如何使用 groupby 将 3 个元组列表转换为元组列表

Question

I have a string below

test = '''AWS-1 - opened at Jan 23 2010 10:30:08AM 
AWS-2 - opened at Jan 23 2010 11:04:56AM 
AWS-2 - closed at Jan 23 2010 1:18:32PM 
AWS-1 - closed at Jan 23 2010 9:43:44PM 
AWS-1 - opened at Feb 1 2010 12:40:28AM
AWS-1 - closed at Jan 23 2010 9:43:44PM
'''

My Code

import re
from itertools import groupby
y = re.findall(r'\b(\w+-\d+)\s+-\s+(\w+[-.\w]+)\s+at\s+(\w+[\s:.\w]+)\n', test)
print (y)
for key, time in groupby(y,lambda z: y[2]):
for thing in y:
    print( (y[1], key))
print (" ")

My Out

(('AWS-2', 'opened', 'Jan 23 2010 11:04:56AM '), ('AWS-2', 'closed', 'Jan 23 2010 1:18:32PM ')) (('AWS-2', 'opened', 'Jan 23 2010 11:04:56AM '), ('AWS-2', 'closed', 'Jan 23 2010 1:18:32PM ')) (('AWS-2', 'opened', 'Jan 23 2010 11:04:56AM '), ('AWS-2', 'closed', 'Jan 23 2010 1:18:32PM ')) (('AWS-2', 'opened', 'Jan 23 2010 11:04:56AM '), ('AWS-2', 'closed', 'Jan 23 2010 1:18:32PM ')) (('AWS-2', 'opened', 'Jan 23 2010 11:04:56AM '), ('AWS-2', 'closed', 'Jan 23 2010 1:18:32PM '))

Expected out does not coming AWS-1 , instead everywhere AWS-2 is coming

(('AWS-1', 'opened', 'Jan 23 2010 10:30:08AM '), ('AWS-1', 'closed', 'Jan 23 2010 9:43:44PM '))
(('AWS-1', 'opened', 'Feb 1 2010 12:40:28AM'), ('AWS-1', 'closed', 'Feb 23 2010 9:43:44PM'))
(('AWS-2', 'opened', 'Jan 23 2010 11:04:56AM '), ('AWS-2', 'closed', 'Jan 23 2010 1:18:32PM '))

Answer 1

Your request is unclear, but it appears you wish to make opened-closed pairs based on parameters.

Given

import re

import dateutil


records = """\
AWS-1 - opened at Jan 23 2010 10:30:08AM
AWS-2 - opened at Jan 23 2010 11:03:56AM
AWS-2 - closed at Jan 23 2010 1:18:32PM 
AWS-1 - closed at Feb 27 2010 9:32:50PM
AWS-1 - opened at Feb 1 2010 12:50:28AM
AWS-1 - closed at Jan 23 2010 9:32:50PM
"""

Code

def splitlines(s: str) -> tuple:
    """Return tuples of parsed lines: id, status, time."""
    res = []

    for line in s.split("\n"):

        if not line:
            continue            
        parsed = tuple(map(str.strip, filter(None, re.split("(\s-\s)|(at)", line))))
        id_, _, status, _, time = parsed 
        data = id_, status, dateutil.parser.parse(time)
        res.append(data)

    return tuple(res)


def pairwise_records(s: str) -> list:
    """Return paired records according to id, status and time."""
    key = lambda x: (x[0], x[2], x[1])

    sorted_recs = ((i, s, str(t)) for i, s, t in sorted(splitlines(s), key=key))

    return list(zip(sorted_recs, sorted_recs))

Demo

pairwise_records(records)

Output

[(('AWS-1', 'opened', '2010-01-23 10:30:08'),
  ('AWS-1', 'closed', '2010-01-23 21:32:50')),
 (('AWS-1', 'opened', '2010-02-01 00:50:28'),
  ('AWS-1', 'closed', '2010-02-27 21:32:50')),
 (('AWS-2', 'opened', '2010-01-23 11:03:56'),
  ('AWS-2', 'closed', '2010-01-23 13:18:32'))]

---

Details

Kudos to the OP on getting a partial answer with an elaborate regex. It turns out you can do this more explicitly with more clear regex and without groupby .

splitlines

We attempt to parse the input string into tuples. We do this with re.split which leaves behind extra elements we don't need. These extras are cleaned up with filter and by unpacking into (id_, status, time) where time is parsed as a datetime object. The results are tuples of parsed lines. Example:

splitlines("AWS-1 - opened at Jan 23 2010 10:30:08AM")
# (('AWS-1', 'opened', datetime.datetime(2010, 1, 23, 10, 30, 8)),)

pairwise_records

We sort the tuples by id and datetime . Sorting by time will naturally align pairs in order. Example, if something opens at 9 AM, it must close at some future time; time is naturally sorted. Finally, we use a "trick" with iterators to pair the results together.