寻找连续重复单词时，Python后视正则表达式“固定宽度模式”错误

Question

I have a text with words separated by . , with instances of 2 and 3 consecutive repeated words:

My.name.name.is.Inigo.Montoya.You.killed.my.father.father.father.Prepare.to.die-

I need to match them independently with regex, excluding the duplicates from the triplicates.

Since there are max. 3 consecutive repeated words, this

r'\\b(\\w+)\\.+\\1\\.+\\1\\b'

successfully catches

father.father.father

However, in order to catch 2 consecutive repeated words, I need to make sure the next and previous words aren't the same. I can do a negative look-ahead

r'\\b(\\w+)\\.+\\1(?!\\.+\\1)\\b'

but my attempts at the negative look-behind

r'(?<!(\\w)\\.)\\b\\1\\.+\\1\\b(?!\\.\\1)'

either return a fixed-width issue (when I keep the + ) or some other issue.

How should I correct the negative look-behind ?

Answer 1

I think that there might be an easier way to capture what you want without the negative look-behind:

r = re.compile(r'\b((\w+)\.+\2\.+\2?)\b')
r.findall(t)

> [('name.name.', 'name'), ('father.father.father', 'father')]

Just making the third repetition optional.

---

A version to capture any number of repetitions of the same word, can look something like this:

r = re.compile(r'\b((\w+)(\.+\2)\3*)\b')
r.findall(t)
> [('name.name', 'name', '.name'), ('father.father.father', 'father', '.father')]

Answer 2

Maybe regexes are not needed at all.

Using itertools.groupby does the job. It's designed to group equal occurrences of consecutive items.

• group by words (after splitting according to dots)
• convert to list and issue a tuple value,count only if length > 1

like this:

import itertools

s = "My.name.name.is.Inigo.Montoya.You.killed.my.father.father.father.Prepare.to.die"

matches = [(l[0],len(l)) for l in (list(v) for k,v in itertools.groupby(s.split("."))) if len(l)>1]

result:

[('name', 2), ('father', 3)]

So basically we can do whatever we want with this list of tuples (filtering it on the number of occurrences for instance)

Bonus (as I misread the question at first, so I'm leaving it in): to remove the duplicates from the sentence - group by words (after splitting according to dots) like above - take only key (value) of the values returned in a list comp (we don't need the values since we don't count) - join back with dot

In one line (still using itertools ):

new_s = ".".join([k for k,_ in itertools.groupby(s.split("."))])

result:

My.name.is.Inigo.Montoya.You.killed.my.father.Prepare.to.die