从单词的开头和结尾删除非字母字符

I need to remove all non-letter characters from the beginning and from the end of a word, but keep them if they appear between two letters.

我需要从单词的开头和结尾删除所有非字母字符,但如果它们出现在两个字母之间,请保留它们。

For example:

'123foo456' --> 'foo'
'2foo1c#BAR' --> 'foo1c#BAR'

I tried using re.sub(), but I couldn't write the regex.

我尝试使用re.sub(),但我无法编写正则表达式。

6 个解决方案

#1

like this?

re.sub('^[^a-zA-z]*|[^a-zA-Z]*$','',s)

s is the input string.

s是输入字符串。

#2

You could use str.strip for this:

您可以使用str.strip:

In [1]: import string

In [4]: '123foo456'.strip(string.digits)
Out[4]: 'foo'

In [5]: '2foo1c#BAR'.strip(string.digits)
Out[5]: 'foo1c#BAR'

As Matt points out in the comments (thanks, Matt), this removes digits only. To remove any non-letter character,

正如Matt在评论中指出的那样(感谢Matt),这只会删除数字。要删除任何非字母字符,

Define what you mean by a non-letter:

用非字母定义你的意思:

In [22]: allchars = string.maketrans('', '')

In [23]: nonletter = allchars.translate(allchars, string.letters)

and then strip:

然后剥离:

In [18]: '2foo1c#BAR'.strip(nonletter)
Out[18]: 'foo1c#BAR'

#3

With your two examples, I was able to create a regex using Python's non-greedy syntax as described here. I broke up the input into three parts: non-letters, exclusively letters, then non-letters until the end. Here's a test run:

通过两个示例,我可以使用Python的非贪婪语法创建一个正则表达式,如此处所述。我将输入分为三部分:非字母,专用字母,然后非字母直到结束。这是一个测试运行:

1:[123]   2:[foo]   3:[456]
1:[2]   2:[foo1c#BAR]   3:[]

Here's the regular expression:

这是正则表达式:

^([^A-Za-z]*)(.*?)([^A-Za-z]*)$

And mo.group(2) what you want, where mo is the MatchObject.

并且mo.group(2)你想要什么,其中mo是MatchObject。

#4

To be unicode compatible:

要兼容unicode:

^\PL+|\PL+$

\PL stands for for not a letter

\ PL代表不是一封信

#5

Try this:

re.sub(r'^[^a-zA-Z]*(.*?)[^a-zA-Z]*$', '\1', string);

The round brackets capture everything between non-letter strings at the beginning and end of the string. The ? makes sure that the . does not capture any non-letter strings at the end, too. The replacement then simply prints the captured group.

圆括号捕获字符串开头和结尾处的非字母字符串之间的所有内容。的?确保。也不会在最后捕获任何非字母字符串。然后替换只是打印捕获的组。

#6

result = re.sub('(.*?)([a-z].*[a-z])(.*)', '\\2', '23WERT#3T67', flags=re.IGNORECASE)

result = re.sub('(。*?)([a-z]。* [a-z])(。*)','\\ 2','23WERT#3T67',flags = re.IGNORECASE)

#1