Question

假设：

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)]
]

在Python中实施：

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

也可以参考https://gist.github.com/mattes/8987332进行更多测试。

问题

实现有效，但我是Python的新手。所以我想知道是否有更好的方法来解决这个问题？或者它没关系？

Answer 1

我喜欢itertools：

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)

结果：

(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)

Answer 2

您可以通过将此answer调整为类似名称的左侧来避免重新发明轮子 Grouping consecutive dates together问题与datetime个对象一起使用：

def parse_date_ranges(dates):

    def group_consecutive(dates):
        dates_iter = iter(sorted(set(dates)))  # de-dup and sort

        run = [next(dates_iter)]
        for d in dates_iter:
            if (d.toordinal() - run[-1].toordinal()) == 1:  # consecutive?
                run.append(d)
            else:  # [start, end] of range else singleton
                yield [run[0], run[-1]] if len(run) > 1 else run[0]
                run = [d]

        yield [run[0], run[-1]] if len(run) > 1 else run[0]

    return list(group_consecutive(dates)) if dates else False

Answer 3

我为你写了另一个解决方案，写了几条评论试图解释代码。

from datetime import datetime, timedelta

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)
]
# Remove duplicates, and sort the dates ascending
sorted_dates = sorted(set(dates))
# Set initial first and last element as the current element
first, last = sorted_dates[0], sorted_dates[0]
date_ranges = []

# Loop over the sorted list from the second value
for d in sorted_dates[1:]:
    # Check if the current date is exactly one day later then the current
    # "last" date
    if d - last != timedelta(days=1):
        date_ranges.append(tuple(sorted({first, last})))
        first, last = d, d
    else:
        last = d

# Handle last element
if first == last:
    date_ranges.append((first,))
else:
    date_ranges.append((first, last))

for dt_pair in date_ranges:
    print dt_pair

输出：

(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))

Answer 4

调整this answer以使用datetime对象。这包括非唯一和非排序的输入，也兼容python3：

utf8

示例：

import itertools
from datetime import datetime, timedelta

def datetimes_to_ranges(iterable):
    iterable = sorted(set(iterable))
    keyfunc = lambda t: t[1] - timedelta(days=t[0])
    for key, group in itertools.groupby(enumerate(iterable), keyfunc):
        group = list(group)
        if len(group) == 1:
            yield group[0][1]
        else:
            yield group[0][1], group[-1][1]

与Python一起将连续日期分组

4 个答案: