Question: I am using this to find the sentences in a text:

`sentences = re.findall("[A-Z].*?[\.!?;]", stripped_value, re.MULTILINE | re.DOTALL | re.UNICODE )`

Is there a way to exclude the sentence boundaries (comma, dot, questionmark, etc) at the same time? (I mean without using strip commands for each character).

Answer: 
Use a capturing group:

```
sentences = re.findall(r'([A-Z].*?)[.!?;]', stripped_value, re.MULTILINE | re.DOTALL | re.UNICODE)

```

`.findall()` returns the contents of a capturing group instead of the whole match, if there is one present in the expression.

Demo:

```
>>> stripped_value = '''Some sentence. And another.
... Multiline text works too! And commas, they are included; but not the semicolon?
... '''
>>> import re
>>> re.findall(r'([A-Z].*?)[.!?;]', stripped_value, re.MULTILINE | re.DOTALL | re.UNICODE)
['Some sentence', 'And another', 'Multiline text works too', 'And commas, they are included']

```

From the [`re.findall()` documentation](http://docs.python.org/2/library/re.html#re.findall):

> 
> If one or more groups are present in the pattern, return a list of groups; this will be a list of tuples if the pattern has more than one group.
> 
> 
> 

Alternatively, you can also use a look-ahead assertion:

```
sentences = re.findall("[A-Z].*?(?=[\.!?;])", stripped_value, re.MULTILINE | re.DOTALL | re.UNICODE)

```

The `(?=..)` positive lookahead assertion acts as an anchor; the pattern matches only if punctiation follows the matched text. Lookaheads can give you faster results as `.findall()` doesn't then have to exctract the matched groups. The output of either alternative is otherwise the same.
You can use capturing groups 

```
regex = "([A-Z].*?)[\.!?;]"

```

... do you searching, get a match object for each match ....

```
sentence = match_obj.groups(1)

```

Also I notice you insist that all sentences begin with capital letters, but you terminate them at the first semicolon. I would say "sentence" usually meant "All ';' separated clauses concatenated". But that if you wanted ';' to be a separator, then I would count each clause as a sentence (since it is, except for the capital letter).