使用 Python 将连续日期分组在一起

Given:

dates = [
  datetime(2014, 10, 11), 

  datetime(2014, 10, 1), 
  datetime(2014, 10, 2),
  datetime(2014, 10, 3),

  datetime(2014, 10, 5), 
  datetime(2014, 10, 5), 
  datetime(2014, 10, 6), 

  datetime(2014, 10, 22), 
  datetime(2014, 10, 20),
  datetime(2014, 10, 21),

  datetime(2014, 10, 9), 

  datetime(2014, 10, 7),
  datetime(2014, 10, 6)
]

预期输出：

expect = [
  [datetime(2014, 10, 1), datetime(2014, 10, 3)],
  [datetime(2014, 10, 5), datetime(2014, 10, 7)],
   datetime(2014, 10, 9),
   datetime(2014, 10, 11),
  [datetime(2014, 10, 20), datetime(2014, 10, 22)]
]

Python 中的实现：

from datetime import datetime, timedelta

def parse_date_ranges(dates):
    if(not dates or not len(dates) > 0):
        return False

    # make sure to order dates
    dates.sort()

    # init values
    result  = []
    tupl    = [dates[0], dates[0]]
    it      = iter(dates)
    date    = True

    def add_tuple_to_result(tuple):
        # if first part of tuple differs from last part -> add full tuple
        # else -> add first part of tuple only
        result.append(tupl if tupl[0] != tupl[1] else tupl[0])

    while date:
        # get next date or False if no next date
        date = next(it, False)

        # filter double dates
        if(date in tupl):
            continue

        elif(date):
            if(date - timedelta(days=1) == tupl[1]):
                # consecutive date, so add date to end of current tuple
                tupl[1] = date
            else:
                # gap larger than 1 day: add current tuple to result
                # and create new tuple
                add_tuple_to_result(tupl)
                tupl = [date, date]

        else:
            # date == false, so this is the last step.
            # add the current tuple to result
            add_tuple_to_result(tupl)

    return result

See https://gist.github.com/mattes/8987332 https://gist.github.com/mattes/8987332还要进行更多测试。

Question

该实现有效，但我是 Python 新手。所以我想知道是否有更好的方法来解决这个问题？还是就这样就好了？

I like 迭代工具 http://docs.python.org/3/library/itertools.html:

from itertools import tee, zip_longest
from datetime import datetime, timedelta

one_day = timedelta(days=1)

def pairwise(iterable):
    a, b = tee(iterable)
    next(b, None)
    return zip_longest(a, b, fillvalue=None)

def collapse_ranges(sorted_iterable, inc):
    pairs = pairwise(sorted_iterable)
    for start, tmp in pairs:
        if inc(start) == tmp:
            for end, tmp in pairs:
                if inc(end) != tmp:
                    break
            yield start, end
        else:
            yield start

# dates = [...]

numbers = [11, 1, 2, 3, 5, 5, 6, 22, 20, 21, 9, 7, 6]

if __name__ == '__main__':
    import pprint
    for each in collapse_ranges(sorted(set(dates)), lambda d: d + one_day):
        pprint.pprint(each)
    for each in collapse_ranges(sorted(set(numbers)), (1).__add__):
        pprint.pprint(each)

Result:

(datetime.datetime(2014, 10, 1, 0, 0), datetime.datetime(2014, 10, 3, 0, 0))
(datetime.datetime(2014, 10, 5, 0, 0), datetime.datetime(2014, 10, 7, 0, 0))
datetime.datetime(2014, 10, 9, 0, 0)
datetime.datetime(2014, 10, 11, 0, 0)
(datetime.datetime(2014, 10, 20, 0, 0), datetime.datetime(2014, 10, 22, 0, 0))
(1, 3)
(5, 7)
9
11
(20, 22)