Regular Expression in Python to match multiple incidents of a string between two known strings

Question

In the following text

My cow always gives milk. Your cow sometimes produces milk.

I want to extract

'always gives', 'sometimes produces'

Using the encapsulating strings "cow" and "milk".

I tried "cow(.*)milk" following Regular Expression to get a string between two strings in Javascript however it only works if the first sentence is alone

My cow always gives milk.

In my case, using that regular expression returns

always gives milk. Your cow sometimes produces

Additionally, I have also tried "(?<=foo)[^bar]*(?=bar)" from Extracting all values between curly braces regex php. And this works great. For example (and this is closer to the actual problem I'm trying to solve)

fooSTRINGbar fooCHARACTERSbar

Returns

'STRING', 'CHARACTERS'

Great! But for some reason if "STRING" contains a character that "bar" has, then the match fails. For example,

fooSTRaINGbar

Doesn't return anything.

score 4 · Accepted Answer · answered May 02 '15 at 14:12

4

Try the non-greedy option:

cow(.*?)milk

The ? qualifier tells the regex engine to stop a match at the earliest opportunity, rather than greedily trying to match as much as possible (which is the problem you saw with cow(.*)milk).

answered May 02 '15 at 14:12

nneonneo

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Alright, this is good. But is there a way I can make it so that if there is a space between the two encapsulating strings, it will ignore the match?. For example, "fooSTRINGbar fooCHAR ACTERSbar" returns only STRING? – Benjamin Saljooghi May 02 '15 at 14:26
You could use `[^ ]` in place of `.`. (If you specify all your constraints ahead of time you'll get more useful answers!) – nneonneo May 02 '15 at 14:31
Awesome! Yeah that last little constraint must've slipped my mind. Thanks. – Benjamin Saljooghi May 02 '15 at 14:47
@Sal `\S` might be better than `[^ ]`. More direct, shorter, and the only difference is that `[^ ]` does match other whitespace, which I can't imagine you want. – Stefan Pochmann May 02 '15 at 15:08

dawg · Answer 2 · 2015-05-02T17:00:44.967

You can use a lookbehind and lookahead as anchors between a non-greedy match as well:

>>> s='My cow always gives milk. Your cow sometimes produces milk.'
>>> re.findall(r'(?<=\bcow\b)(.*?)(?=\bmilk\b)', s)
[' always gives ', ' sometimes produces ']

If you want to strip the surrounding spaces:

>>> re.findall(r'(?<=\bcow\b)\s*(.*?)\s*(?=\bmilk\b)', s)
['always gives', 'sometimes produces']

You can also use a lookahead to validate that the match does not contain spaces:

>>> s='My cow gives milk. Your cow sometimes produces milk.'
>>> re.findall(r'(?<=\bcow\b)(?=\s*\S+\s*milk\b)\s*(.*?)\s*(?=\bmilk\b)', s)
['gives']

Or:

>>> re.findall(r'(?<=\bcow\b)\s*(\S+)\s*(?=\bmilk\b)', s)
['gives']

Same method works with your second example:

>>> s='fooSTRINGbar fooCHARACTERSbar'
>>> re.findall(r'(?<=foo)([A-Z]+)(?=bar)', s)
['STRING', 'CHARACTERS']

Regular Expression in Python to match multiple incidents of a string between two known strings

2 Answers2