（？for | AT | of） - 对于，或
\ s+ - 一个或多个whitepsepaces
（（？：（？），或的的总体 yat a 尽可能多的事件，整个单词char sequence
$ 代码> - 字符串的结尾。

注意，您还可以使用re.Search，因为您期望一个匹配：

You can use

re.findall(r'.*\b(?:for|at|of)\s+(.*)', text)

See the regex demo. Details:

.* - any zero or more chars other than line break chars, as many as possible
\b - a word boundary
(?:for|at|of) - for, at or of
\s+ - one or more whitespaces
(.*) - Group 1: any zero or more chars other than line break chars, as many as possible.

Another regex that will fetch the same results is

re.findall(r'\b(?:for|at|of)\s+((?:(?!\b(?:for|at|of)\b).)*)
Details:

\b - a word boundary
(?:for|at|of) - for, at or of
\s+ - one or more whitespaces
((?:(?!\b(?:for|at|of)\b).)*) - Group 1: any char, other than a line break char, zero or more but as many as possible, occurrences, that does not start a for, at or of as a whole word char sequence
$ - end of string.

Note you can also use re.search since you expect a single match:
match = re.search(r'.*\b(?:for|at|of)\s+(.*)', text)
if match:
    print(match.group(1))

, text)

Details:

\b - a word boundary
(?:for|at|of) - for, at or of
\s+ - one or more whitespaces
((?:(?!\b(?:for|at|of)\b).)*) - Group 1: any char, other than a line break char, zero or more but as many as possible, occurrences, that does not start a for, at or of as a whole word char sequence
$ - end of string.

Note you can also use re.search since you expect a single match:

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离旧人 2025-02-20 04:12:10

尝试此re.split可以使用此操作。

您的问题尚不完全清楚，给出了更多的输入和输出示例。

import re
s = 'Manager of Medical Threat Devlop at Micro'
s = re.split(r'at |for |of ',s)[-1:]
print(s)

输出

                 IN                         :  OUTPUT
'Manager of Medical Threat Devlop at Micro' : ['Micro']
'Threat Devlop at Canno Matt'               : ['Canno Matt']

还有另一种方法可以执行此操作（使用re.finditer）。

import re
string = 'Threat Devlop at Canno Matt'
s = re.finditer(r'(at | for | of )',string,)
last_index = list(s)[-1].end()
print(string[last_index:])

我在re中都不擅长。（但是我明白了）

是的。


import re
string = 'Threat Devlop at Canno of Matjkasa'
s = re.findall(r'.*(?:at|for|of)\s+', string)

print(string.replace(*s,''))

Try this re.split would work this.

Your question is not fully clear give some more input and output examples.

import re
s = 'Manager of Medical Threat Devlop at Micro'
s = re.split(r'at |for |of ',s)[-1:]
print(s)

OUTPUT

                 IN                         :  OUTPUT
'Manager of Medical Threat Devlop at Micro' : ['Micro']
'Threat Devlop at Canno Matt'               : ['Canno Matt']

THERE IS ANOTHER METHOD TO DO THIS (USING re.finditer).

import re
string = 'Threat Devlop at Canno Matt'
s = re.finditer(r'(at | for | of )',string,)
last_index = list(s)[-1].end()
print(string[last_index:])

I am not good in re at all.(But I get it)

Yeah there is another to do this.(Using re.findall)


import re
string = 'Threat Devlop at Canno of Matjkasa'
s = re.findall(r'.*(?:at|for|of)\s+', string)

print(string.replace(*s,''))

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转瞬即逝 2025-02-20 04:12:10

如果您想用正则表达式执行此操作，那么这就是这样做的方法。

用空字符串替换以下等级的匹配：

.*\b(?:for|at|of)\b\s?

这将匹配：

*：任何字符（从本质上，此模式都尽可能匹配
）对于| at | of）\ b ：边界符号之间的热词
\ s？：可选空间

检查演示在这里

If you want to do it with a regex, then here's the way to do it.

Replace matches of the following regex with the empty string:

.*\b(?:for|at|of)\b\s?

This will match:

.*: any character (by its nature, this pattern will match as most characters as possible)
\b(?:for|at|of)\b: your hotwords between boundary symbols
\s?: an optional space

Check the demo here

回复收藏 0 原文

~没有更多了~