CODEWITHSUNDEEP
pythonregexsequences
dbr
dbr

Reputation: 169563

Regex and a sequences of patterns?

Is there a way to match a pattern (e\d\d) several times, capturing each one into a group? For example, given the string..

blah.s01e24e25

..I wish to get four groups:

1 -> blah
2 -> 01
3 -> 24
4 -> 25

The obvious regex to use is (in Python regex:

import re
re.match("(\w+).s(\d+)e(\d+)e(\d+)", "blah.s01e24e25").groups()

..but I also want to match either of the following:

blah.s01e24
blah.s01e24e25e26

You can't seem to do (e\d\d)+, or rather you can, but it only captures the last occurrence:

>>> re.match("(\w+).s(\d+)(e\d\d){2}", "blah.s01e24e25e26").groups()
('blah', '01', 'e25')
>>> re.match("(\w+).s(\d+)(e\d\d){3}", "blah.s01e24e25e26").groups()
('blah', '01', 'e26')

I want to do this in a single regex because I have multiple patterns to match TV episode filenames, and do not want to duplicate each expression to handle multiple episodes:

\w+\.s(\d+)\.e(\d+) # matches blah.s01e01
\w+\.s(\d+)\.e(\d+)\.e(\d+) # matches blah.s01e01e02
\w+\.s(\d+)\.e(\d+)\.e(\d+)\.e(\d+) # matches blah.s01e01e02e03

\w - \d+x\d+ # matches blah - 01x01
\w - \d+x\d+\d+ # matches blah - 01x01x02
\w - \d+x\d+\d+\d+ # matches blah - 01x01x02x03

..and so on for numerous other patterns.

Another thing to complicate matters - I wish to store these regexs in a config file, so a solution using multiple regexs and function calls is not desired - but if this proves impossible I'll just allow the user to add simple regexs

Basically, is there a way to capture a repeating pattern using regex?

Upvotes: 4

Views: 2547

Answers (5)

tzot
tzot

Reputation: 95921

Perhaps something like that?

def episode_matcher(filename):
    m1= re.match(r"(?i)(.*?)\.s(\d+)((?:e\d+)+)", filename)
    if m1:
        m2= re.findall(r"\d+", m1.group(3))
        return m1.group(1), m1.group(2), m2
    # auto return None here

>>> episode_matcher("blah.s01e02")
('blah', '01', ['02'])
>>> episode_matcher("blah.S01e02E03")
('blah', '01', ['02', '03'])

Upvotes: 0

dbr
dbr

Reputation: 169563

After thinking about the problem, I think I have a simpler solution, using named groups.

The simplest regex a user (or I) could use is:

(\w+\).s(\d+)\.e(\d+)

The filename parsing class will take the first group as the show name, second as season number, third as episode number. This covers a majority of files.

I'll allow a few different named groups for these:

(?P<showname>\w+\).s(?P<seasonnumber>\d+)\.e(?P<episodenumber>\d+)

To support multiple episodes, I'll support two named groups, something like startingepisodenumber and endingepisodenumber to support things like showname.s01e01-03:

(?P<showname>\w+\)\.s(?P<seasonnumber>\d+)\.e(?P<startingepisodenumber>\d+)-(?P<endingepisodenumber>e\d+)

And finally, allow named groups with names matching episodenumber\d+ (episodenumber1, episodenumber2 etc):

(?P<showname>\w+\)\.
s(?P<seasonnumber>\d+)\.
e(?P<episodenumber1>\d+)
e(?P<episodenumber2>\d+)
e(?P<episodenumber3>\d+)

It still requires possibly duplicating the patterns for different amounts of e01s, but there will never be a file with two non-consecutive episodes (like show.s01e01e03e04), so using the starting/endingepisodenumber groups should solve this, and for weird cases users come across, they can use the episodenumber\d+ group names

This doesn't really answer the sequence-of-patterns question, but it solves the problem that led me to ask it! (I'll still accept another answer that shows how to match s01e23e24...e27 in one regex - if someone works this out!)

Upvotes: 0

Adrian Panasiuk
Adrian Panasiuk

Reputation: 7343

non-grouping parentheses: (?:asdfasdg)

which do not have to appear: (?:adsfasdf)?

c = re.compile(r"""(\w+).s(\d+)
                       (?:
                            e(\d+)
                            (?:
                                  e(\d+)
                            )?
                        )?
               """, re.X)

or

c = re.compile(r"""(\w+).s(\d+)(?:e(\d+)(?:e(\d+))?)?""", re.X)

Upvotes: 1

Rorick
Rorick

Reputation: 8953

Number of captured groups equal to number of parentheses groups. Look at findall or finditer for solving your problem.

Upvotes: 1

RichieHindle
RichieHindle

Reputation: 281455

Do it in two steps, one to find all the numbers, then one to split them:

import re

def get_pieces(s):
    # Error checking omitted!
    whole_match = re.search(r'\w+\.(s\d+(?:e\d+)+)', s)
    return re.findall(r'\d+', whole_match.group(1))

print get_pieces(r"blah.s01e01")
print get_pieces(r"blah.s01e01e02")
print get_pieces(r"blah.s01e01e02e03")

# prints:
# ['01', '01']
# ['01', '01', '02']
# ['01', '01', '02', '03']

Upvotes: 5

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