Replacing end numbers in a string python

Question

Basically, I have scraped some tennis scores from a website and dumped them into a dictionary. However, the scores of tiebreak sets are returned like "63" and "77". I want to be able to add a starting parenthesis after the 6 or 7 and close the parentheses at the end of the entire number. So a pseudo code example would be:

>>>s = '613'
>>>s_new = addParentheses(s)
>>>s_new
6(13)

The number after the 6 or 7 can be a one or two digit number from 0 onwards (extremely unlikely that it will be a 3 digit number, so I am leaving that possibility out). I tried reading through some of the regex documentation on the Python website but am having trouble incorporating it in this problem. Thanks.

Hey btw, I notice you haven't yet voted on StackOverflow answers. For any of the answers you find helpful, please consider voting up as this is how the reputation system works. Of course no obligation to do so. Thanks for listening to my 10-second SO rep tutorial. :) — zx81
– zx81, Commented Jun 8, 2014 at 4:54
Yeah, I just got 15 rep so I voted up all the answers. They are all great solutions, but I wanted to use regular expressions for this so I chose yours. Thanks again. — user3685742
– user3685742, Commented Jun 8, 2014 at 5:04

zx81 · Accepted Answer · 2014-06-08 04:36:57Z

4

If the strings always start with 6 or 7, you're looking for something like:

result = re.sub(r"^([67])(\d{1,2})$", r"\1(\2)", subject)

Explain Regex

^                        # the beginning of the string
(                        # group and capture to \1:
  [67]                   #   any character of: '6', '7'
)                        # end of \1
(                        # group and capture to \2:
  \d{1,2}                #   digits (0-9) (between 1 and 2 times
                         #   (matching the most amount possible))
)                        # end of \2
$                        # before an optional \n, and the end of the
                         # string

The replacement string "\1(\2) concatenates capture Group 1 and capture Group 2 between parentheses.

answered Jun 8, 2014 at 4:36

zx81

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1 Comment

zx81 Over a year ago

@CTZhu Thanks, nice to see you again. :)

Akavall · Accepted Answer · 2014-06-08 04:39:47Z

1

What about something like this:

def addParentheses(s):
    if not s[0] in ('6','7'):
        return s
    else:
        temp = [s[0], '(']
        for ele in s[1:]:
            temp.append(ele)
        else:
            temp.append(')')
    return ''.join(temp)

Demo:

>>> addParentheses('613')
'6(13)'
>>> addParentheses('6163')
'6(163)'
>>> addParentheses('68')
'6(8)'
>>> addParentheses('77')
'7(7)'
>>> addParentheses('123')
'123'

answered Jun 8, 2014 at 4:39

Akavall

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Comments

CT Zhu · Accepted Answer · 2014-06-08 04:58:14Z

1

Just add two parentheses at the 2nd and last position? Seems too easy:

In [42]:

s = '613'
def f(s):
    L=list(s)
    L.insert(1,'(')
    return "".join(L)+')'
f(s)
Out[42]:
'6(13)'

Or just ''.join([s[0],'(',s[1:],')'])

If you situation is simple, go for a simple solution, which will be faster. The more general a solution is, the slower it is likely to be:

In [56]:

%timeit ''.join([s[0],'(',s[1:],')'])
100000 loops, best of 3: 1.88 µs per loop
In [57]:

%timeit f(s)
100000 loops, best of 3: 4.97 µs per loop
In [58]:

%timeit addParentheses(s)
100000 loops, best of 3: 5.82 µs per loop
In [59]:

%timeit re.sub(r"^([67])(\d{1,2})$", r"\1(\2)", s)
10000 loops, best of 3: 22 µs per loop

edited Jun 8, 2014 at 4:58

answered Jun 8, 2014 at 4:43

CT Zhu

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