如何删除字符串中最后一个数字之后的所有内容（某些字符除外）

Question

This is a follow-up of this question.

There I learned how to remove all characters after the last number in a string; so I can turn

w = 'w123 o456 t789-- --'

into

w123 o456 t789

Now I might have strings like this:

w = 'w123 o456 (t789)'

In this case,

re.sub(r'\D+$', '', w)

would give me

w123 o456 (t789

So I have then actually two closely related questions:

1) How can I modify the command re.sub(r'\\D+$', '', w) in a way that certain characters are kept (eg parenthesis)?

2) How can I modify the command re.sub(r'\\D+$', '', w) so that only certain characters are removed (eg dashes and white spaces)?

EDIT

@Martin Bonner's answer gets very close but eg for

w='w123 -o456 t789--) --'

the command

 re.sub('[- ]+$', '', w)

gives me w123 -o456 t789--) but it should also get rid of the remaining dashes.

Answer 1

To keep certain characters ( and ) use:

re.sub('[^0-9()]+$', '', w)

to remove only certain characters from the end of the line:

re.sub('[- ]+$', '', w)

In square brackets, you can list the characters you want to match. If the first character is ^ then everything except the specified characters are matched. The only minor niggle is that - usually specifies a range (so we can specify, eg, all digits without having to list all 10 of them). That means that if we are going to specify - as one of the characters to match, it needs to go first . (If you want to specify ^ , then escape it with \\ and go back to raw strings.)

From the comment, I think you actually meant the second challenge to be "remove all the dashes and spaces from the string that lie between the last digit, and the end of the line". That may be possible with a regular expression, but somebody who comes back to maintain the code in three months time will hate you (and it may well be you). Just remember the Jamie Zawinski quote:

Some people, when confronted with a problem, think “I know, I'll use regular expressions.” Now they have two problems.

Answer 2

You may use another re.sub in the callback as the replacement pattern.

re.sub(r'\D+$', lambda m: re.sub(r'[^()]+','',m.group(0)), s)

Here, you match all symbols other than digits at the end of the string, pass that value to the callback, and all symbols other than ( and ) are removed from that value.

Answer 3

If there is always 3 groups of characters and each group start's with a single letter and has 3 digits after that, and only the last group might have brackets, this expression might be just what you need:

w = 'w123 o456 (t789)'
clean = re.sub(r'^.*(\w\d{3})[ -]+(\w\d{3})[ -]+(\(?\w\d{3}\)?).*$', r'\1 \2 \3', w)

clean now prints 'w123 o456 (t789)' even if there are some other characters at the beginning or end of string.

This expression look's for 3 groups of characters each consisting of a letter and 3 digits. For the last group there are optional brackets - \\(? and \\)? . All characters before and after the 3 groups are matched with ^.* and .*$ . Then we replace everything with just the 3 captured groups - \\1 \\2 \\3

Answer 4

Instead of Regex, why not use list comprehension (this auto keeps letters and digits if you don't want certain letters or digits we can change it too):

w = 'w123 o456 t789-- --'
list_to_keep =[' ']
print(''.join([x for x in w if x.isalnum() or x in list_to_keep]))
>> w123 o456 t789 

w = 'w123 o456 (t789)'
list_to_keep =[' '] # add to me
print(''.join([x for x in w if x.isalnum() or x in list_to_keep]))
>> w123 o456 t789 

and for example:

w = 'w123 o456 (t789)'
list_to_keep =[' ', '('] # add to me (I added '(' to keep for example)
print(''.join([x for x in w if x.isalnum() or x in list_to_keep]))
>> w123 o456 (t789

and it works against what you edited saying Martin doesn't work:

w='w123 -o456 t789--) --'
list_to_keep =[' '] # add to me (I added '(' to keep for example)
print(''.join([x for x in w if x.isalnum() or x in list_to_keep]))
>> w123 o456 t789