从字符串中提取出现在关键字之前的单词/句子 - Python

Question

I have a string like this,

my_str ='·in this match, dated may 1, 2013 (the "the match") is between brooklyn centenniel, resident of detroit, michigan ("champion") and kamil kubaru, the challenger from alexandria, virginia ("underdog").'

Now, I want to extract the current champion and the underdog using keywords champion and underdog .

What is really challenging here is both contender's names appear before the keyword inside parenthesis. I want to use regular expression and extract information.

Following is what I did,

champion = re.findall(r'("champion"[^.]*.)', my_str)
print(champion)

>> ['"champion") and kamil kubaru, the challenger from alexandria, virginia ("underdog").']


underdog = re.findall(r'("underdog"[^.]*.)', my_str)
print(underdog)

>>['"underdog").']

However, I need the results, champion as :

brooklyn centenniel, resident of detroit, michigan

and the underdog as:

kamil kubaru, the challenger from alexandria, virginia

How can I do this using regular expression? (I have been searching, if I could go back couple or words from the keyword to get the result I want, but no luck yet) Any help or suggestion would be appreciated.

Answer 1

You can use named captured group to capture the desired results:

between\s+(?P<champion>.*?)\s+\("champion"\)\s+and\s+(?P<underdog>.*?)\s+\("underdog"\)

• between\\s+(?P<champion>.*?)\\s+\\("champion"\\) matches the chunk from between to ("champion") and put the desired portion in between as the named captured group champion

• After that, \\s+and\\s+(?P<underdog>.*?)\\s+\\("underdog"\\) matches the chunk upto ("underdog") and again get the desired portion from here as named captured group underdog

Example:

In [26]: my_str ='·in this match, dated may 1, 2013 (the "the match") is between brooklyn centenniel, resident of detroit, michigan ("champion") and kamil kubaru, the challenger from alexandria, virginia 
    ...: ("underdog").'

In [27]: out = re.search(r'between\s+(?P<champion>.*?)\s+\("champion"\)\s+and\s+(?P<underdog>.*?)\s+\("underdog"\)', my_str)

In [28]: out.groupdict()
Out[28]: 
{'champion': 'brooklyn centenniel, resident of detroit, michigan',
 'underdog': 'kamil kubaru, the challenger from alexandria, virginia'}

Answer 2

There will be a better answer than this, and I don't know regex at all, but I'm bored, so here's my 2 cents.

Here's how I would go about it:

words = my_str.split()
index = words.index('("champion")')
champion = words[index - 6:index]
champion = " ".join(champion)

for the underdog, you will have to change the 6 to a 7, and '("champion")' to '("underdog").'

Not sure if this will solve your problem, but for this particular string, this worked when I tested it.

You could also use str.strip() to remove punctuation if that trailing period on underdog is a problem.